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Let $X_k = 1$ with probability $0.5$ and $X_k = -1$ with probability 0.5, and let $X_k$ be independent random variables $k = (1,2,...,n)$.

I was able to prove that the characteristic function $\Omega_X (\omega)$ of the random variable $\sum_{i=1}^n \frac{X_i}{\sqrt{n}}$ equals $\left(\cos{\frac{\omega}{\sqrt{n}}}\right)^n$ without too much trouble.

However I'm struggling to prove the next part of the question, which asks me to prove that the limit of that characteristic function approaches $e^{-\frac{\omega^2}{2}}$as n approaches infinity.

There is a hint provided which says that I should take the logarithm of the characteristic function and then apply L'Hopital's rule, but I'm struggling at the LHOP part. I'm sure it's a weakness in my calc skills, but I seem to be stuck unfortunately. Any help would be much appreciated!

Problem is from Schaum's Outline of Probability and Statistics, 4th edition. P 3.74.

Davide Giraudo
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David Loungani
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    I would use a taylor series expansion and show co-efficents match. Go on desmos and see for your self what happens as you add more terms to the sequence. – OskarSzarowicz Sep 05 '21 at 20:53
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    If you put $\tau=\omega/\sqrt{n}$ then $\tau \rightarrow0$ as $n\rightarrow \infty$ and you can say $$\lim_{n\rightarrow \infty}\bigg[\cos\Big(\frac{\omega}{\sqrt{n}}\Big)\bigg]^n=\lim_{\tau \rightarrow 0}\bigg[\big(\cos(\tau)\big)^{1/\tau^2}\bigg]^{\omega^2}$$ Can you take it from here? – Matthew H. Sep 05 '21 at 21:35
  • Yep, got it! Thank you! – David Loungani Sep 06 '21 at 14:32

2 Answers2

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$$\lim_{t \to 0} \frac {\ln (\cos (\omega \sqrt t)} t$$ $$=\lim_{t \to 0} \frac {-\omega \sin (\omega (\sqrt t)\frac 1 2 t^{-1/2}} {{\cos (\omega (\sqrt t)}}$$ $$=-\frac {\omega^{2}} 2$$ using the fact that $\frac {\sin x} x \to 1$ as $ x \to 0$. Put $t=\frac 1n$ to see that the required limit is $e^{-\frac {\omega^{2}} 2}$.

Kavi Rama Murthy
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Recall that $$\cos z=\sum^\infty_{k=0}\frac{(-1)^kz^{2k}}{(2k)!}=1-\frac{z^2}{2}(1+g(z))$$ where $\lim_{z\rightarrow0}g(z)=0$.

Also, recall that for any sequence of complex numbers $c_n\xrightarrow{n\rightarrow\infty}c$, $$\lim_{n\rightarrow\infty}\Big(1+\tfrac{c_n}{n}\Big)^n=e^c$$

Then $$\begin{align} \Big(\cos\frac{\omega}{\sqrt{n}}\Big)^n&=\left(1-\frac{\omega^2}{2n}\Big(1+g\big(\frac{\omega^2}{n}\big)\Big)\right)^n\\ &=\left(1+\frac{-\tfrac{\omega^2}{2}(1+g\big(\tfrac{\omega^2}{n}\big))}{n}\right)^n\xrightarrow{n\rightarrow\infty}e^{-\tfrac{\omega^2}{2}} \end{align}$$

Mittens
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