Chapter 4.6, Problem 100E

a.

To determine

Prove that $P\left(X_1=0\right)>P\left(X_2=0\right)$.

Answer to Problem 100E

It is proved that $P\left(X_1=0\right)>P\left(X_2=0\right)$.

Explanation of Solution

It is given that $X_1\text{ and }{X}_2$ are two Poisson variables with means ${\lambda }_1\text{ and }{\lambda }_2$ respectively. Also it is given that ${\lambda }_2>{\lambda }_1$.

Poisson distribution:

Let X be a discrete random variable and is said to follow Poisson distribution if its probability mass function is given by,

$P\left(X=k\right)=\{\begin{array}{l}\frac{e^{-\lambda }{\lambda }^k}{k!}\text{ ; }k=0,1,2,\cdots \\ 0\text{ ; Otherwise}\text{.}\end{array}$ 

Consider,

$\begin{array}{c}P\left(X_1=0\right)=\frac{e^{-{\lambda }_1}{\lambda }_1^0}{0!}\\ =e^{-{\lambda }_1}\end{array}$

$\begin{array}{c}P\left(X_2=0\right)=\frac{e^{-{\lambda }_2}{\lambda }_2^0}{0!}\\ =e^{-{\lambda }_2}\end{array}$

It is known that,

$\begin{array}{c}{\lambda }_2>{\lambda }_1\\ e^{-{\lambda }_2}<e^{-{\lambda }_1}\\ P\left(X_2=0\right)<P\left(X_1=0\right)\\ P\left(X_1=0\right)>P\left(X_2=0\right)\end{array}$

Thus, it is proved that $P\left(X_1=0\right)>P\left(X_2=0\right)$.

b.

To determine

Prove that $P\left(X_1\le k\right)=P\left(Y>{\lambda }_1\right)$ and $P\left(X_2\le k\right)=P\left(Y>{\lambda }_2\right)$.

Explanation of Solution

Calculation:

It is given that for $\alpha$ is a positive integer, $\beta =1$ and $y>0$.

$P\left(Y>y\right)={\displaystyle \sum _{x=0}^{\alpha -1}\frac{y^x{e}^{-y}}{x!}}$ 

From the PMF given in Part (a),

$\begin{array}{c}P\left(X_1\le k\right)={\displaystyle \sum _{j=1}^k\frac{e^{-{\lambda }_1}{\lambda }_1^j}{j!}}\\ ={\displaystyle \sum _{j-1=0}^{\alpha -1}\frac{{\lambda }_1{}^{j-1}{e}^{-{\lambda }_1}}{\left(j-1\right)!}}\\ ={\displaystyle \sum _{x=0}^{\alpha -1}\frac{{\lambda }_1{}^x{e}^{-{\lambda }_1}}{x!}}\text{, where }\alpha =k+1\\ =P\left(Y>{\lambda }_1\right)\end{array}$

Where Y follows Gamma distribution with $\alpha -1=k$ and $\beta =1$.

Similarly,

$\begin{array}{c}P\left(X_2\le k\right)={\displaystyle \sum _{j=1}^k\frac{e^{-{\lambda }_2}{\lambda }_2^j}{j!}}\\ ={\displaystyle \sum _{j-1=0}^{\alpha -1}\frac{{\lambda }_2{}^{j-1}{e}^{-{\lambda }_2}}{\left(j-1\right)!}}\\ ={\displaystyle \sum _{x=0}^{\alpha -1}\frac{{\lambda }_2{}^x{e}^{-{\lambda }_2}}{x!}}\text{, where }\alpha =k+1\\ =P\left(Y>{\lambda }_2\right)\end{array}$

Where Y follows Gamma distribution with $\alpha -1=k$ and $\beta =1$.

Hence, it is proved that $P\left(X_1\le k\right)=P\left(Y>{\lambda }_1\right)$ and $P\left(X_2\le k\right)=P\left(Y>{\lambda }_2\right)$.

c.

To determine

Prove that $P\left(X_1\le k\right)>P\left(X_2\le k\right)$.

Explain the answer.

Explanation of Solution

From Part (b), it is obtained that that $P\left(X_1\le k\right)=P\left(Y>{\lambda }_1\right)$ and $P\left(X_2\le k\right)=P\left(Y>{\lambda }_2\right)$ and it is also given that ${\lambda }_2>{\lambda }_1.$

Now,

$\begin{array}{c}{\lambda }_2>{\lambda }_1\\ e^{-{\lambda }_2}<e^{-{\lambda }_1}\\ {\displaystyle \sum _{x=0}^{\alpha -1}\frac{{\lambda }_2{}^x{e}^{-{\lambda }_2}}{x!}}<{\displaystyle \sum _{x=0}^{\alpha -1}\frac{{\lambda }_1{}^x{e}^{-{\lambda }_1}}{x!}}\\ P\left(Y>{\lambda }_2\right)<P\left(Y>{\lambda }_1\right)\\ {\displaystyle \sum _{j=1}^k\frac{e^{-{\lambda }_2}{\lambda }_2^j}{j!}}<{\displaystyle \sum _{j=1}^k\frac{e^{-{\lambda }_1}{\lambda }_1^j}{j!}}\text{, where }\alpha =k+1\text{and }k>0\\ P\left(X_2\le k\right)<P\left(X_1\le k\right)\end{array}$

Hence, it is proved that $P\left(X_1\le k\right)>P\left(X_2\le k\right)$.

d.

To determine

Interpret the result of Part (c).

Explanation of Solution

From Part (c), it is proved that $P\left(X_1\le k\right)>P\left(X_2\le k\right)$.

Hence, it can be said that the Poisson distribution which has greater mean has the smaller cumulative probability value for any constant k, such that $k=0,1,2,\mathrm{...}$ .