AP Statistics Curriculum 2007 Distrib Dists
From Socr
m (→Negative Multinomial Distribution (NMD)) 
(→Normal approximation to Negative Binomial distribution) 

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*Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is <math>P(X = x) = (1  p)^{x1} \times p</math>, for x = 1, 2, 3, 4,....  *Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is <math>P(X = x) = (1  p)^{x1} \times p</math>, for x = 1, 2, 3, 4,....  
  * Expectation: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar  Expected Value]] of a geometrically distributed random variable ''X'' is <math>{1\over p}.</math>  +  * Expectation: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar  Expected Value]] of a geometrically distributed random variable ''X'' is <math>{1\over p}.</math> This is because [http://en.wikipedia.org/wiki/Geometric_progression geometric series have this property]: 
+  :<math>\sum_{k=0}^{n} p(1p)^k = p(1p)^0+p(1p)^1+p(1p)^2+p(1p)^3+\cdots+p(1p)^n.</math>  
+  : Let r=(1p), then p=(1r) and <math>\sum_{k=0}^{n} p(1p)^k = \begin{align}  
+  (1r) \sum_{k=0}^{n} r^k & = (1r)(r^0 + r^1+r^2+r^3+\cdots+r^n) \\  
+  & = r^0 + r^1+r^2+r^3+\cdots+r^n \\  
+  & ( r^1+r^2+r^3+\cdots +r^n + r^{n+1}) \\  
+  & = r^0  r^{n+1} = 1  r^{n+1}.  
+  \end{align}</math>  
+  : Thus: <math>\sum_{k=0}^{n} p(1p)^k = \frac{p  pr^{n+1}}{1r} = 1pr^{n+1},</math> which converges to 1, as <math>n\longrightarrow \infty,</math>, and hence the above geometric density is well defined.  
+  
+  : Denote the geometric expectation by E = E(X) = <math>\sum_{k=0}^{\infty} kpr^k</math>, where r=1p. Then <math>pE = E  (1p)E = \sum_{k=0}^{\infty} kpr^k  (\sum_{k=0}^{\infty} kpr^{k+1})=</math> <math>\sum_{k=0}^{\infty} pr^k = 1</math>. Therefore, <math>E = \frac{1}{p}</math>.  
*Variance: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar  Variance]] is <math>{1p\over p^2}.</math>  *Variance: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar  Variance]] is <math>{1p\over p^2}.</math>  
*Example: See [[SOCR_EduMaterials_Activities_Binomial_Distributions  this SOCR Geometric distribution activity]].  *Example: See [[SOCR_EduMaterials_Activities_Binomial_Distributions  this SOCR Geometric distribution activity]].  
+  
+  * The Geometric distribution gets its name because its probability mass function is a [http://en.wikipedia.org/wiki/Geometric_progression geometric progression]. It is the discrete analogue of the Exponential distribution and is also known as Furry distribution.  
===HyperGeometric===  ===HyperGeometric===  
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====Examples====  ====Examples====  
  * SOCR Activity: The [[SOCR_EduMaterials_Activities_BallAndRunExperiment  SOCR Ball and Urn Experiment]] provides a handson demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N  R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can  +  * SOCR Activity: The [[SOCR_EduMaterials_Activities_BallAndRunExperiment  SOCR Ball and Urn Experiment]] provides a handson demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N  R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can vary with scroll bars. 
<center>[[Image:SOCR_Activities_BallAndUrnExperiment_SubTopic_Chui_050307_Fig2.JPG500px]]</center>  <center>[[Image:SOCR_Activities_BallAndUrnExperiment_SubTopic_Chui_050307_Fig2.JPG500px]]</center>  
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<center>[[Image:SOCR_EBook_Dinov_RV_HyperGeom_013008_Fig9.jpg500px]]</center>  <center>[[Image:SOCR_EBook_Dinov_RV_HyperGeom_013008_Fig9.jpg500px]]</center>  
  * Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the samplesize (n) is < 5% of the population size(N), we can use [[AP_Statistics_Curriculum_2007_Limits_Bin2HyperG binomial approximation to hypergeometric]]. Thus if the sample of n=100 fish had 5 tagged, the sampleproportion (estimate of the population proportion) will be <math>\hat{p}={5\over 100}=0.05</math>. Thus, we can estimate that <math>0.05=\hat{p}={200\over N}</math>, and <math>N\approx 4,000</math>, as shown  +  * Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish, we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the samplesize (n) is < 5% of the population size(N), we can use [[AP_Statistics_Curriculum_2007_Limits_Bin2HyperG binomial approximation to hypergeometric]]. Thus if the sample of n=100 fish had 5 tagged, the sampleproportion (estimate of the population proportion) will be <math>\hat{p}={5\over 100}=0.05</math>. Thus, we can estimate that <math>0.05=\hat{p}={200\over N}</math>, and <math>N\approx 4,000</math>, as shown in the figure below. 
<center>[[Image:SOCR_EBook_Dinov_Prob_HyperG_041108_Fig9a.jpg500px]]</center>  <center>[[Image:SOCR_EBook_Dinov_Prob_HyperG_041108_Fig9a.jpg500px]]</center>  
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====Application====  ====Application====  
  Suppose Jane is promoting and fundraising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and 70% chance that she'll fail.  +  Suppose Jane is promoting and fundraising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and a 70% chance that she'll fail. 
* ''What's the probability mass function of the number of failures (''k=nr'') to get ''r=6'' successes''?''  * ''What's the probability mass function of the number of failures (''k=nr'') to get ''r=6'' successes''?''  
  : In other words, ''  +  : In other words, ''what's the probability mass function that the last 6<sup>th</sup> state she succeeds to secure all electoral votes happens to be at the ''n''<sup>th</sup> state she campaigns in?'' 
NegBin(''r'', ''p'') distribution describes the probability of ''k'' failures and ''r'' successes in ''n''=''k''+''r'' Bernoulli(''p'') trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is ''n''=''k+6''. The random variable we are interested in is '''X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}'''. So, ''n'' = ''k+6'', and <math>X\sim NegBin(r=6, p=0.3)</math>. Thus, for <math>n \geq 6</math>, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:  NegBin(''r'', ''p'') distribution describes the probability of ''k'' failures and ''r'' successes in ''n''=''k''+''r'' Bernoulli(''p'') trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is ''n''=''k+6''. The random variable we are interested in is '''X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}'''. So, ''n'' = ''k+6'', and <math>X\sim NegBin(r=6, p=0.3)</math>. Thus, for <math>n \geq 6</math>, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:  
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<center>[[Image:SOCR_EBook_Dinov_RV_NegBinomial_013008_Fig5.jpg500px]]</center>  <center>[[Image:SOCR_EBook_Dinov_RV_NegBinomial_013008_Fig5.jpg500px]]</center>  
  * Suppose the success of getting all electoral votes within a state is reduced to only 10%, then '''X~NegBin(r=6, p=0.1)'''. Notice that the shape and domain the NegativeBinomial distribution significantly chance now (see image below)  +  * Suppose the success of getting all electoral votes within a state is reduced to only 10%, then '''X~NegBin(r=6, p=0.1)'''. Notice that the shape and domain the NegativeBinomial distribution significantly chance now (see image below). 
: ''What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?''  : ''What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?''  
:<math> P(X\geq 50) = 0.632391</math>  :<math> P(X\geq 50) = 0.632391</math>  
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* SOCR Activity: If you want to see an interactive NegativeBinomial Graphical calculator you can go to [http://socr.ucla.edu/htmls/SOCR_Experiments.html this applet (select Negative Binomial)] and see [[SOCR_EduMaterials_Activities_NegativeBinomial this activity]].  * SOCR Activity: If you want to see an interactive NegativeBinomial Graphical calculator you can go to [http://socr.ucla.edu/htmls/SOCR_Experiments.html this applet (select Negative Binomial)] and see [[SOCR_EduMaterials_Activities_NegativeBinomial this activity]].  
+  
+  ====Normal approximation to Negative Binomial distribution====  
+  The [[AP_Statistics_Curriculum_2007_Limits_CLTcentral limit theorem]] provides the foundation for approximation of negative binomial distribution by [[AP_Statistics_Curriculum_2007_Normal_Std Normal distribution]]. Each negative binomial random variable, \(V_k \sim NB(k,p)\), may be expressed as a sum of '''k''' independent, identically distributed ([[AP_Statistics_Curriculum_2007_Distrib_Dists#Geometricgeometric]]) random variables \(\{X_i\}\), i.e., \( V_k = \sum_{i=1}^k{X_i}\), where [[AP_Statistics_Curriculum_2007_Distrib_Dists \( X_i \sim Geometric(p)\)]]. In various scientific applications, given a large '''k''', the distribution of \(V_k\) is approximately normal with mean and variance given by \(\mu=k\frac{1}{p}\) and \(\sigma^2=k\frac{1p}{p^2}\), as \(k \longrightarrow \infty\). Depending on the parameter '''p''', '''k''' may need to be rather large for the approximation to work well. Also, when using the normal approximation, we should remember to use the continuity correction, since the negative binomial and Normal distributions are discrete and continuous, respectively.  
+  
+  In the above example, \(P(X\le 8)\), \(V_k \sim NegBin(k=r=6, p=0.3)\), the normal distribution approximation, \(N(\mu=\frac{k}{p}=20, \sigma=\sqrt{k\frac{1p}{p^2}}=6.83)\), is shown it the following image and table:  
+  
+  <center>[[Image:SOCR_EBook_Dinov_RV_NegBinomial_013008_Fig4a.png500px]]</center>  
+  
+  The probabilities of the real [http://socr.ucla.edu/htmls/dist/NegativeBinomial_Distribution.html Negative Binomial] and [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html approximate Normal] distributions (on the range [2:4]) are not identical but are sufficiently close.  
+  
+  <center>  
+  { class="wikitable" style="textalign:center; width:75%" border="1"  
+    
+  ! Summary [http://socr.ucla.edu/htmls/dist/NegativeBinomial_Distribution.html \(NegativeBinomial(k=6,p=0.3)\) ]  [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html \(Normal(\mu=20, \sigma=6.83)\) ]  
+    
+   Mean20.020.0  
+    
+   Median19.020.0  
+    
+   Variance46.66666746.6489  
+    
+   Standard Deviation6.8313016.83  
+    
+   Max Density 0.0624390.058410  
+    
+  ! colspan=3Probability Areas  
+    
+   \(\le 8\) .011292 0.039433  
+    
+   >8 .9887080.960537  
+  }  
+  </center>  
===Negative Multinomial Distribution (NMD)===  ===Negative Multinomial Distribution (NMD)===  
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::''X=Total # of experiments (n) to get r successes'' (and therefore nr failures);  ::''X=Total # of experiments (n) to get r successes'' (and therefore nr failures);  
: <math>X \sim Negative Multinomial(k_0,\{p_0,p_1\})</math>,  : <math>X \sim Negative Multinomial(k_0,\{p_0,p_1\})</math>,  
  :: ''X=Total # of experiments (n) to get <math>k_0</math> (  +  :: ''X=Total # of experiments (n) to get <math>k_0</math> (default variable, <math>X_o</math>) and <math>nk_0</math> outcomes of the other possible outcome (<math>X_1</math>)''. 
====Negative Multinomial Summary====  ====Negative Multinomial Summary====  
  * Probability Mass Function: <math> P(  +  * Probability Mass Function: <math> P(k_1, \cdots, k_mk_0,\{p_1,\cdots,p_m\}) = \left (\sum_{i=0}^m{k_i}1\right)!\frac{p_0^{k_0}}{(k_01)!} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}</math>, or equivalently: 
  : <math> P(  +  : <math> P(k_1, \cdots, k_mk_0,\{p_1,\cdots,p_m\}) = \Gamma\left(\sum_{i=1}^m{k_i}\right)\frac{p_0^{k_0}}{\Gamma(k_0)} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}</math>, where <math>\Gamma(x)</math> is the [http://en.wikipedia.org/wiki/Gamma_function Gamma function]. 
  * Mean (vector): <math>\mu=E(X_1,\cdots,X_m)= (\mu_1=E(X_1), \cdots, \mu_m=E(X_m)) = \  +  * Mean (vector): <math>\mu=E(X_1,\cdots,X_m)= (\mu_1=E(X_1), \cdots, \mu_m=E(X_m)) = \left ( \frac{k_0p_1}{p_0}, \cdots, \frac{k_0p_m}{p_0} \right)</math>. 
  * VarianceCovariance (matrix): <math>Cov(X_i,X_j)= \{cov[i,j]\}</math>, where <math> cov[i,j] = \begin{cases} \frac{k_0  +  * VarianceCovariance (matrix): <math>Cov(X_i,X_j)= \{cov[i,j]\}</math>, where 
  \frac{k_0  +  : <math> cov[i,j] = \begin{cases} \frac{k_0 p_i p_j}{p_0^2},& i\not= j,\\ 
+  \frac{k_0 p_i (p_i + p_0)}{p_0^2},& i=j.\end{cases}</math>.  
====Cancer Example====  ====Cancer Example====  
  The [[AP_Statistics_Curriculum_2007_Prob_Rules Probability Theory Chapter]] of the [[EBook]] shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as in the Table below.  +  The [[AP_Statistics_Curriculum_2007_Prob_Rules Probability Theory Chapter]] of the [[EBook]] shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as shown in the Table below. 
<center>  <center>  
{ class="wikitable" style="textalign:center; width:75%" border="1"  { class="wikitable" style="textalign:center; width:75%" border="1"  
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The sites (locations) of the cancer may be independent, but there may be positive dependencies of the type of cancer for a given location (site). For example, localized exposure to radiation implies that elevated level of one type of cancer (at a given location) may indicate higher level of another cancer type at the same location. We want to use the Negative Multinomial distribution to model the sites cancer rates and try to measure some of the cancer type dependencies within each location.  The sites (locations) of the cancer may be independent, but there may be positive dependencies of the type of cancer for a given location (site). For example, localized exposure to radiation implies that elevated level of one type of cancer (at a given location) may indicate higher level of another cancer type at the same location. We want to use the Negative Multinomial distribution to model the sites cancer rates and try to measure some of the cancer type dependencies within each location.  
  Let's denote by <math>  +  Let's denote by <math>x_{i,j}</math> the cancer rates for each site (<math>0\leq i \leq 2</math>) and each type of cancer (<math>0\leq j \leq 3</math>). For each (fixed) site (<math>0\leq i \leq 2</math>), the cancer rates are independent Negative Multinomial distributed random variables. That is, for each column index (site) the columnvector X has the following distribution: 
  : <math>X=\{  +  : <math>X=\{X_1, X_2, X_3\} \sim NMD(k_0,\{p_1,p_2,p_3\})</math>. 
  Different columns (sites) are considered to be different instances of the random multinomially distributed  +  Different columns (sites) are considered to be different instances of the random negativemultinomially distributed vector, X. Then we have the following estimates: 
* [[AP_Statistics_Curriculum_2007_Estim_MOM_MLE MLE estimate]] of the Mean: is given by:  * [[AP_Statistics_Curriculum_2007_Estim_MOM_MLE MLE estimate]] of the Mean: is given by:  
  : <math>\hat{\mu}_{i,j} = \frac{x_{i,.}\times x_{.,j}}{x_{.,.}}</math>  +  : <math>\hat{\mu}_{i,j} = \frac{x_{i,.}\times x_{.,j}}{x_{.,.}}</math> 
:: <math>x_{i,.} = \sum_{j=0}^{3}{x_{i,j}}</math>  :: <math>x_{i,.} = \sum_{j=0}^{3}{x_{i,j}}</math>  
:: <math>x_{.,j} = \sum_{i=0}^{2}{x_{i,j}}</math>  :: <math>x_{.,j} = \sum_{i=0}^{2}{x_{i,j}}</math>  
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:: Example: <math>\hat{\mu}_{1,1} = \frac{x_{1,.}\times x_{.,1}}{x_{.,.}}=\frac{34\times 68}{400}=5.78</math>  :: Example: <math>\hat{\mu}_{1,1} = \frac{x_{1,.}\times x_{.,1}}{x_{.,.}}=\frac{34\times 68}{400}=5.78</math>  
  * VarianceCovariance: For a single column vector, <math>X=\{  +  * VarianceCovariance: For a single column vector, <math>X=\{X_1, X_2, X_3\} \sim NMD(k_0,\{p_1,p_2,p_3\})</math>, covariance between any pair of Negative Multinomial counts (<math>X_i</math> and <math>X_j</math>) is: 
  : <math> cov[X_i,X_j] = \begin{cases}k_0  +  : <math> cov[X_i,X_j] = \begin{cases} \frac{k_0 p_i p_j}{p_0^2},& i\not= j,\\ 
  k_0  +  \frac{k_0 p_i (p_i + p_0)}{p_0^2},& i=j.\end{cases}</math>. 
  :: ''Example'': For the first site (Head and Neck,  +  :: ''Example'': For the first site (Head and Neck, j=0), suppose that <math>X=\left \{X_1=5, X_2=1, X_3=5\right \}</math> and <math>X \sim NMD(k_0=10, \{p_1=0.2, p_2=0.1, p_3=0.2 \})</math>. Then: 
  :: <math>NMD(Xk_0,\{  +  :: <math>p_0 = 1  \sum_{i=1}^3{p_i}=0.5</math> 
  :: <math>cov[X_1,X_3] = \frac{  +  :: <math>NMD(Xk_0,\{p_1, p_2, p_3\})= 0.00465585119998784 </math> 
+  :: <math>cov[X_1,X_3] = \frac{10 \times 0.2 \times 0.2}{0.5^2}=1.6</math>  
+  :: <math>\mu_2=\frac{k_0 p_2}{p_0} = \frac{10\times 0.1}{0.5}=2.0</math>  
+  :: <math>\mu_3=\frac{k_0 p_3}{p_0} = \frac{10\times 0.2}{0.5}=4.0</math>  
+  :: <math>corr[X_2,X_3] = \left (\frac{\mu_2 \times \mu_3}{(k_0+\mu_2)(k_0+\mu_3)} \right )^{\frac{1}{2}}</math> and therefore, <math>corr[X_2,X_3] = \left (\frac{2 \times 4}{(10+2)(10+4)} \right )^{\frac{1}{2}} = 0.21821789023599242. </math>  
:: You can also use the interactive [http://socr.ucla.edu/htmls/dist/NegativeMultinomial_Distribution.html SOCR negative multinomial distribution calculator] to compute these quantities, as shown on the figure below.  :: You can also use the interactive [http://socr.ucla.edu/htmls/dist/NegativeMultinomial_Distribution.html SOCR negative multinomial distribution calculator] to compute these quantities, as shown on the figure below.  
  <center>[[Image:  +  <center>[[Image:SOCR_EBook_Dinov_RV_NegMultinomial_Fig8.png500px]]</center> 
* There is no [[AP_Statistics_Curriculum_2007_Estim_MOM_MLEMLE estimate]] for the NMD <math>k_0</math> parameter ([http://books.google.com/books?id=V7w7dDEKfuoC&pg=PA12&lpg=PA12&dq=example+%22negative+multinomial%22&source=bl&ots=fFRF5X3Dug&sig=7qbOZhv5ysHI_QVdVDxYZPL8Eg&hl=en&ei=g2jzStaMEo3QtgPF36QD&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBQQ6AEwAw#v=onepage&q=example%20%22negative%20multinomial%22&f=false see this reference]). However, there are approximate protocols for estimating the <math>k_0</math> parameter, see the example below.  * There is no [[AP_Statistics_Curriculum_2007_Estim_MOM_MLEMLE estimate]] for the NMD <math>k_0</math> parameter ([http://books.google.com/books?id=V7w7dDEKfuoC&pg=PA12&lpg=PA12&dq=example+%22negative+multinomial%22&source=bl&ots=fFRF5X3Dug&sig=7qbOZhv5ysHI_QVdVDxYZPL8Eg&hl=en&ei=g2jzStaMEo3QtgPF36QD&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBQQ6AEwAw#v=onepage&q=example%20%22negative%20multinomial%22&f=false see this reference]). However, there are approximate protocols for estimating the <math>k_0</math> parameter, see the example below.  
  * Correlation: correlation between any pair of Negative  +  * Correlation: correlation between any pair of Negative Multinomial counts (<math>X_i</math> and <math>X_j</math>) is: 
  : <math> Corr[X_i,X_j] = \begin{cases} \left (\frac{\mu_i \times \mu_j}{(k_0+\mu_i)(k_0+\mu_j)} \right )^{\frac{1}{2}}, & i\not= j, \\  +  : <math> Corr[X_i,X_j] = \begin{cases} \left (\frac{\mu_i \times \mu_j}{(k_0+\mu_i)(k_0+\mu_j)} \right )^{\frac{1}{2}} = 
+  \left (\frac{p_i p_j}{(p_0+p_i)(p_0+p_j)} \right )^{\frac{1}{2}}, & i\not= j, \\  
1, & i=j.\end{cases}</math>.  1, & i=j.\end{cases}</math>.  
  * The [http://en.wikipedia.org/wiki/Marginal_distribution marginal distribution] of each of the <math>X_i</math> variables is [[AP_Statistics_Curriculum_2007_Distrib_Dists#Negative_Binomial negative binomial]], as the <math>X_i</math> count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of <math>X=\{  +  * The [http://en.wikipedia.org/wiki/Marginal_distribution marginal distribution] of each of the <math>X_i</math> variables is [[AP_Statistics_Curriculum_2007_Distrib_Dists#Negative_Binomial negative binomial]], as the <math>X_i</math> count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of <math>X=\{X_1,\cdots,X_m\}</math> is negative multinomial, i.e., <math>X \sim NMD(k_0,\{p_1,\cdots,p_m\})</math> . 
  Notice that the pairwise NMD correlations are always positive, where as the correlations between [[AP_Statistics_Curriculum_2007_Distrib_Multinomial multinomail counts]] are always negative. Also note that as the parameter <math>k_0</math> increases, the paired correlations go to zero! Thus, for large <math>k_0</math>, the Negative Multinomial counts <math>X_i</math> behave as ''independent'' [[AP_Statistics_Curriculum_2007_Distrib_Poisson Poisson random variables]] with respect to their means  +  Notice that the pairwise NMD correlations are always positive, where as the correlations between [[AP_Statistics_Curriculum_2007_Distrib_Multinomial multinomail counts]] are always negative. Also note that as the parameter <math>k_0</math> increases, the paired correlations go to zero! Thus, for large <math>k_0</math>, the Negative Multinomial counts <math>X_i</math> behave as ''independent'' [[AP_Statistics_Curriculum_2007_Distrib_Poisson Poisson random variables]] with respect to their means <math>\left ( \mu_i= k_0\frac{p_i}{p_0}\right )</math>. 
  ====  +  ====Parameter estimation==== 
  +  * Estimation of the mean (expected) frequency counts (<math>\mu_j</math>) of each outcome (<math>X_j</math>):  
  : <math>\Chi^2 = \sum_i{\frac{(x_i\mu_i)^2}{\mu_i}}</math>  +  : The MLE estimates of the NMD mean parameters <math>\mu_j</math> are easy to compute. 
  we can replace the expectedmeans (<math>\mu_i</math>) by  +  :: If we have a single observation vector <math>\{x_1, \cdots,x_m\}</math>, then <math>\hat{\mu}_i=x_i.</math> 
+  :: If we have several observation vectors, like in this case we have the cancer type frequencies for 3 different sites, then the MLE estimates of the mean counts are <math>\hat{\mu}_j=\frac{x_{j,.}}{I}</math>, where <math>0\leq j \leq J</math> is the cancertype index and the summation is over the number of observed (sampled) vectors (I).  
+  :: For the cancer data above, we have the following MLE estimates for the expectations for the frequency counts:  
+  ::: Hutchinson's melanomic freckle type of cancer (<math>X_0</math>) is <math>\hat{\mu}_0 = 34/3=11.33</math>.  
+  ::: Superficial type of cancer (<math>X_1</math>) is <math>\hat{\mu}_1 = 185/3=61.67</math>.  
+  ::: Nodular type of cancer (<math>X_2</math>) is <math>\hat{\mu}_2 = 125/3=41.67</math>.  
+  ::: Indeterminant type of cancer (<math>X_3</math>) is <math>\hat{\mu}_3 = 56/3=18.67</math>.  
+  
+  * Estimation of the <math>k_0</math> (gamma) parameter:  
+  : There is no MLE for the <math>k_0</math> parameter; however, there is a protocol for estimating <math>k_0</math> using the [[AP_Statistics_Curriculum_2007_Contingency_Fit chisquared goodness of fit statistic]]. In the usual chisquared statistic:  
+  : <math>\Chi^2 = \sum_i{\frac{(x_i\mu_i)^2}{\mu_i}}</math>, we can replace the expectedmeans (<math>\mu_i</math>) by their estimates, <math>\hat{\mu_i}</math>, and replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data:  
: <math>\Chi^2(k_0) = \sum_{i}{\frac{(x_i\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}</math>.  : <math>\Chi^2(k_0) = \sum_{i}{\frac{(x_i\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}</math>.  
  Now we can derive a simple method for estimating the <math>k_0</math> parameter by varying the  +  : Now we can derive a simple method for estimating the <math>k_0</math> parameter by varying the values of <math>k_0</math> in the expression <math>\Chi^2(k_0)</math> and matching the values of this statistic with the corresponding asymptotic chisquared distribution. The following protocol summarizes these steps using the cancer data above: 
  values of <math>k_0</math> in the expression <math>\Chi^2(k_0)</math> and matching the values of this statistic with the corresponding asymptotic chisquared distribution. The following protocol summarizes these steps using the cancer data above:  +  
* ''DF'': The [[AP_Statistics_Curriculum_2007_Contingency_Indep#Calculations degree of freedom for the Chisquare distribution]] in this case is:  * ''DF'': The [[AP_Statistics_Curriculum_2007_Contingency_Indep#Calculations degree of freedom for the Chisquare distribution]] in this case is:  
Line 179:  Line 238:  
* ''Median'': The [http://socr.ucla.edu/htmls/dist/ChiSquare_Distribution.html median of a chisquared random variable with 6 df] is 5.261948.  * ''Median'': The [http://socr.ucla.edu/htmls/dist/ChiSquare_Distribution.html median of a chisquared random variable with 6 df] is 5.261948.  
  * ''Mean Counts Estimates'': The mean counts estimates (<math>\  +  * ''Mean Counts Estimates'': The mean counts estimates (<math>\mu_j</math>) for the 4 different cancer types are: 
  ::  +  ::<math>\hat{\mu}_1 = 185/3=61.67</math>; <math>\hat{\mu}_2 = 125/3=41.67</math>; and <math>\hat{\mu}_3 = 56/3=18.67</math>. 
+  
+  * Thus, we can solve the equation above <math>\Chi^2(k_0) = 5.261948</math> for the single variable of interest  the unknown parameter <math>k_0</math>. Suppose we are using the same example as before, <math>x=\{x_1=5,x_2=1,x_3=5\}</math>. Then the solution is an asymptotic chisquared distribution driven estimate of the parameter <math>k_0</math>.  
+  <math>\Chi^2(k_0) = \sum_{i=1}^3{\frac{(x_i\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}</math>.  
+  <math>\Chi^2(k_0) = \frac{(561.67)^2}{61.67(1+61.67/k_0)}+\frac{(141.67)^2}{41.67(1+41.67/k_0)}+\frac{(518.67)^2}{18.67(1+18.67/k_0)}=5.261948.</math> Solving this equation for <math>k_0</math> provides the desired estimate for the last parameter.  
+  :: [http://www.mathematica.com/ Mathematica] provides 3 distinct (<math>k_0</math>) solutions to this equation: {'''50.5466''', 21.5204, '''2.40461'''}. Since <math>k_0>0</math> there are 2 candidate solutions.  
+  
+  * '''Estimates of Probabilities''': Assume <math>k_0=2</math> and <math>\frac{\mu_i}{k_0}p_0=p_i</math>, we have:  
+  : <math>\frac{61.67}{k_0}p_0=31p_0=p_1</math>  
+  : <math>20p_0=p_2</math>  
+  : <math>9p_0=p_3</math>  
+  : Hence, <math>1p_0=p_1+p_2+p_3=60p_0</math>. Therefore, <math>p_0=\frac{1}{61}</math>, <math>p_1=\frac{31}{61}</math>, <math>p_2=\frac{20}{61}</math> and <math>p_3=\frac{9}{61}</math>.  
+  : Therefore, the best model distribution for the observed sample <math>x=\{x_1=5,x_2=1,x_3=5\}</math> is <math>X \sim NMD\left (2, \left \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\right\} \right ).</math>  
  +  : Notice that in this calculation, we explicitly used the complete cancer data table, not only the sample <math>x=\{x_1=5,x_2=1,x_3=5\}</math>, as we need multiple samples (multiple sites or columns) to estimate the <math>k_0</math> parameter.  
====SOCR Negative Multinomial Distribution Calculator====  ====SOCR Negative Multinomial Distribution Calculator==== 
Current revision as of 19:35, 23 June 2012
Contents 
General AdvancePlacement (AP) Statistics Curriculum  Geometric, HyperGeometric, Negative Binomial Random Variables and Experiments
Geometric
 Definition: The Geometric Distribution is the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set {1, 2, 3, ...}. The name geometric is a direct derivative from the mathematical notion of geometric series.
 Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is , for x = 1, 2, 3, 4,....
 Expectation: The Expected Value of a geometrically distributed random variable X is This is because geometric series have this property:
 Let r=(1p), then p=(1r) and
 Thus: which converges to 1, as , and hence the above geometric density is well defined.
 Denote the geometric expectation by E = E(X) = , where r=1p. Then . Therefore, .
 Variance: The Variance is
 Example: See this SOCR Geometric distribution activity.
 The Geometric distribution gets its name because its probability mass function is a geometric progression. It is the discrete analogue of the Exponential distribution and is also known as Furry distribution.
HyperGeometric
The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. An experimental design for using Hypergeometric distribution is illustrated in this table:
Type  Drawn  NotDrawn  Total 
Defective  k  mk  m 
NonDefective  nk  N+knm  Nm 
Total  n  Nn  N 
 Explanation: Suppose there is a shipment of N objects in which m are defective. The Hypergeometric Distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective.
 Mass function: The random variable X follows the Hypergeometric Distribution with parameters N, m and n, then the probability of getting exactly k successes is given by
This formula for the Hypergeometric Mass Function may be interpreted as follows: There are possible samples (without replacement). There are ways to obtain k defective objects and there are ways to fill out the rest of the sample with nondefective objects.
The mean and variance of the hypergeometric distribution have the following closed forms:
 Mean:
 Variance:
Examples
 SOCR Activity: The SOCR Ball and Urn Experiment provides a handson demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N  R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can vary with scroll bars.
 A lake contains 1,000 fish; 100 are randomly caught and tagged. Suppose that later we catch 20 fish. Use SOCR Hypergeometric Distribution to:
 Compute the probability mass function of the number of tagged fish in the sample of 20.
 Compute the expected value and the variance of the number of tagged fish in this sample.
 Compute the probability that this random sample contains more than 3 tagged fish.
 Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish, we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the samplesize (n) is < 5% of the population size(N), we can use binomial approximation to hypergeometric. Thus if the sample of n=100 fish had 5 tagged, the sampleproportion (estimate of the population proportion) will be . Thus, we can estimate that , and , as shown in the figure below.
 You can also see a manual calculation example using the hypergeometric distribution here.
Negative Binomial
The family of Negative Binomial Distributions is a twoparameter family; p and r with 0 < p < 1 and r > 0. There are two (identical) combinatorial interpretations of Negative Binomial processes (X or Y).
X=Trial index (n) of the r^{th} success, or Total # of experiments (n) to get r successes
 Probability Mass Function: , for n = r,r+1,r+2,.... (n=trial number of the r^{th} success)
 Mean:
 Variance:
Y = Number of failures (k) to get r successes
 Probability Mass Function: , for k = 0,1,2,.... (k=number of failures before the r^{th} successes)
 , the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial.
 Mean: .
 Variance: .
 Note that X = Y + r, and E(X) = E(Y) + r, whereas VAR(X)=VAR(Y).
SOCR Negative Binomial Experiment
Application
Suppose Jane is promoting and fundraising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and a 70% chance that she'll fail.
 What's the probability mass function of the number of failures (k=nr) to get r=6 successes?
 In other words, what's the probability mass function that the last 6^{th} state she succeeds to secure all electoral votes happens to be at the n^{th} state she campaigns in?
NegBin(r, p) distribution describes the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is n=k+6. The random variable we are interested in is X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}. So, n = k+6, and . Thus, for , the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:
 What's the probability that Jane finishes her campaign in the 10^{th} state?
 Let , then
 What's the probability that Jane finishes campaigning on or before reaching the 8^{th} state?
 Suppose the success of getting all electoral votes within a state is reduced to only 10%, then X~NegBin(r=6, p=0.1). Notice that the shape and domain the NegativeBinomial distribution significantly chance now (see image below).
 What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?
 SOCR Activity: If you want to see an interactive NegativeBinomial Graphical calculator you can go to this applet (select Negative Binomial) and see this activity.
Normal approximation to Negative Binomial distribution
The central limit theorem provides the foundation for approximation of negative binomial distribution by Normal distribution. Each negative binomial random variable, \(V_k \sim NB(k,p)\), may be expressed as a sum of k independent, identically distributed (geometric) random variables \(\{X_i\}\), i.e., \( V_k = \sum_{i=1}^k{X_i}\), where \( X_i \sim Geometric(p)\). In various scientific applications, given a large k, the distribution of \(V_k\) is approximately normal with mean and variance given by \(\mu=k\frac{1}{p}\) and \(\sigma^2=k\frac{1p}{p^2}\), as \(k \longrightarrow \infty\). Depending on the parameter p, k may need to be rather large for the approximation to work well. Also, when using the normal approximation, we should remember to use the continuity correction, since the negative binomial and Normal distributions are discrete and continuous, respectively.
In the above example, \(P(X\le 8)\), \(V_k \sim NegBin(k=r=6, p=0.3)\), the normal distribution approximation, \(N(\mu=\frac{k}{p}=20, \sigma=\sqrt{k\frac{1p}{p^2}}=6.83)\), is shown it the following image and table:
The probabilities of the real Negative Binomial and approximate Normal distributions (on the range [2:4]) are not identical but are sufficiently close.
Summary  \(NegativeBinomial(k=6,p=0.3)\)  \(Normal(\mu=20, \sigma=6.83)\) 

Mean  20.0  20.0 
Median  19.0  20.0 
Variance  46.666667  46.6489 
Standard Deviation  6.831301  6.83 
Max Density  0.062439  0.058410 
Probability Areas  
\(\le 8\)  .011292  0.039433 
>8  .988708  0.960537 
Negative Multinomial Distribution (NMD)
The Negative Multinomial Distribution is a generalization of the twoparameter Negative Binomial distribution (NB(r,p)) to outcomes. Suppose we have an experiment that generates possible outcomes, , each occurring with probability , respectively, where with 0 < p_{i} < 1 and . That is, . If the experiment proceeds to generate independent outcomes until occur exactly times, then the distribution of the mtuple is Negative Multinomial with parameter vector . Notice that the degreeoffreedom here is actually m, not (m+1). That is why we only have a probability parameter vector of size m, not (m+1), as all probabilities add up to 1 (so this introduces one relation). Contrast this with the combinatorial interpretation of Negative Binomial (special case with m=1):
 X˜NegativeBinomial(NumberOfSuccesses = r,ProbOfSuccess = p),
 X=Total # of experiments (n) to get r successes (and therefore nr failures);
 X˜NegativeMultinomial(k_{0},{p_{0},p_{1}}),
 X=Total # of experiments (n) to get k_{0} (default variable, X_{o}) and n − k_{0} outcomes of the other possible outcome (X_{1}).
Negative Multinomial Summary
 Probability Mass Function: , or equivalently:
 , where Γ(x) is the Gamma function.
 Mean (vector): .
 VarianceCovariance (matrix): Cov(X_{i},X_{j}) = {cov[i,j]}, where
 .
Cancer Example
The Probability Theory Chapter of the EBook shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as shown in the Table below.
Type  Site  Totals  
Head and Neck  Trunk  Extremities  
Hutchinson's melanomic freckle  22  2  10  34 
Superficial  16  54  115  185 
Nodular  19  33  73  125 
Indeterminant  11  17  28  56 
Column Totals  68  106  226  400 
The sites (locations) of the cancer may be independent, but there may be positive dependencies of the type of cancer for a given location (site). For example, localized exposure to radiation implies that elevated level of one type of cancer (at a given location) may indicate higher level of another cancer type at the same location. We want to use the Negative Multinomial distribution to model the sites cancer rates and try to measure some of the cancer type dependencies within each location.
Let's denote by x_{i,j} the cancer rates for each site () and each type of cancer (). For each (fixed) site (), the cancer rates are independent Negative Multinomial distributed random variables. That is, for each column index (site) the columnvector X has the following distribution:
 X = {X_{1},X_{2},X_{3}}˜NMD(k_{0},{p_{1},p_{2},p_{3}}).
Different columns (sites) are considered to be different instances of the random negativemultinomially distributed vector, X. Then we have the following estimates:
 MLE estimate of the Mean: is given by:

 Example:
 VarianceCovariance: For a single column vector, X = {X_{1},X_{2},X_{3}}˜NMD(k_{0},{p_{1},p_{2},p_{3}}), covariance between any pair of Negative Multinomial counts (X_{i} and X_{j}) is:
 .
 Example: For the first site (Head and Neck, j=0), suppose that and X˜NMD(k_{0} = 10,{p_{1} = 0.2,p_{2} = 0.1,p_{3} = 0.2}). Then:
 NMD(X  k_{0},{p_{1},p_{2},p_{3}}) = 0.00465585119998784
 and therefore,
 You can also use the interactive SOCR negative multinomial distribution calculator to compute these quantities, as shown on the figure below.
 There is no MLE estimate for the NMD k_{0} parameter (see this reference). However, there are approximate protocols for estimating the k_{0} parameter, see the example below.
 Correlation: correlation between any pair of Negative Multinomial counts (X_{i} and X_{j}) is:
 .
 The marginal distribution of each of the X_{i} variables is negative binomial, as the X_{i} count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of is negative multinomial, i.e., .
Notice that the pairwise NMD correlations are always positive, where as the correlations between multinomail counts are always negative. Also note that as the parameter k_{0} increases, the paired correlations go to zero! Thus, for large k_{0}, the Negative Multinomial counts X_{i} behave as independent Poisson random variables with respect to their means .
Parameter estimation
 Estimation of the mean (expected) frequency counts (μ_{j}) of each outcome (X_{j}):
 The MLE estimates of the NMD mean parameters μ_{j} are easy to compute.
 If we have a single observation vector , then
 If we have several observation vectors, like in this case we have the cancer type frequencies for 3 different sites, then the MLE estimates of the mean counts are , where is the cancertype index and the summation is over the number of observed (sampled) vectors (I).
 For the cancer data above, we have the following MLE estimates for the expectations for the frequency counts:
 Hutchinson's melanomic freckle type of cancer (X_{0}) is .
 Superficial type of cancer (X_{1}) is .
 Nodular type of cancer (X_{2}) is .
 Indeterminant type of cancer (X_{3}) is .
 Estimation of the k_{0} (gamma) parameter:
 There is no MLE for the k_{0} parameter; however, there is a protocol for estimating k_{0} using the chisquared goodness of fit statistic. In the usual chisquared statistic:
 , we can replace the expectedmeans (μ_{i}) by their estimates, , and replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data:
 .
 Now we can derive a simple method for estimating the k_{0} parameter by varying the values of k_{0} in the expression Χ^{2}(k_{0}) and matching the values of this statistic with the corresponding asymptotic chisquared distribution. The following protocol summarizes these steps using the cancer data above:
 DF: The degree of freedom for the Chisquare distribution in this case is:
 df = (# rows – 1)(# columns – 1) = (31)*(41) = 6
 Median: The median of a chisquared random variable with 6 df is 5.261948.
 Mean Counts Estimates: The mean counts estimates (μ_{j}) for the 4 different cancer types are:
 ; ; and .
 Thus, we can solve the equation above Χ^{2}(k_{0}) = 5.261948 for the single variable of interest  the unknown parameter k_{0}. Suppose we are using the same example as before, x = {x_{1} = 5,x_{2} = 1,x_{3} = 5}. Then the solution is an asymptotic chisquared distribution driven estimate of the parameter k_{0}.
. Solving this equation for k_{0} provides the desired estimate for the last parameter.
 Mathematica provides 3 distinct (k_{0}) solutions to this equation: {50.5466, 21.5204, 2.40461}. Since k_{0} > 0 there are 2 candidate solutions.
 Estimates of Probabilities: Assume k_{0} = 2 and , we have:
 20p_{0} = p_{2}
 9p_{0} = p_{3}
 Hence, 1 − p_{0} = p_{1} + p_{2} + p_{3} = 60p_{0}. Therefore, , , and .
 Therefore, the best model distribution for the observed sample x = {x_{1} = 5,x_{2} = 1,x_{3} = 5} is
 Notice that in this calculation, we explicitly used the complete cancer data table, not only the sample x = {x_{1} = 5,x_{2} = 1,x_{3} = 5}, as we need multiple samples (multiple sites or columns) to estimate the k_{0} parameter.
SOCR Negative Multinomial Distribution Calculator
Problems
References
 NegativeBinomial Activity
 Le Gall, F. The modes of a negative multinomial distribution, Statistics & Probability Letters, 2005.
 Johnson et al., 1997 Johnson, N.L., Kotz, S., Balakrishnan, N., 1997. Discrete Multivariate Distributions. Wiley Series in Probability and Mathematical Statistics.
 Kotz and Johnson, 1982 In: S. Kotz and N.L. Johnson, Editors, Encyclopedia of Statistical Sciences, Wiley, New York (1982).
 SOCR Home page: http://www.socr.ucla.edu
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