Question about limiting distribution for standard normal distribution

Question

I am studying CLT and meet a question but I have no idea how to solve it, could you please show me how to prove this question (b)?

Thank you so much! Wang

Answer 1

Hint for (b):

• In this case you can separate the sum into $\dfrac{\sum\limits_{i=1}^n Z_i}{\sqrt{n}}+\dfrac{\sum\limits_{i=1}^n \frac1n}{\sqrt{n}}$

• Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $\sum\limits_{i=1}^n Z_i$ and so of $\dfrac{\sum\limits_{i=1}^n Z_i}{\sqrt{n}}$, and thus find the distribution to which this converges in the limit

• $\dfrac{\sum\limits_{i=1}^n \frac1n}{\sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit

• Finally you can combine these to find the limiting distribution of $\dfrac{\sum\limits_{i=1}^n \left(Z_i+\frac1n\right)}{\sqrt{n}}$

Answer 2

(a) $P(|1/n\sum (X_i-p_i)|>\epsilon)\leq 1/(n^2\epsilon^2)Var(\sum X_i)=1/(n^2\epsilon^2)\sum p_i(1-p_i)\leq 1/(n^2\epsilon^2)\sum 1/4=\frac{1}{4n\epsilon^2}\rightarrow 0.$ (b) $\sum (Z_i+1/n)\sim N(1,n)\Rightarrow 1/\sqrt{n}\sum (Z_i+1/n)\sim N(1/\sqrt{n},1)\Rightarrow1/\sqrt{n}\sum (Z_i+1/n)\rightarrow_d N(0,1)$