# The poisson....

• April 26th 2009, 06:29 AM
roresc
The poisson....
The problem as stated is Let X have a poisson distribution w/ lambda = n.

If Y= (X-n)/Sqrt(n), find the MGF for Y.

I know that M(t) for X is given by e^n((e^t)-1). I cannot figure out how to make the jump to find the MGF for Y. I have tried finding the expected value of e^ty but what I get makes no sense. Help.....
• April 26th 2009, 06:38 AM
mr fantastic
Quote:

Originally Posted by roresc
The problem as stated is Let X have a poisson distribution w/ lambda = n.

If Y= (X-n)/Sqrt(n), find the MGF for Y.

I know that M(t) for X is given by e^n((e^t)-1). I cannot figure out how to make the jump to find the MGF for Y. I have tried finding the expected value of e^ty but what I get makes no sense. Help.....

$Y = \frac{1}{\sqrt{n}} X - \sqrt{n}$.

You're expected to know that if $Y = aX + b$ then $m_Y(t) = e^{bt} M_X(at)$.

This theorem follows easily from the linearity of expectations.
• April 26th 2009, 06:43 AM
Moo
Hello,
Quote:

Originally Posted by roresc
The problem as stated is Let X have a poisson distribution w/ lambda = n.

If Y= (X-n)/Sqrt(n), find the MGF for Y.

I know that M(t) for X is given by e^n((e^t)-1). I cannot figure out how to make the jump to find the MGF for Y. I have tried finding the expected value of e^ty but what I get makes no sense. Help.....

Go back to the definition of the mgf :
$M_Y(t)=\mathbb{E}(e^{tY})=\mathbb{E}(e^{\frac{Xt}{ \sqrt{n}}-\frac{1}{\sqrt{n}}})\mathbb{E}(e^{\frac{Xt}{\sqrt{ n}}} e^{-\frac{1}{\sqrt{n}}})$

But $e^{-\frac{1}{\sqrt{n}}}$ is a constant. And since for any constant a, $\mathbb{E}(aX)=a \mathbb{E}(X)$, we have :

$M_Y(t)=e^{-\frac{1}{\sqrt{n}}} \cdot \mathbb{E}(e^{X \cdot \frac{t}{\sqrt{n}}})=e^{-\frac{1}{\sqrt{n}}} \cdot M\left(\tfrac{t}{\sqrt{n}}\right)$

Edit : woops too late...Again ><
• April 26th 2009, 06:49 AM
roresc
Thanks to both.
Thanks to both for the help. Forgot my rules of exponents here. Things are now clear. Thanks again...