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2nd year stats hw

Q: Suppose you have a sequence $X_1, X_2, ...$ of iid random variables with mean $E(X_1)=\mu_X$ and variance $Var(X_1)=\sigma^2_X$ and another sequence $Y_1, Y_2, ...$ of iid random variables with mean $E(Y_1)=\mu_Y$ and variance $Var(Y_1)=\sigma^2_Y$. For each $n=1,2,...$ let $A_n$ be the random variable $$\frac{\sqrt n}{\sqrt {\sigma^2_X+\sigma^2_Y}}[\bar X_n - \bar Y_n - (\mu_X - \mu_Y)]$$ where $\bar X_n = \sum_{i=1}^n \frac{X_i}{n}$ and $\bar Y_n = \sum_{i=1}^n \frac{Y_i}{n}$.

Show that, in distribution, $A_n$ converges to $N(0,1)$ as $n \to \infty$.


I know that this will require use of the central limit theorem and when I asked my lecturer for help he just reminded me that the $X$ variables are independent to the $Y$ variables, but I don't know how to apply this. Please help - even if its just pointing me in the right direction!

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  • $\begingroup$ Independence of $(X_i)$ and $(Y_i)$ (from each other) has to be in the hypothesis. Othrwise the result is false. $\endgroup$ Commented Aug 30, 2020 at 0:16
  • $\begingroup$ @BrianMoehring Ah yes I meant $(\mu_X - \mu_Y)$. Good pick up! $\endgroup$
    – Viv4660
    Commented Aug 30, 2020 at 0:39
  • $\begingroup$ @KaviRamaMurthy Yeah that's what I thought and so I am a bit confused why he reminded me that - it doesn't seem much help. $\endgroup$
    – Viv4660
    Commented Aug 30, 2020 at 0:41

1 Answer 1

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Defining $Z_n = X_n - Y_n$, by linearity of expectation $E[Z_n] = E[X_n] - E[Y_n] = \mu_x - \mu_y$. Using the properties of variance, we also have $$Var(Z_n) = E[(X_n - Y_n)^2] - E[X_n-Y_n]^2 = E[{X_n}^2] - 2 E[X_n Y_n] + E[{Y_n}^2] - (E[X_n]^2 - 2E[X_n]E[Y_n] + E[Y_n]^2) = (E[{X_n}^2] - E[X_n]^2) + (E[{Y_n}^2] - E[Y_n]^2) + 2(E[X_n]E[Y_n] - E[X_n Y_n]) = \sigma_x^2 + \sigma_y^2 + 2(E[X_n]E[Y_n] - E[X_n Y_n])$$ Because $X_n$ and $Y_n$ are independent, $E[X_n Y_n] = E[X_n]E[Y_n]$ so the variance simplifies to $Var(Z_n) = \sigma_x^2 + \sigma_y^2$. The sample mean of $Z_n$ is $\frac{1}{n}\sum_{i} (X_i - Y_i) = \overline{X_n} - \overline{Y_n}$, so $A_n$ can be rewritten as $$A_n = \frac{\sqrt{n}}{\sqrt{Var(Z_n)}}(\overline{Z_n} - E[Z_n])$$ which by the CLT converges in distribution to $N(0,1)$.

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  • $\begingroup$ Thanks this was really helpful :) $\endgroup$
    – Viv4660
    Commented Aug 30, 2020 at 3:41
  • $\begingroup$ What if $X_n$'s and $Y_n$'s are not independent? $\endgroup$ Commented May 20, 2022 at 11:44

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