# Proving Convergence in Distribution

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.

• How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$? – Michael Dec 2 '18 at 4:59
• 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. – Michael Dec 2 '18 at 5:07
• @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) – clovis Dec 2 '18 at 5:31
• $\{W_n\}$ 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. – Michael Dec 2 '18 at 14:41