Chapter 7, Problem 100SE

a.

To determine

Prove that the moment-generating function of $Y=\left(X-\lambda \right)/\sqrt{\lambda }$ is given by $m_Y\left(t\right)=\exp \left(\lambda e^{t/\sqrt{\lambda }}-\sqrt{\lambda }t-\lambda \right)$.

Explanation of Solution

From the given information, X follows Poisson distribution with parameter $\lambda$ and $Y=\left(X-\lambda \right)/\sqrt{\lambda }$.

The moment generating function of X is $M_X\left(t\right)=e^{\lambda \left(e^t-1\right)}$.

Then,

$\begin{array}{c}{m}_Y\left(t\right)=m_{\left(X-\lambda \right)/\sqrt{\lambda }}\left(t\right)\\ =E\left(e^{t\left(X-\lambda \right)/\sqrt{\lambda }}\right)\\ =E\left(e^{\frac{tX}{\sqrt{\lambda }}}\right)E\left(e^{-\frac{t\lambda }{\sqrt{\lambda }}}\right)\\ =e^{-t\sqrt{\lambda }}E\left(e^{\frac{tX}{\sqrt{\lambda }}}\right)\\ =e^{-t\sqrt{\lambda }}{m}_X\left(\frac{t}{\sqrt{\lambda }}\right)\\ =e^{-t\sqrt{\lambda }}{e}^{\lambda \left(e^{\frac{t}{\sqrt{\lambda }}}-1\right)}\\ =e^{-t\sqrt{\lambda }}{e}^{\lambda \left(e^{\frac{t}{\sqrt{\lambda }}}\right)-\lambda }\\ =\exp \left(-t\sqrt{\lambda }+\lambda e^{\frac{t}{\sqrt{\lambda }}}-\lambda \right)\end{array}$

Hence proved

b.

To determine

Prove that $m_Y\left(t\right)=e^{t^2/2}$ as $\lambda \to \infty$.

Explanation of Solution

From the given information, the expansion is $e^{t/\sqrt{\lambda }}={\displaystyle \sum _{i=0}^{\infty }\frac{{\left(t/\sqrt{\lambda }\right)}^i}{i!}}$

From the part a $m_Y\left(t\right)=\exp \left(-t\sqrt{\lambda }+\lambda e^{\frac{t}{\sqrt{\lambda }}}-\lambda \right)$

Then,

$\begin{array}{c}{m}_Y\left(t\right)=\exp \left(-t\sqrt{\lambda }+\lambda {\displaystyle \sum _{i=0}^{\infty }\frac{{\left(t/\sqrt{\lambda }\right)}^i}{i!}}-\lambda \right)\\ =\exp \left(-t\sqrt{\lambda }+\lambda \left({\displaystyle \sum _{i=0}^{\infty }\frac{{\left(t/\sqrt{\lambda }\right)}^i}{i!}}-1\right)\right)\\ =\exp \left(-t\sqrt{\lambda }+\lambda \left(1+\frac{t}{\sqrt{\lambda }}+\frac{t^2}{2\lambda }+\frac{t^3}{6{\lambda }^{3/2}}+\mathrm{..........}-1\right)\right)\\ =\exp \left(-t\sqrt{\lambda }+\lambda \left(\frac{t}{\sqrt{\lambda }}+\frac{t^2}{2\lambda }+\frac{t^3}{6{\lambda }^{3/2}}+\mathrm{........}\right)\right)\\ =\exp \left(-t\sqrt{\lambda }+t\sqrt{\lambda }+\frac{t^2}{2}+\frac{t^3}{6{\lambda }^{1/2}}+\mathrm{.........}\right)\\ =\exp \left(\frac{t^2}{2}+\frac{t^3}{6{\lambda }^{1/2}}+\mathrm{..........}\right)\end{array}$

As $\lambda \to \infty$

$m_Y\left(t\right)=\exp \left(\frac{t^2}{2}\right)$

Hence proved.

c.

To determine

Prove that the distribution function of Y converges to a standard normal distribution function as $\lambda \to \infty$ by using theorem 7.5.

Explanation of Solution

From the theorem 7.5, if Y and $Y_1,Y_2,Y_3,\mathrm{..........}$ are the random variables with the moment generating functions $m\left(t\right)$, $m_1\left(t\right),m_2\left(t\right),m_3\left(t\right),\mathrm{........}$ and as $n\to \infty$ $m_n\left(t\right)=m\left(t\right)$, then the distribution function of $Y_n$ tends to distribution function of Y.

From the part b, as $\lambda \to \infty$

$m_Y\left(t\right)=\exp \left(\frac{t^2}{2}\right)$

This is the moment generating function of the standard normal distribution.

By using theorem 7.5, $Y_n$ follows a standard normal distribution function.

Hence proved.