# Thread: Find the Limiting Distribution of X_n/n With a given MGF

1. ## Find the Limiting Distribution of X_n/n With a given MGF

I am having extreme difficulty solving the following problem;

Let X_n have mgf given by

M_xn (t) = (1 - 2 at - nat^2 + na^2t^3)^-n/2, t<1/a , a>0

Find the limiting distribution for X_n?n.

Any help on this one would be appreciated. Not even sure where to really begin on this....

2. Hello,
Originally Posted by roresc
I am having extreme difficulty solving the following problem;

Let X_n have mgf given by

M_xn (t) = (1 - 2 at - nat^2 + na^2t^3)^-n/2, t<1/a , a>0

Find the limiting distribution for X_n?n.

Any help on this one would be appreciated. Not even sure where to really begin on this....
So a method would be to find the mgf of $\frac{X_n}{n}$ (let's call it $Y_n$) and then find its limit as n goes to infinity.

\begin{aligned}
M_{Y_n}(t)
&=\mathbb{E}\left(e^{tY_n}\right)=\mathbb{E}\left( e^{t \cdot \frac{X_n}{n}}\right)=\mathbb{E}\left(e^{X_n \cdot \frac tn}\right)=M_{X_n}\left(\tfrac tn\right) \\
&=\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)^{-n/2}
\end{aligned}

That's some start

Hint :
Spoiler:
take the logarithm of $M_{Y_n}(t)$

If you want more (not necessarily with a method you would know) :
Spoiler:

$\ln(M_{Y_n}(t))=-\frac n2 \ln\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)$

Since there's a coefficient n in front of the logarithm, it is sufficient to get a $o\left(\frac 1n\right)$ (asymptotic equivalence)

But we know that $\ln(1+x)=x+o(x)$

So $\ln\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)=-2a \frac tn-a \frac{t^2}{n}+o\left(\frac 1n\right)$ (the $\frac{t^3}{n^2}$ can be omitted, because the power of n is smaller than 1/n)

So $\ln(M_{Y_n}(t))=-\frac n2 \left(-2a \frac tn-a \frac{t^2}{n}+o\left(\frac 1n\right)\right)=at+a\frac{t^2}{2}+o(1)$

Hence, as n goes to infinity, $\ln(M_{Y_n}(t)) \to at+a \frac{t^2}{2}$

This gives you the limit for $M_{Y_n}(t)$