# Convergence of $U_n=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right)$ - central limit theorem

Suppose that $$U_n=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right)$$, where $$X_1,X_2,\ldots$$ and $$Y_i,Y_2, \ldots$$ are i.i.d. sequences of random variables with mean $$\mu$$ and variance $$\sigma^2$$ and the sequences $$\{X_i\}$$ and $$\{Y_i\}$$ are independent.

Then, for all real numbers $$x$$, show that, as $$n\to \infty$$,

$$\mathbb{P}(U_n\leq x) \to \int_{-\infty}^{x} \frac{1}{(\sqrt(2\pi))} \:e^{-\frac 12{u^2}}\mathrm du$$ Approach

I tried to carry out the integration on the right, but this does not integrate nicely. It ends up involving an error function, so I figured this was the wrong was to go.

Should I be looking at convergence in probability to answer this question? Is Chebyshev's inequality relevant here? If so, how and how does that relate to the integral?

• Are the sums in denominator? or outside? what do we know about the first moments of X and Y? – t.f Dec 2 '18 at 14:24
• No the sums aren't in denominator. I've clarified above and also added the mean and variance that we know. – StatisticsPersonInTraining Dec 2 '18 at 14:44
• Update: this is actually to do with central limit theorem, yes? – StatisticsPersonInTraining Dec 2 '18 at 14:55
• I have edited your question to clean up the notation a little, You can revert it back to the previous version if you don't like the changes or if I have changed the meaning of the question. But, as stated originally or in the edited question, I don't think that what you are trying to prove is true. $\Sigma X_n - \Sigma Y_n$ is a zero-mean random variable with variance $2n\sigma^2$ and so $U_n$ cannot converge to anything: its variance is increasing. The CLT is not applicable to the result you are trying to prove. You need to fix the definition of $U_n$ to make the CLT work. – Dilip Sarwate Dec 2 '18 at 15:28
• I've corrected the definition of $U_n$. – StatisticsPersonInTraining Dec 2 '18 at 18:22

\begin{align} U_n&=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right) \\ & = {\sqrt n}\left(\sum_{j=1}^n \frac {X_j - Y_j}{n\sqrt {2\sigma^2}}\right) \\ & = {\sqrt n}\left(\sum_{j=1}^n \frac {Z_j}{n}\right) \\ & = {\sqrt n}\bar Z \end{align}
where $$Z_j=\frac {X_j - Y_j}{\sqrt {2\sigma^2}}$$. So $$E(Z_i) = 0$$ and $$Var(Z_i) = 1$$. Following the central limit theorem (CLT), $$U_n$$ converges in distribution to standard normal distribution. Then $$\mathbb{P}(U_n\leq x) \to \int_{-\infty}^{x} \frac{1}{(\sqrt(2\pi))} \:e^{-\frac 12{u^2}}\mathrm du$$ is obvious.
• Thank you. Why does using the substitution $Z_j=\frac {X_j - Y_j}{\sqrt {2\sigma^2}}$ make sense? – StatisticsPersonInTraining Dec 2 '18 at 20:41
• Because $Z_j$ has the properties that are needed to use CTL and convert to standard normal. – user158565 Dec 2 '18 at 20:43