# How to solve the below limit by means of L'H$\hat{\text{o}}$pital rule?

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? $$$$\lim\limits_{n\rightarrow\infty}\mathbb{E}\{e^{iu\frac{S_n}{\sqrt{n}}}\}$$$$

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

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 $$\lim_{x\to\infty} \frac{u^2/(2x^2)}{1 - u^2/(2x)}\cdot\frac{1}{-1/x^2} = \lim_{x\to\infty} \frac{-u^2}{2(1 - u^2/(2x))} = -\frac{u^2}{2}.$$

• 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 May 28, 2020 at 10:02
• 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. May 28, 2020 at 11:28
• Thank you a lot for your time and help!!!! Fantastic explanation @Rik93 May 28, 2020 at 11:34