sampling distribution of mean

Question

Suppose we have a binomial population with parameters $n$ and $p$ so that the mean in $np$. How to find the sampling distribution of mean from the population?

Answer 1

Using characteristic functions (the Fourier transform of PDF):

$X_{1},\ldots,X_{N} \sim B(n,p) \implies \phi_{X_{i}}(t)=(q+pe^{it})^{n}$

$\phi_{X_{1}+\ldots+X_{N}}(t)=\phi_{X_{1}}(t) \times \ldots \times \phi_{X_{N}}(t) =(q+pe^{it})^{nN}$

$\therefore \quad X_{1}+\ldots+X_{N} \sim B(nN,p)$

But for sample mean $\bar{X}$,

$\displaystyle \phi_{\bar{X}}(t)= \phi_{X_{1}} \left(\frac{t}{N} \right) \times \ldots \times \phi_{X_{N}} \left(\frac{t}{N} \right) =(q+pe^{it/N})^{nN}$,

$\displaystyle f_{\bar{X}} \left( \frac{k}{N} \right)= \binom{nN}{k}p^{k}q^{nN-k} \quad$ for $\quad k=0,1, \ldots, nN$