Can anyone show me the correct working out to find the variance for $Y_n$, My variance seems to be $\frac{p(1-p)}{n}$
Asked
Active
Viewed 82 times
2
-
Didn't you forget the scaling factor $\sqrt n$ ? – Jun 07 '16 at 14:37
-
What does the scaling factor do? – HueHue Jun 07 '16 at 14:42
-
Hem, it... scales. – Jun 07 '16 at 14:48
-
@HueHue Does the answer from Gorden help or not ? A reaction would be kindful. – callculus42 Jun 07 '16 at 15:34
1 Answers
3
The variance is $\frac{p(1-p)}{n}$: \begin{align*} Var(Y_n) &= E\left((Y_n-E(Y_n))^2\right)\\ &=\frac{1}{n^2}E\left(\left[\sum_{i=1}^n\big(X_i-E(X_i)\big)\right]^2\right)\\ &= \frac{1}{n^2}\sum_{i=1}^n E\left((X_i-E(X_i))^2\right) \\ &= \frac{1}{n}\left[p(1-p)^2+p^2(1-p) \right]\\ &=\frac{p(1-p)}{n}. \end{align*} Then, \begin{align*} \frac{\sqrt{n}(Y_n-p)}{\sqrt{p(1-p)}}\rightarrow N(0, 1) \end{align*} in distribution, and, consequently, \begin{align*} \sqrt{n}(Y_n-p)\rightarrow N(0, p(1-p)) \end{align*} in distribution.
Gordon
- 4,461