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?

  • $\begingroup$ Are the sums in denominator? or outside? what do we know about the first moments of X and Y? $\endgroup$ – t.f Dec 2 '18 at 14:24
  • $\begingroup$ No the sums aren't in denominator. I've clarified above and also added the mean and variance that we know. $\endgroup$ – StatisticsPersonInTraining Dec 2 '18 at 14:44
  • $\begingroup$ Update: this is actually to do with central limit theorem, yes? $\endgroup$ – StatisticsPersonInTraining Dec 2 '18 at 14:55
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    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Dec 2 '18 at 15:28
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    $\begingroup$ I've corrected the definition of $U_n$. $\endgroup$ – 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.

  • $\begingroup$ Thank you. Why does using the substitution $Z_j=\frac {X_j - Y_j}{\sqrt {2\sigma^2}}$ make sense? $\endgroup$ – StatisticsPersonInTraining Dec 2 '18 at 20:41
  • $\begingroup$ Because $Z_j$ has the properties that are needed to use CTL and convert to standard normal. $\endgroup$ – user158565 Dec 2 '18 at 20:43

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