Question

An expedition is sent to the Himalayas with the objective of catching a pair of wild yaks for breeding. Assume yaks are loners and roam about Himalayas at random. The probability \( p\in(0,1) \) that a given trapped yak is male is independent of prior outcomes. Let N be the number of yaks that must be caught until a pair is obtained.

(a) Show that the expected value of N is 1 + p/q + q/p, where q = 1-p.

(b) Find the variance of N.

 

Answer 1

Solution

(a) Show that the expected value of N is 1 + p/q + q/p, where q = 1-p.

case1 : \( MM...F\implies P(N=n)=P^{n-1}q, n\geq 2 \)

case2 : \( FFF...M\implies P(N=n)=q^{n-1}P, n\geq 2 \)

\( \implies P(N=n)=P^{n-1}q+q^{n-1}P \)

\( \implies E(N)=q\sum _{n=2}^{\infty }nP^{n-1}+P\sum _{n=2}^{\infty }nq^{n-1} \)

By Geometric series for  \( |P|<1 \)

\( \sum _{n=1}^{\infty }P^n=\frac{1}{1-P}\implies \sum _{n=1}^{\infty }nP^{n-1}=\frac{1}{\left(1-P\right)^2} \)

\( \implies \sum _{n=2}^{\infty }nP^{n-1}=\frac{1-\left(1-P\right)^2}{\left(1-P\right)^2}=\frac{1-q^2}{q^2} \)

Similarly  \( \sum _{n=2}^{\infty }nq^{n-1}=\frac{1-\left(1-q\right)^2}{\left(1-q\right)^2}=\frac{1-P^2}{P^2} \)

\( \implies E(N)=\frac{1-\left(1-P\right)^2}{q}+\frac{1-\left(1-q\right)^2}{P}=\frac{1}{q}+\frac{1}{P}-q-p \)

                     \( =\frac{1}{q}+\frac{1}{P}-1=1+\frac{p}{q}+\frac{q}{p} \)

Therefore. \( E(Y)=1+\frac{p}{q}+\frac{q}{p} \)

(b) Find the variance of N.

\( \implies V(N)=E(N^2)-\Big[E(N)\Big]^2 \)

\( E(N^2)=q\sum _{n=2}^{\infty }n^2P^{n-1}+P\sum _{n=2}^{\infty }n^2q^{n-1} \)

we have \( \sum _{n=2}^{\infty }nP^{n-1}=\frac{1-q^2}{q^2}=\frac{1}{q^2}-1=\frac{1}{(1-P)^2}-1 \)

\( \implies \sum _{n=2}^{\infty }n(n-1)P^{n-2}=\frac{2(1-P)}{(1-P)^4}=\frac{2}{(1-P)^3} \)

\( \implies \sum _{n=2}^{\infty }n^2P^{n-1}=\frac{2P}{(1-P)^3}+\sum_{n=2}^{\infty}nP^{n-1}=\frac{2P+q-q^3}{q^3} \)

Similarly  \( \sum _{n=2}^{\infty }n^2P^{n-1}=\frac{2P+q-q^3}{q^3} \)

\( \implies E(N^2)=\frac{2P+q-q^3}{q^2}+\frac{2q+P-P^3}{P^2} \)

\( \implies \Big[E(N)\Big]^2=\Big(\frac{1}{q}+\frac{q}{P}\Big)^2=\Big(\frac{1}{q^2}+\frac{2}{P}+\frac{q^2}{P^2}\Big) \)

\( \implies V(N)=\frac{2P}{q^2}+\frac{1}{q}-q+\frac{2q}{P^2}+\frac{1}{P}-P-\frac{1}{q^2}-\frac{2}{P}-\frac{q^2}{P^2} \)

                     \( =\frac{2P-1}{q^2}-\frac{1}{P}+\frac{1}{q}-q-P+\frac{2q-q^2}{P^2} \)

                     \( =\frac{2P-1}{q^2}+\frac{1}{Pq}-1+\frac{2q-q^2}{P^2} \)

 

Therefore. 

a). \( E(Y)=1+\frac{p}{q}+\frac{q}{p} \)

b). \( V(N)=\frac{2P-1}{q^2}+\frac{1}{Pq}-1+\frac{2q-q^2}{P^2} \)