Sum of n standardized iid Geometric(pi) random variables

Question

$X_1,...,X_n$ are iid $Geometric(\pi)$

therefore $E[X_i]=1/\pi$ and $var(X_i)=(1-\pi)/\pi^2$

let $$Z_i = \frac {X_i - 1/\pi}{\sqrt{(1-\pi)/\pi^2}}$$

also, let $Y$ be the sample mean of $X_1,...,X_n$

then $E[Y] = 1/\pi$ and $var(Y) = (1-\pi)/(n\pi^2)$

Please explain why if $$Z = \frac {Y - 1/\pi}{\sqrt{(1-\pi)/(n\pi^2)}}$$

then the following is also true $$Z = \frac{Z_1+...+Z_n}{\sqrt{n}}$$

Answer 1

So first you substityte $Y$ as the sample mean of $X_1,\dots,X_n$: $$Z=\frac{Y-\frac{1}{\pi}}{\sqrt{\frac{1-\pi}{n\pi^2}}}=\frac{\frac{X_1+\dots+X_n}{n}-\frac{1}{\pi}}{\sqrt{\frac{1-\pi}{n\pi^2}}}.$$

Then, rearranging terms, $$Z=\frac{\frac{X_1+\dots+X_n}{n}-\frac{1}{\pi}}{\sqrt{\frac{1-\pi}{n\pi^2}}}=\frac{\frac{X_1+\dots+X_n}{n}-\frac{n}{n\pi}}{\sqrt{\frac{1-\pi}{n\pi^2}}}=\frac{X_1+\dots+X_n-n\frac{1}{\pi}}{n\sqrt{\frac{1-\pi}{n\pi^2}}}=\frac{X_1+\dots+X_n-\frac{1}{\pi}-\dots-\frac{1}{\pi}}{\frac{n}{\sqrt{n}}\sqrt{\frac{1-\pi}{\pi^2}}}=\frac{\left(X_1-\frac{1}{\pi}\right)+\dots+\left(X_n-\frac{1}{\pi}\right)}{\sqrt{n}\sqrt{\frac{1-\pi}{\pi^2}}}=\frac{1}{\sqrt{n}}\left[\frac{\left(X_1-\frac{1}{\pi}\right)}{\sqrt{\frac{1-\pi}{\pi^2}}}+\dots+\frac{\left(X_n-\frac{1}{\pi}\right)}{\sqrt{\frac{1-\pi}{\pi^2}}}\right]=\frac{1}{\sqrt{n}}(Z_1+\dots+Z_n)=\frac{Z_1+\dots+Z_n}{\sqrt{n}}.$$

All the steps are based on rearranging terms, taking common denominator, common factor, etc. Please let me know if something is unclear.