Question #185767

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

(a).h(w)/h(w1)=p(nw+1)/(1p)w,w>0h(w)/h(w-1)=p(n-w+1)/(1-p)w,w>0

(b).E[W(W1)]=n(n1)pE[W(W-1)]=n(n-1)p2

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

1
Expert's answer
2021-05-07T10:35:03-0400

Solution:

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

(a) h(w1)=nCw1pw1(1p)nw+1h(w-1)=^nC_{w-1}p^{w-1}(1-p)^{n-w+1}

Now, h(w)h(w1)=nCwpw(1p)nwnCw1pw1(1p)nw+1\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}}

=n!w!(nw)!pn!(w1)!(nw+1)(1p)=\dfrac{\dfrac{n!}{w!(n-w)!}p}{\dfrac{n!}{(w-1)!(n-w+1)}(1-p)}

=(w1)!(nw+1)!p(w!)(nw)!(1p)=(w1)!(nw+1)(nw)!pw(w1)!(nw)!(1p)=\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)}

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

(b) E[W(W1)]=E[W2W]=E(W2)E(W)E[W(W-1)]=E[W^2-W]=E(W^2)-E(W) ...(i)

Now, E(W)=np,Var(W)=E(W2)[E(W)]2=np(1p)E(W)=np, Var(W)=E(W^2)-[E(W)]^2=np(1-p)

So, E(W2)=[E(W)]2+np(1p)=n2p2+np(1p)E(W^2)=[E(W)]^2+np(1-p)=n^2p^2+np(1-p)

Putting these values in (i)

E[W(W1)]=E(W2)E(W)=n2p2+np(1p)npE[W(W-1)]=E(W^2)-E(W)=n^2p^2+np(1-p)-np

=n2p2+npnp2np=n2p2np2=np2(n1)=n(n1)p2=n^2p^2+np-np^2-np \\=n^2p^2-np^2 \\=np^2(n-1)=n(n-1)p^2

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

=Σw=0n(1w+1)×nCwpw(1p)nw=\Sigma_{w=0}^n(\dfrac1{w+1})\times ^nC_wp^w(1-p)^{n-w}

=1p(n+1)Σw=0n.n+1Cw+1.pw1(1p)nw=\dfrac{1}{p(n+1)}\Sigma_{w=0}^n .^{n+1}C_{w+1}.p^{w-1}(1-p)^{n-w}

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



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS