# Thread: Finding a moment-generating function of a continuous r.v. given its pdf

1. ## Finding a moment-generating function of a continuous r.v. given its pdf

Hi,

I'm trying to make sense of a problem we saw in my probability class. mr. fantastic helped me dust off my calculus skills for the integration part but now I'm having trouble understanding the very last part. Here's the problem and its solution:

The probability density of a continuous random variable Y is given by
$f(y) = \begin{cases}\alpha{e^{-\alpha y}}&\text{ for 0} < y < \infty \text{ for some integer k},\\0&\text{otherwise},\end{cases}$
where $\alpha >0$

Prove that the moment generating function, m(t), of Y is given by
$
m(t) = \frac{\alpha}{\alpha - t}$
for $t<\alpha$

Solution:

$m(t) = E[e^{ty}] = \int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$= \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy
$

$
= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)$

$
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}
$

Can someone please explain
$
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}
$
to me? I'm confused by the very end. If k is the one being taken to $\infty$ then why should the outcome depend on t? Please explain how we get $\frac{\alpha}{\alpha -t}$ and $\infty$.

Thanks

2. Originally Posted by Canadian0469
Hi,

I'm trying to make sense of a problem we saw in my probability class. mr. fantastic helped me dust off my calculus skills for the integration part but now I'm having trouble understanding the very last part. Here's the problem and its solution:

The probability density of a continuous random variable Y is given by
$f(y) = \begin{cases}\alpha{e^{-\alpha y}}&\text{ for 0} < y < \infty \text{ for some integer k},\\0&\text{otherwise},\end{cases}$
where $\alpha >0$

Prove that the moment generating function, m(t), of Y is given by
$
m(t) = \frac{\alpha}{\alpha - t}$
for $t<\alpha$

Solution:

$m(t) = E[e^{ty}] = \int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$= \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy
$

$
= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)$

$
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}
$

Can someone please explain
$
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}
$
to me? I'm confused by the very end. If k is the one being taken to $\infty$ then why should the outcome depend on t? Please explain how we get $\frac{\alpha}{\alpha -t}$ and $\infty$.

Thanks
In my opinion, it should be [and it makes more sense this way] $
\frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < {\color{red}\alpha} ,\\\infty &\text{if t} \geq {\color{red}\alpha},\end{cases}
$

If $t\geq\alpha$, then $t-\alpha\geq0$ and the limit would diverge for $t>\alpha$, making $M(t)=\infty$ and $M(t)=0$ if $t=\alpha$. [which really doesn't make sense].

If $t<\alpha$, then $t-\alpha<0$ and the limit would converge to -1, causing $M(t)=-\frac{\alpha}{t-\alpha}=\frac{\alpha}{\alpha-t}$.

Does this help?

--Chris

3. If $t\geq\alpha$, then $t-\alpha\geq0$ and the limit would diverge for $t>\alpha$, making $M(t)=\infty$ and $M(t)=0$ if $t=\alpha$. [which really doesn't make sense].
When you say "which really doesn't make sense", what are you referring to? Are you suggesting that there ought to be a third case for $t=\alpha$?

BTW, I should have noticed the correction you made myself... The question actually states "for t < $\alpha$"

Thanks Chris!

4. Originally Posted by Canadian0469
When you say "which really doesn't make sense", what are you referring to?
I was referring to the moment generating functions...

It doesn't make sense to have $M(t)=\infty$ or $M(t)=0$....

--Chris