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Let $(X_n)_{n\geq1}$ be a sequence of random variables satisfying:

i) $\mathbb{E}\{X_n|\mathcal{F}_{n-1}\}=0$;
ii) $\mathbb{E}\{X_n^2|\mathcal{F}_{n-1}\}=1$;
iii) $\mathbb{E}\{|X_n|^3|\mathcal{F}_{n-1}\}\leq K<\infty$

Let $S_n=\sum\limits_{i=1}^nX_i$, $S_0=0$, $u\in\mathbb{R}$, $i$ denote imaginary unit and $\mathbb{E}\{\cdot\cdot\cdot\}$ denote the expectation of a random variable.

Could you please help me show step-by-step how the below limit can be solved by means of L'H$\hat{\text{o}}$pital rule? \begin{equation} \lim\limits_{n\rightarrow\infty}\mathbb{E}\{e^{iu\frac{S_n}{\sqrt{n}}}\} \end{equation}

Even if I am not sure it is correct, I was thinking of first applying Monotone Convergence Theorem so as to interchange $\lim$ and $\mathbb{E}$, getting $\mathbb{E}\lim\limits_{n\rightarrow\infty}\{e^{iu\frac{S_n}{\sqrt{n}}}\}$. However, I have no clue about how to go on with L'Hopital.


The result to get is: $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{e^{iu\frac{S_n}{\sqrt{n}}}\}=e^{-\frac{u^2}{2}}$

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I don't immediately see how to apply L'hopitals rule directly to the given limit, but I have an approach where it shows up in the end if you wish.

By conditioning on $\mathcal{F}_{n-1}$, we have $$ \mathbb{E}\left(e^{iuS_n/\sqrt n}\right) = \mathbb{E}\left(\mathbb{E}\left(e^{iuS_n/\sqrt n}\middle|\mathcal{F}_{n-1}\right)\right) = \mathbb{E}\left(e^{iuS_{n-1}/\sqrt n}\mathbb{E}\left(e^{iuX_n/\sqrt n}\middle|\mathcal{F}_{n-1}\right)\right). $$

To use the assumptions on $X_n$ which are given, we use a Taylor expansion of the exponential. This gives us that there is some constant $C > 0$ such that $$ \left|\mathbb{E}\left(e^{iuX_n/\sqrt n}\middle|\mathcal{F}_{n-1}\right) - \mathbb{E}\left(1 + \frac{iu}{\sqrt n}X_n - \frac{u^2}{2n}X_n^2\middle|\mathcal{F}_{n-1}\right)\right| \leq \frac{C}{n^{3/2}}\mathbb{E}(|X_n^3||\mathcal{F}_{n-1}). $$

If we now use the assumptions on $X_n$ and put everything together, we find that $$ \left|\mathbb{E}\left(e^{iuS_n/\sqrt n}\right) - \left(1 - \frac{u^2}{2n}\right)\mathbb{E}\left(e^{iuS_{n-1}/\sqrt n}\right)\right| \leq \frac{C}{n^{3/2}}\mathbb{E}(|X_n^3||\mathcal{F}_{n-1}) \leq \frac{CK}{n^{3/2}}. $$

Repeating this and using the triangle inequality, we obtain that $$ \left|\mathbb{E}\left(e^{iuS_n/\sqrt n}\right) - \left(1 - \frac{u^2}{2n}\right)^n\right| \leq \frac{CK}{n^{1/2}}. $$

Since the upper bound goes to 0, we find that $$ \lim_{n\to\infty} \mathbb{E}\left(e^{iuS_n/\sqrt n}\right) = \lim_{n\to\infty} \left(1 - \frac{u^2}{2n}\right)^n = e^{-u^2/2}. $$

Here, the latter is a standard limit, but you can compute it using L'hopitals rule by first taking the logarithm.

EDIT: To see how we can use L'hopital's rule to compute the last limit, consider the function $f(x) = x\log\left(1 - \frac{u^2}{2x}\right)$. The limit we need to compute is $$ \lim_{n\to\infty} e^{f(n)} = e^{\lim_{n\to\infty} f(n)}, $$ where we used the continuity of the exponential. Now $$ \lim_{n\to\infty} f(n) = \lim_{x\to\infty} x\log\left(1 - \frac{u^2}{2x}\right) = \lim_{x\to\infty} \frac{\log\left(1 - \frac{u^2}{2x}\right)}{1/x} $$

By L'hopital's rule, this is equal to $$ \lim_{x\to\infty} \frac{u^2/(2x^2)}{1 - u^2/(2x)}\cdot\frac{1}{-1/x^2} = \lim_{x\to\infty} \frac{-u^2}{2(1 - u^2/(2x))} = -\frac{u^2}{2}. $$

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  • $\begingroup$ First of all, thank you a lot! The point is that I already know the approach you have shown since I am studying this part from Jacod-Protter. However, my doubt was properly related to the explicit computation of that limit by means of L'hopital rule, I am interested in that @Rik93 $\endgroup$ May 28, 2020 at 10:02
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    $\begingroup$ I found the specific part, and see that they exactly use the above argument. They suggest to use L'hopital's rule exactly for this last step. I have edited how to do this more explicitly. $\endgroup$
    – Rik93
    May 28, 2020 at 11:28
  • $\begingroup$ Thank you a lot for your time and help!!!! Fantastic explanation @Rik93 $\endgroup$ May 28, 2020 at 11:34

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