Say $X_n \sim binomial(n,p)$ and $Y_n =\frac{X_n}{n}$. How would I go about showing $Z_n = \frac{\sqrt n (Y_n - p)}{\sqrt {Y_n(1-Y_n)}} \xrightarrow{D} Z$ if $Z \sim N(0,1)$

I was thinking maybe I should try and show convergence in probability first using some combination of the Central Limit Theorem and Slutsky theorem, but I'm mostly confused on how I would implement them.

Thanks in advance!

  • $\begingroup$ How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$? $\endgroup$ – Michael Dec 2 '18 at 4:59
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    $\begingroup$ Assuming $$Z_n = \left\{ \begin{array}{ll} \frac{\sqrt{n}(Y_n-p)}{\sqrt{Y_n(1-Y_n)}} &\mbox{ if $Y_n \notin \{0,1\}$} \\ 0 & \mbox{ otherwise} \end{array} \right.$$ Then you can note $Y_n$ has the same distribution as $\frac{1}{n}\sum_{i=1}^n W_i$ with $\{W_i\}$ iid Bernoulli, and Slutsky directly applies. $\endgroup$ – Michael Dec 2 '18 at 5:07
  • $\begingroup$ @Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem) $\endgroup$ – clovis Dec 2 '18 at 5:31
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    $\begingroup$ $\{W_n\}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem. $\endgroup$ – Michael Dec 2 '18 at 14:41

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