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during exam preparation I found the following question which I could not solve: $$\lim_{n \rightarrow \infty}\left(m_X\left(\frac{t}{n}\right)\right)^n = \exp\left(t\cdot\mathbb{E}(X)\right)$$ where $m_X(t)$ is the moment generating function.
As hint was given to use L'Hopital and to interchange limit and expected value without argumentation. However, I have no idea how to apply L'Hospital since there is no fraction.

I tried to verify the formula for an exponentially distributed random variable but it turned out that either I have a calculation error or the formula is wrong.

Would really appreciate if anyone can help me through the exercise or give me a hint how to start!

notimportant
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1 Answers1

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First notice that, as $n\to \infty$, you have $m_X(t/n) \to m_X(0) = 1$. Hence the limit you want to evaluate is of the form $\lim_{n\to\infty} f(n)^{g(n)}$ where $$\begin{cases}\lim_{n\to\infty} f(n) = 1\\ \lim_{n\to\infty} g(n) = +\infty\end{cases} $$

By a known identity, we have in this setting that $$\lim_{n\to\infty} f(n)^{g(n)} = \exp\left(\lim_{n\to\infty}g(n)\cdot(f(n) - 1)\right) $$

Plugging this back in the limit we're interested in yields : $$\begin{align*}\lim_{n \rightarrow \infty}\left(m_X\left(\frac{t}{n}\right)\right)^n &= \exp\left(\lim_{n\to\infty}n\cdot\left(m_X\left(\frac{t}{n}\right)-1\right)\right) \\ &=\exp\left(\lim_{n\to\infty}\frac{m_X\left(\frac{t}{n}\right)-1}{1/n}\right)\\ &=\exp\left(\lim_{h\to0}\frac{m_X\left(t\cdot h\right)-1}{h}\right)\end{align*} $$

Can you finish ?

Stratos supports the strike
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    In concept, this is exactly what I thought. However, $m_X(t/n)\to 1$ and $m'_X(t/n)\to m'_X(0)$ do not seem to me exactly immediate since we need conditions on $X$ to have continuity, differentiability etc of the MGF and OP probably is not aware. Would you agree? – Snoop Apr 09 '23 at 18:09
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    Yeah @Snoop, we definitely need to make sure that $m_X$ is sufficiently well behaved for this proof to work ! In practice, I believe it is sufficient for $m_X$ to be finite on some interval $(-\epsilon,\epsilon)$ to guarantee that it is $C^\infty$ on some neighbourhood of $0$. OP didn't specify what assumption they're working with so I swept the details under the rug :p – Stratos supports the strike Apr 09 '23 at 18:24
  • Thank you so much! Writing $1=m_X(0)$, multiplying by $\frac{t}{t}$, seeing a difference quotient and finally using $\mathbb{E}(X)=m'_X(0)$ does it. And yes, you are absolutely right, I did not think about the conditions under which this equality holds and thus it should not be surprising that it does not hold for the exponential distribution which is not differentiable in 0. – notimportant Apr 09 '23 at 20:25
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    @notimportant the mgf of the exponential distribution is differentiable at $0$ so you may be confused – Snoop Apr 09 '23 at 21:50
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    @notimportant Also $(\frac{\lambda}{\lambda-t/n})^n=(\frac{1}{1-t/(\lambda n)})^n\to e^{t/\lambda}$ so the statement holds for $t<\lambda$. – Snoop Apr 09 '23 at 21:53