1
$\begingroup$

If $Y_n$ is a Poisson random variable with mean $n^{1/2}$

$S_n = \left(\frac{\sqrt{2} Y_n - \sqrt{2n}}{n^{1/4}}\right)$

If we consider a sequence $S_1, S_2, . . . S_n$, provide the the limiting distribution.

Attempt

Mgf of $Y_n$ is as follows:

$M_{Y_n}(t) = exp\sqrt{n}(e^{t} -1)$

So for all $t \in \mathbb{R}$, we have

$M_{\left(\frac{\sqrt{2}}{n^{1/4}}\right)Y_n - \left(\frac{\sqrt{2}n}{n^{1/4}}\right)}(t) = exp\sqrt{n}(e^{\left(\frac{\sqrt{2}}{n^{1/4}}\right)(t-n )-1} )$

Taking the $\lim_{x\to \infty}$ $M_{T_n}(t) = $ will lead to a lot of mess that, I think, will be a Maclaurin Series expansion. Any nudge in the right direction or an easier approach would be appreciated.

$\endgroup$
  • $\begingroup$ You should check your last expression for the mgf of $S_n$, it doesn't look right. $\endgroup$ – kjetil b halvorsen May 24 '18 at 14:42
1
$\begingroup$

First, I will simplify by removing a factor of $\sqrt{2}$ from your definition of $S_n$, leading to $$ S_n = \frac{Y_n-\sqrt{n}}{n^{1/4}} $$ Then $S_n$ is standardized with expectation 0 and variance $1$, for all $n$ and we should really expect a limiting standard normal distribution. Using that the mgf (moment generating function) of $Y_n$ is $M_{Y_n}(t) = e^{\sqrt{n}(e^t - 1)}$ we find that $$ M_{S_n}(t)=\exp\left( -n^{1/4}t - \sqrt{n} + \sqrt{n}e^{n^{-1/4}t} \right) $$ (note that the mgf calculation in the question seems to be wrong) and the cgf (cumulant generating function) is its logarithm $$ K_{S_n}(t)= -n^{1/4}t - \sqrt{n} + \sqrt{n}e^{n^{-1/4}t} $$ Taking the limit when $n \to \infty$ gives $$ \lim_{n \to \infty} K_{S_n}(t) = t^2/2, $$ which is the cgf of the standard normal distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.