# AP Statistics Curriculum 2007 Distrib Dists

(Difference between revisions)
 Revision as of 17:26, 6 November 2009 (view source)IvoDinov (Talk | contribs) (→Negative Multinomial Distribution (NMD): added estimation of k_0)← Older edit Revision as of 17:27, 6 November 2009 (view source)IvoDinov (Talk | contribs) m (→k_0 Parameter estimation)Newer edit → Line 165: Line 165: Notice that the pair-wise 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 $k_0$ increases, the paired correlations go to zero! Thus, for large $k_0$, the Negative Multinomial counts $X_i$ behave as ''independent'' [[AP_Statistics_Curriculum_2007_Distrib_Poisson |Poisson random variables]] with respect to their means ($\mu_i$). Notice that the pair-wise 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 $k_0$ increases, the paired correlations go to zero! Thus, for large $k_0$, the Negative Multinomial counts $X_i$ behave as ''independent'' [[AP_Statistics_Curriculum_2007_Distrib_Poisson |Poisson random variables]] with respect to their means ($\mu_i$). - ====k_0 Parameter estimation==== + ====$k_0$ Parameter estimation==== We already discussed that for the NMD the MLE estimates of the mean parameters $\mu_i$ are easy to compute. There is no MLE for the $k_0$ parameter, however, there is a protocol for estimating $k_0$ using the [[AP_Statistics_Curriculum_2007_Contingency_Fit |chi-squared goodness of fit statistic]]. In the usual chi-squared statistic: We already discussed that for the NMD the MLE estimates of the mean parameters $\mu_i$ are easy to compute. There is no MLE for the $k_0$ parameter, however, there is a protocol for estimating $k_0$ using the [[AP_Statistics_Curriculum_2007_Contingency_Fit |chi-squared goodness of fit statistic]]. In the usual chi-squared statistic: - : $\Chi^2 = \sum_{i}{\frac{x_i − \mu_i)^2}{\mu_i}$ + : $\Chi^2 = \sum_{i}{\frac{x_i − \mu_i)^2}{\mu_i}}$ we can replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data: we can replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data: - : $\Chi^2(k_0) = \sum_{i}{\frac{x_i − \hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}$. + : $\Chi^2(k_0) = \sum_{i}{\frac{x_i − \hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$. ====SOCR Negative Multinomial Distribution Calculator==== ====SOCR Negative Multinomial Distribution Calculator====

## General Advance-Placement (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 $P(X = x) = (1 - p)^{x-1} \times p$, for x = 1, 2, 3, 4,....
• Expectation: The Expected Value of a geometrically distributed random variable X is ${1\over p}.$
• Variance: The Variance is ${1-p\over p^2}.$

### 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 Not-Drawn Total Defective k m-k m Non-Defective n-k N+k-n-m N-m Total n N-n 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
$P(X=k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}.$

This formula for the Hypergeometric Mass Function may be interpreted as follows: There are ${{N}\choose{n}}$ possible samples (without replacement). There are ${{m}\choose{k}}$ ways to obtain k defective objects and there are ${{N-m}\choose{n-k}}$ ways to fill out the rest of the sample with non-defective objects.

The mean and variance of the hypergeometric distribution have the following closed forms:

Mean: $n \times m\over N$
Variance: ${ {nm\over N} ( 1-{m\over N} ) (N-n)\over N-1}$

#### Examples

• SOCR Activity: The SOCR Ball and Urn Experiment provides a hands-on 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 be varied 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 sample-size (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 sample-proportion (estimate of the population proportion) will be $\hat{p}={5\over 100}=0.05$. Thus, we can estimate that $0.05=\hat{p}={200\over N}$, and $N\approx 4,000$, as shown on the figure below.

### Negative Binomial

The family of Negative Binomial Distributions is a two-parameter 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 rth success, or Total # of experiments (n) to get r successes

• Probability Mass Function: $P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} \!$, for n = r,r+1,r+2,.... (n=trial number of the rth success)
• Mean: $E(X)= {r \over p}$
• Variance: $Var(X)= {r(1-p) \over p^2}$

#### Y = Number of failures (k) to get r successes

• Probability Mass Function: $P(Y=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k \!$, for k = 0,1,2,.... (k=number of failures before the rth successes)
• $Y \sim NegBin(r, p)$, the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial.
• Mean: $E(Y)= {r(1-p) \over p}$.
• Variance: $Var(Y)= {r(1-p) \over p^2}$.
• Note that X = Y + r, and E(X) = E(Y) + r, whereas VAR(X)=VAR(Y).

#### Application

Suppose Jane is promoting and fund-raising 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.

• What's the probability mass function of the number of failures (k=n-r) to get r=6 successes?
In other words, What's the probability mass function that the last 6th state she succeeds to secure all electoral votes happens to be the at the nth 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 $X\sim NegBin(r=6, p=0.3)$. Thus, for $n \geq 6$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:

$P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} = {n - 1 \choose r-1} \cdot 0.3^6 \cdot 0.7^{n-r}$
• What's the probability that Jane finishes her campaign in the 10th state?
Let $X\sim NegBin(r=6, p=0.3)$, then $P(X=10) = {10-1 \choose 6-1}\cdot 0.3^6 \cdot 0.7^{10-6} = 0.022054.$
• What's the probability that Jane finishes campaigning on or before reaching the 8th state?
$P(X\leq 8) = 0.011292$
• 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 Negative-Binomial 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)?
$P(X\geq 50) = 0.632391$

### Negative Multinomial Distribution (NMD)

The Negative Multinomial Distribution is a generalization of the two-parameter Negative Binomial distribution (NB(r,p)) to $m\ge 1$ outcomes. Suppose we have an experiment that generates $m\ge 1$ possible outcomes, $\{X_0,\cdots,X_m\}$, each occurring with probability $\{p_0,\cdots,p_m\}$, respectively, where with 0 < pi < 1 and $\sum_{i=0}^m{p_i}=1$. That is, $p_0 = 1-\sum_{i=1}^m{p_i}$. If the experiment proceeds to generate independent outcomes until $\{X_0, X_1, \cdots, X_m\}$ occur exactly $\{k_0, k_1, \cdots, k_m\}$ times, the distribution of the (m+1)-tuple $\{k_0, k_1, \cdots, k_m\}$ is Negative Multinomial with parameter vector $(k_0,\{p_0,\cdots,p_m\})$. 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;
X˜NegativeMultinomial(k0,{p0,p1}),
X=Total # of experiments (n) to get k0 and nk0 outcomes of each of the m+1=2 possible outcomes.

#### Negative Multinomial Summary

• Probability Mass Function: $P(k_0, \cdots, k_m) = \left (\sum_{i=0}^m{k_i}-1\right)!\frac{p_0^{k_0}}{(k_0-1)!} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, or equivalently:
$P(k_0, \cdots, k_m) = \Gamma\left(\sum_{i=1}^m{k_i}\right)\frac{p_0^{k_0}}{\Gamma(k_0)} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, where Γ(x) is the Gamma function.
• Mean (vector): $E(k_0,\cdots,k_m)= (E(k_0), \cdots, E(k_m)) = (k_0*p_0, \cdots, k_m*p_m)$.
• Variance-Covariance (matrix): $Cov(k_0,\cdots,k_m)= \{cov[i,j]\}$, where $cov[i,j] = \begin{cases} \frac{k_0 * p[i] * p[j]}{p_0 * p_0},& i\not= j,\\ \frac{k_0* p[i] * (p[i] + p_0)}{p_0 * p_0},& i=j.\end{cases}$.

#### 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 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 yi,j the cancer rates for each site ($0\leq i \leq 2$) and each type of cancer ($0\leq j \leq 3$). For each (fixed) site ($0\leq i \leq 2$), the cancer rates are independent Negative Multinomial distributed random variables. That is, for each column index (site) the column-vector X

X = {X0,X1,X2,X3NMD(k0,{p0,p1,p2,p3}).

Different columns (sites) are considered to be different instances of the random multinomially distributed X vector. Then we have the following estimates:

$\hat{\mu}_{i,j} = \frac{x_{i,.}\times x_{.,j}}{x_{.,.}}$, where
$x_{i,.} = \sum_{j=0}^{3}{x_{i,j}}$
$x_{.,j} = \sum_{i=0}^{2}{x_{i,j}}$
$x_{.,.} = \sum_{i=0}^{2}\sum_{j=0}^{3}{{x_{i,j}}}$
Example: $\hat{\mu}_{1,1} = \frac{x_{1,.}\times x_{.,1}}{x_{.,.}}=\frac{34\times 68}{400}=5.78$
• Variance-Covariance: For a single column vector, X = {X0,X1,X2,X3NMD(k0,{p0,p1,p2,p3}), covariance between any pair of Negative Multinomial counts (Xi and Xj) is:
$cov[X_i,X_j] = \begin{cases}k_0 * p[X_i] * p[X_j] / (p_0 * p_0),& i\not= j,\\ k_0* p[X_i] * (p[X_i] + p_0) / (p_0 * p_0),& i=j.\end{cases}$.
Example: For the first site (Head and Neck, i=0), suppose that $X=\left \{X_0=3, X_1=5, X_2=7, X_3=10\right \}$ and X˜NMD(k0 = 20,{p0 = 0.3,p1 = 0.2,p2 = 0.1,p3 = 0.2}). Then:
NMD(X | k0,{p0,p1,p2,p3}) = 1.5395445930162726E − 9
$cov[X_1,X_3] = \frac{20 * 0.2 * 0.2}{0.3^2}=8.89$
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 k0 parameter (see this reference). However, there are approximate protocols for estimating the k0 parameter, see the example below.
• Correlation: correlation between any pair of Negative Binomial counts (Xi and Xj) is:
$Corr[X_i,X_j] = \begin{cases} \left (\frac{\mu_i \times \mu_j}{(k_0+\mu_i)(k_0+\mu_j)} \right )^{\frac{1}{2}}, & i\not= j, \\ 1, & i=j.\end{cases}$.
• The marginal distribution of each of the Xi variables is negative binomial, as the Xi count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of $X=\{X_0,\cdots,X_m\}$ is negative multinomial.

Notice that the pair-wise NMD correlations are always positive, where as the correlations between multinomail counts are always negative. Also note that as the parameter k0 increases, the paired correlations go to zero! Thus, for large k0, the Negative Multinomial counts Xi behave as independent Poisson random variables with respect to their means (μi).

#### k0 Parameter estimation

We already discussed that for the NMD the MLE estimates of the mean parameters μi are easy to compute. There is no MLE for the k0 parameter, however, there is a protocol for estimating k0 using the chi-squared goodness of fit statistic. In the usual chi-squared statistic:

Failed to parse (lexing error): \Chi^2 = \sum_{i}{\frac{x_i − \mu_i)^2}{\mu_i}}

we can replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data:

Failed to parse (lexing error): \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 )}}

.