# Finding the limit of $\frac{S_n}{\sigma\sqrt{n}}-\frac{S_{2n}}{\sigma\sqrt{2n}}$ using the central limit theorem

Background Knowledge:

Central Limit Theorem: Assume that $$X_1,\dots,X_n$$ are i.i.d random variables with mean $$\mu$$ and variance $$\sigma^2$$. >Then, $$\lim_{n\to\infty} \sqrt{n}(\frac{\bar{X}_n-\mu}{\sigma})$$ is the standard normal distribution (where $$\bar{X}_n$$ is the sample mean). (taken from Wikipedia)

Question:

Assume that $$X_1,X_2,\dots,X_n$$ are i.i.d random variables having mean $$0$$ and variance $$\sigma^2$$. Define:

$$S_n:=X_1+X_2+\dots+X_n$$ $$Y_n:=\frac{S_n}{\sigma\sqrt{n}}-\frac{S_{2n}}{\sigma\sqrt{2n}}$$

Using the central limit theorem, find the value of $$\lim_{n\to\infty}Y_n$$.

My try:

Applying the central limit theorem, we have: $$\lim_{n\to\infty} \sqrt{n}(\frac{\bar{X}_n}{\sigma})$$ is the normal distribution. I do not get it. How should I reach for calculating the actual limit in the question?!

Finding the limit of $$\frac{S_n}{\sigma\sqrt{n}}-\frac{S_{2n}}{\sigma\sqrt{2n}}$$ using the central limit theorem

• Hint: In the limit you're substracting to RVs with normal distribution. Are they independent? What does that imply for your conclusion. Commented Apr 19, 2021 at 17:29
• @PedroIgnacioMartinezBruera Sorry but I'm new to probability theory ... I don't get what you're implying. Would you please elaborate on it? (Maybe as an answer?) Commented Apr 19, 2021 at 17:50
• Almost a duplicate of math.stackexchange.com/questions/1320265/… Commented Apr 19, 2021 at 18:03
• @GabrielRomon Is the answer the same as the question you mentioned? I am a little bit confused ... Commented Apr 19, 2021 at 18:16

We can write $$\frac{S_n}{\sigma\sqrt{n}}- \frac{S_{2n}}{\sigma\sqrt{2n}}=-\frac{1}{\sqrt{2}}\frac{1}{\sigma\sqrt{n}}\sum_{i=n+1}^{2n} X_i + \left(1-\frac{1}{\sqrt{2}}\right)\frac{1}{\sigma\sqrt{n}}\sum_{i=1}^n X_i.$$ Let $$Y_n:=\frac{1}{\sigma\sqrt{n}}\sum_{i=n+1}^{2n}X_i$$ and $$Z_n:=\frac{1}{\sigma\sqrt{n}}\sum_{i=1}^n X_i.$$ These are independent and converge in distribution to standard normal by the CLT. So what does $$-\frac{1}{\sqrt{2}} Y_n + \left(1-\frac{1}{\sqrt 2}\right) Z_n$$ converge in distribution to?
• Is the final answer just $\mathcal{N}(0,\sigma^2)$ ? Commented Apr 19, 2021 at 18:17
• @AshtonMath No. (In fact, it doesn't even depend on $\sigma$.) Commented Apr 19, 2021 at 18:18
• @AshtonMath Whatever the distribution of $(1-1/\sqrt{2})Z -Y/\sqrt{2}$ is for $Y$ and $Z$ independent N(0,1)'s. Normal with mean zero and a variance that I'm not going to calculate for you. Commented Apr 19, 2021 at 18:21