Chapter 6.5, Problem 53E

Let $Y_1,Y_2,\dots ,Y_n$ be independent binomial random variable with n_i, trials and probability of success given by $p_i$, $i=1,2,\dots ,n$.

1. a If all of the $n_i$’s are equal and all of the p’s are equal, find the distribution of ${\displaystyle {\sum }_{i=1}^n{Y}_i}$.
2. b If all of the $n_i$’s are different and all of the p’s are equal, find the distribution of ${\displaystyle {\sum }_{i=1}^n{Y}_i}$.
3. c If all of the $n_i$’s are different and all of the p’s are equal, find the conditional distribution Y₁ given ${\displaystyle {\sum }_{i=1}^n{Y}_i=m}$.
4. d If all of the $n_i$’s are different and all of the p’s are equal, find the conditional distribution $Y_1+Y_2$ given ${\displaystyle {\sum }_{i=1}^n{Y}_i=m}$.
5. e If all of the p’s are different, does the method of moment-generating functions work well to find the distribution of ${\displaystyle {\sum }_{i=1}^n{Y}_i}$? Why?

To determine

Find the distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$ if all of the $n_i{}^{\prime }s$ are equal and all of the p’s are equal.

Answer to Problem 53E

The distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$ is Binomial distribution with m(n) trails and the probability of success p.

Explanation of Solution

Calculation:

From the given information, $Y_1,Y_2,\mathrm{...},Y_n$ are independent binomial random variables with $n_i$ trails and $p_i,i=1,2,\mathrm{...},n$, respectively.

The moment generating function for $Y_i$ with $n_i$ trails and the probability of success $p_i$, is $m_{Y_i}\left(t\right)={\left[1-p_i+p_i{e}^t\right]}^{n_i}$.

Consider,

$\begin{array}{c}{m}_{U={\displaystyle \sum _{i=1}^n{Y}_i}}\left(t\right)=m_{Y_1}\left(t\right)\times m_{Y_2}\left(t\right)\times \mathrm{...}\times m_{Y_n}\left(t\right)\\ ={\displaystyle \prod _i{\left[1-p_i+p_i{e}^t\right]}^{n_i}}\end{array}$

Given that $p_i=p$ and $n_i=m$ for all i.

Therefore,

$\begin{array}{c}{m}_U\left(t\right)={\displaystyle \prod _{i=1}^n{\left[1-p+p{e}^t\right]}^m}\\ ={\left[1-p+p{e}^t\right]}^{mn}\end{array}$

This is the moment generating function of Binomial distribution with m(n) trails and the probability of success p. Thus, the random variable ${\displaystyle \sum _{i=1}^n{Y}_i}$ has Binomial distribution with m(n) trails and the probability of success p.

To determine

Find the distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$ if all of the $n_i{}^{\prime }s$ are different and all of the p’s are equal.

Answer to Problem 53E

The distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$ is Binomial distribution with ${\displaystyle \sum n_i}$ trails and the probability of success p.

Explanation of Solution

Calculation:

Given that $p_i=p$ and $n_i{}^{\prime }s$ trails for all i.

Therefore,

$\begin{array}{c}{m}_U\left(t\right)={\displaystyle \prod _i{\left[1-p+p{e}^t\right]}^{n_i}}\\ ={\left[1-p+p{e}^t\right]}^{{\displaystyle \sum n_i}}\end{array}$

This is the moment generating function of Binomial distribution with ${\displaystyle \sum n_i}$ trails and the probability of success p. Thus, the random variable ${\displaystyle \sum _{i=1}^n{Y}_i}$ has Binomial distribution with ${\displaystyle \sum n_i}$ trails and the probability of success p.

To determine

Find the conditional distribution of Y₁ given ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ if all of the $n_i{}^{\prime }s$ are different and all of the p’s are equal.

Answer to Problem 53E

The conditional distribution of Y₁ given ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ if all of the $n_i{}^{\prime }s$ are different and all of the p’s are equal has hypergeometric distribution with $r=n_i$ and $N={\displaystyle \sum _i{n}_i}$.

Explanation of Solution

Calculation:

From Exercise 5-40, the conditional distribution of Y₁ given ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ is hypergeometric distribution with $r=n_i$ and $N=n_1+n_2+\mathrm{...}+n_m$.

Thus, the conditional distribution of Y₁ given ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ if all of the $n_i{}^{\prime }s$ are different and all of the p’s are equal is hypergeometric distribution with $r=n_i$ and $N={\displaystyle \sum _i{n}_i}$.

To determine

Find the conditional probability function of $Y_1+Y_2$, given that ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ if all of the $n_i{}^{\prime }s$ are different and all of the p’s are equal.

Answer to Problem 53E

The conditional probability function of $Y_1+Y_2$, given that ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ has a hypergeometric distribution with $r=n_1+n_2$.

Explanation of Solution

Calculation:

Consider,

$\begin{array}{c}P\left(Y_1+Y_2=k|{\displaystyle \sum _{i=1}^n{Y}_i}\right)=\frac{P\left(Y_1+Y_2=k,{\displaystyle \sum _{i=1}^n{Y}_i}=m\right)}{P\left({\displaystyle \sum _{i=1}^n{Y}_i}=m\right)}\\ =\frac{P\left(Y_1+Y_2=k,{\displaystyle \sum _{i=3}^n{Y}_i}=m-k\right)}{P\left({\displaystyle \sum _{i=1}^n{Y}_i}=m\right)}\\ =\frac{P\left(Y_1+Y_2=k\right)P\left({\displaystyle \sum _{i=3}^n{Y}_i}=m-k\right)}{P\left({\displaystyle \sum _{i=1}^n{Y}_i}=m\right)}\\ =\frac{\left(\begin{array}{l}{n}_1+n_2\\ k\end{array}\right)\left(\begin{array}{l}{\displaystyle \sum _{i=3}^n{n}_i}\\ m-k\end{array}\right)}{\left(\begin{array}{l}{\displaystyle \sum _{i=1}^n{n}_i}\\ m\end{array}\right)}\end{array}$

Therefore, the conditional probability function of $Y_1+Y_2$, given that ${\displaystyle \sum _{i=1}^n{Y}_i}=m$ has a hypergeometric distribution with $r=n_1+n_2$.

To determine

Observe whether the method of moment-generating functions work well to find the distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$ or not.

Answer to Problem 53E

No. The method of moment-generating functions does not work well for finding the distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$.

Explanation of Solution

Calculation:

If all p’s are different, then the moment generating function is,

$\begin{array}{c}{m}_{U={\displaystyle \sum _{i=1}^n{Y}_i}}\left(t\right)=m_{Y_1}\left(t\right)\times m_{Y_2}\left(t\right)\times \mathrm{...}\times m_{Y_n}\left(t\right)\\ ={\displaystyle \prod _i{\left[1-p_i+p_i{e}^t\right]}^{n_i}}\end{array}$

Here, the moment generating function for U does not divided into individual factors of Y_i’s. That is, the mgf of U does not simplified into recognizable form. Hence, the moment-generating functions does not work well for finding the distribution of ${\displaystyle \sum _{i=1}^n{Y}_i}$.