Answer to Question #185767 in Statistics and Probability for Mohammed Moro

Question

Given that the random $W$ is binomial distribution with $n$ trials and success probability $p$ in each trial and $P(W=w)=h(w)$,show that

(a).$h(w)/h(w-1)=p(n-w+1)/(1-p)w,w>0$

(b).$E[W(W-1)]=n(n-1)p$²

(c).$E(1/W+1)=1-(1-p)$$n+1$ /(n+1)p

Answer 1

Solution:

$P(W=w)=h(w)=^nC_wp^w(1-p)^{n-w}$

(a) $h(w-1)=^nC_{w-1}p^{w-1}(1-p)^{n-w+1}$

Now, $\dfrac{h(w)}{h(w-1)}=\dfrac{^nC_wp^w(1-p)^{n-w}}{^nC_{w-1}p^{w-1}(1-p)^{n-w+1}}$

$=\dfrac{\dfrac{n!}{w!(n-w)!}p}{\dfrac{n!}{(w-1)!(n-w+1)}(1-p)}$

$=\dfrac{(w-1)!(n-w+1)!p}{(w!)(n-w)!(1-p)}=\dfrac{(w-1)!(n-w+1)(n-w)!p}{w(w-1)!(n-w)!(1-p)}$

$=\dfrac{(n-w+1)p}{w(1-p)}$

(b) $E[W(W-1)]=E[W^2-W]=E(W^2)-E(W)$ ...(i)

Now, $E(W)=np, Var(W)=E(W^2)-[E(W)]^2=np(1-p)$

So, $E(W^2)=[E(W)]^2+np(1-p)=n^2p^2+np(1-p)$

Putting these values in (i)

$E[W(W-1)]=E(W^2)-E(W)=n^2p^2+np(1-p)-np$

$=n^2p^2+np-np^2-np \\=n^2p^2-np^2 \\=np^2(n-1)=n(n-1)p^2$

(c) $E[\dfrac1{W+1}]=\Sigma(\dfrac1{w+1})P(W=w)$

$=\Sigma_{w=0}^n(\dfrac1{w+1})\times ^nC_wp^w(1-p)^{n-w}$

$=\dfrac{1}{p(n+1)}\Sigma_{w=0}^n .^{n+1}C_{w+1}.p^{w-1}(1-p)^{n-w}$

$=\dfrac{1}{p(n+1)}.(1-(1-p)^{n+1})$