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.


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.

  • $\begingroup$ 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

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.


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