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I ran into a homework problem to derive the mgf for a standardized sum of random variables. Having already solved for when $\mu$ = 0, I am now asked to generalize for when $\mu$ =/= 0. Furthermore, it is required to use L'Hopital rule to solve the question. My issue is with the highlighted portion, where I do not know how to proceed.

$S_n^* = \frac{S_n - n\mu}{\sqrt{n} \sigma} = \frac{X_1 + ... + X_n - n\mu}{\sqrt{n}\sigma}$

$M_{S_n^*}(t) = E[e^t(\frac{X_1 + ... + X_n - n\mu}{\sqrt{n}\sigma})] = M_x(\frac{t}{\sqrt{n}\sigma})^n.e^{t(\frac{-\sqrt{n}\mu}{\sigma})}$

$log[M_{S_n^*}(t)] = n log[M_x(\frac{t}{\sqrt{n}\sigma})] - \frac{t\mu}{\sigma(1/n)}$

Let $\theta$ = $\frac{1}{\sqrt{n}}$. When n tends to $\infty$, $\theta$ tends to 0.

$\lim_{n \to \infty} log[M_{S_n^*}(t)]$ = $\lim_{\theta \to 0} \frac{log[M_x(\frac{t}{\sigma}\theta)]}{\theta^2}$ - $\lim_{\theta \to 0} \frac{t\mu}{\sigma \theta}$.

I do not know how to manipulate the second term. Ideally, I want to make the term disappear (i.e. limit tends to 0), but taking limits makes the highlighted term $\infty$.

Image of workings thus far

The answer should be the mgf of a standard normal variable i.e. $e^{t^2/2}$.

codinglearner21
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1 Answers1

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You can't separate the limit into a difference if it is of the form $\infty-\infty$. If you want to use L'Hopital's rule, you ought to combine the terms into a single fraction:

$$ \lim_{\theta\to0}\frac{\ln M_X(\frac{t}{\sigma}\theta)-\frac{t\mu}{\sigma}\theta}{\theta^2}. $$

This limit is of the form $0/0$, and you can use the rule. If you differentiate the top and bottom, it will still be of the form $0/0$, but then you can use the rule again and you're good!

Keep in mind $M_X(0)=1$, $M_X'(0)=\mu_X$, $M_X''(0)=\mu_{X^2}$.

anon
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