Chapter 4.7, Problem 135E

a.

To determine

Use the binomial formula to prove that $P\left(Y_1=0\right)>P\left(Y_2=0\right)$ subjected to the condition is $p_1<p_2$.

Answer to Problem 135E

It is proved that $P\left(Y_1=0\right)>P\left(Y_2=0\right)$.

Explanation of Solution

The random variable $Y_1$ follows a binomial distribution with parameters $n$ and $p_1$. The probability mass function for $Y_1$ is given below:

$P\left(Y_2=y_1\right)=\left(\begin{array}{l}n\\ y_1\end{array}\right)p_2{}^{y_1}{\left(1-p_2\right)}^{n-y_1},y_1=0,1,2,\mathrm{....}n$

The random variable $Y_2$ follows a binomial distribution with parameters $n$ and $p_2$. The probability mass function for $Y_2$ is given below:

$P\left(Y_2=y_2\right)=\left(\begin{array}{l}n\\ y_2\end{array}\right)p_2{}^{y_2}{\left(1-p_2\right)}^{n-y_2},y_2=0,1,2,\mathrm{....}n$

Consider,

$\begin{array}{c}P\left(Y_1=0\right)=\left(\begin{array}{l}n\\ 0\end{array}\right)p_1{}^0{\left(1-p_1\right)}^{n-0}\\ ={\left(1-p_1\right)}^n\end{array}$

$\begin{array}{c}P\left(Y_2=0\right)=\left(\begin{array}{l}n\\ 0\end{array}\right)p_2{}^0{\left(1-p_2\right)}^{n-0}\\ ={\left(1-p_2\right)}^n\end{array}$

It is noted that parameter $n$ is same for the two variables and subjected to the condition is $p_1<p_2$. This implies that,

$\begin{array}{l}{\left(1-p_1\right)}^n>{\left(1-p_2\right)}^n\\ P\left(Y_1=0\right)>P\left(Y_2=0\right),p_1<p_2\end{array}$

Therefore, it is proved that $P\left(Y_1=0\right)>P\left(Y_2=0\right)$.

b.

To determine

Use the relationship between the beta distribution function and sum of binomial probabilities given in Exercise 4.134. If k is an integer between 1 and $n-1$, deduce as follows:

$P\left(Y_1\le k\right)={\displaystyle \sum _{i=0}^n\left(\begin{array}{l}n\\ i\end{array}\right)}{p}_1{}^i{\left(1-p_1\right)}^{n-i}={\displaystyle \underset{p_1}{\overset{1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}$

Answer to Problem 135E

It is deduced that $P\left(Y_1\le k\right)={\displaystyle \underset{p_1}{\overset{1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}$.

Explanation of Solution

Consider L.H.S:

$\begin{array}{c}P\left(Y_1\le k\right)=1-P\left(Y_1\ge k+1\right)\\ =1-{\displaystyle \sum _{i=k+1}^n\left(\begin{array}{l}n\\ i\end{array}\right)}{p}_1{}^i{\left(1-p_1\right)}^{n-i}\\ =1-{\displaystyle \underset{0}{\overset{p_1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}\\ =1-P\left(X\le p_1\right)\\ =P\left(X>p_1\right)\\ ={\displaystyle \underset{p_1}{\overset{1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}\\ =R.H.S\end{array}$

Where X is a beta distribution with parameters $k+1$ and $n-k$.

Thus, it is deduced that $P\left(Y_1\le k\right)={\displaystyle \underset{p_1}{\overset{1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}$.

c.

To determine

If k is any integer between 1 and $n-k$, show that $P\left(Y_1\le k\right)>P\left(Y_2\le k\right)$

Interpret the results.

Answer to Problem 135E

It is proved that $P\left(Y_1\le k\right)>P\left(Y_2\le k\right)$.

Explanation of Solution

One has to prove that $P\left(Y_1\le k\right)>P\left(Y_2\le k\right)$.

Consider,

$P\left(Y_1\le k\right)={\displaystyle \underset{p_1}{\overset{1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}$

Again,

$P\left(Y_2\le k\right)={\displaystyle \underset{p_2}{\overset{1}{\int }}\frac{t^k{\left(1-t\right)}^{n-k-1}}{B\left(k+1,n-k\right)}dt}$

It is noted that the integrands for $P\left(Y_1\le k\right)$ and $P\left(Y_2\le k\right)$ are same but the regions are different because the condition is $p_1<p_2$. Thus, $Y_1$ is stochastically greater than $Y_2$.

Hence, it is proved that $P\left(Y_1\le k\right)>P\left(Y_2\le k\right)$