Chapter 3.8, Problem 142E

a

To determine

Show that $\frac{p\left(y\right)}{p\left(y-1\right)}=\frac{\lambda }{y},\text{for}y=1,2,\mathrm{...}$

Explanation of Solution

Calculation:

The Poisson distribution with parameter $\lambda$ for the random variable Y is,

$p\left(y\right)=\frac{{\lambda }^y}{y!}{e}^{-\lambda },y=0,1,2,\mathrm{...},\lambda >0$

The Poisson distribution with parameter $\lambda$ for the random variable $Y-1$ is,

$p\left(y-1\right)=\frac{{\lambda }^{y-1}}{\left(y-1\right)!}{e}^{-\lambda },y=0,1,2,\mathrm{...},\lambda >0$

Taking the ratio of two probability functions,

$\begin{array}{c}\frac{p\left(y\right)}{p\left(y-1\right)}=\frac{\left(\frac{{\lambda }^y}{y!}{e}^{-\lambda }\right)}{\left(\frac{{\lambda }^{y-1}}{\left(y-1\right)!}{e}^{-\lambda }\right)}\\ =\frac{\left(\frac{{\lambda }^y}{y\left(y-1\right)!}\right)}{\left(\frac{{\lambda }^y{\lambda }^{-1}}{\left(y-1\right)!}\right)}\\ =\frac{\left(\frac{1}{y}\right)}{\left(\frac{{\lambda }^{-1}}{1}\right)}\\ =\frac{\lambda }{y}\end{array}$

Hence, it is showed that $\frac{p\left(y\right)}{p\left(y-1\right)}=\frac{\lambda }{y},\text{for}y=1,2,\mathrm{...}$

b

To determine

Find the values of y for which $p\left(y\right)>p\left(y-1\right)$.

Answer to Problem 142E

The values of y for which $p\left(y\right)>p\left(y-1\right)$ is $y<\lambda$.

Explanation of Solution

Calculation:

From part (a), $\frac{p\left(y\right)}{p\left(y-1\right)}=\frac{\lambda }{y},\text{for}y=1,2,\mathrm{...}$.

When $\frac{p\left(y\right)}{p\left(y-1\right)}>0$ then $p\left(y\right)>p\left(y-1\right)$ holds, then

$\begin{array}{c}\frac{\lambda }{y}>0\\ \lambda >y\\ y<\lambda \end{array}$

Hence, the values of y for which $p\left(y\right)>p\left(y-1\right)$ is $y<\lambda$.

c

To determine

Show that $p\left(y\right)$ is maximized when y denotes the greatest integer less than or equal to λ.

Explanation of Solution

Calculation:

Suppose that $\lambda$ is non-integer, then from part (b) $p\left(y\right)$ is maximized when, $p\left(y\right)\ge p\left(y-1\right)$ and $p\left(y+1\right)<p\left(y\right)$.

From part (a),

$\begin{array}{c}p\left(y\right)\ge p\left(y-1\right)\\ \frac{p\left(y\right)}{p\left(y-1\right)}\ge 1\\ \frac{\lambda }{y}\ge 1\\ \lambda \ge y\\ y\le \lambda \end{array}$

Similarly,

$\begin{array}{c}p\left(y+1\right)<p\left(y\right)\\ \frac{p\left(y\right)}{p\left(y+1\right)}<1\\ \frac{\lambda }{y+1}<1\\ \lambda <y+1\\ y>\lambda -1\end{array}$

This shows that, $p\left(y\right)$ is maximized for both the values of $\lambda$ and $\lambda -1$ if $\lambda$ is integer.