Question

) Prove that for any discrete bivariate random variable (X, N) for which the first

moments of X and N exists,

E(X) = E [E (X|N)]

(b) The number N of customers entering the University of Ghana book-shop each day

is a random variable. Suppose that each customer has, independently of other

customers, a probability θ of buying at least one book. Let X denote the number

of customers that buy at least one book each day.

Describe without proof the distribution of X conditional on N = n. Hence use the

results in (a) to evaluate the expectation of X if N has the distribution.

i. P(N = k) = M

k
θk

(1 − θ)M−k

, k = 0, 1, · · · , M

ii. P(N = k) = θ(1 − θ)k

, k = 0, 1, 2, · · · ,

iii. P(N = k) =

e−θθk

k!

, k = 0, 1, 2, · · ·

iv. P(N = k) = θ(1 − θ)k−1

, k = 1, 2, · · ·

Find the probability distribution of X if N has the distribution in (b) i-iv.

Answer 1

$a:\\E\left( X|N=n \right) =\frac{EXI\left\{ N=n \right\}}{P\left\{ N=n \right\}}=\frac{\sum{mP\left\{ X=m,N=n \right\}}}{P\left\{ N=n \right\}}\\E\left[ E\left( X|N \right) \right] =\sum{E\left( X|N=n \right) P\left\{ N=n \right\}}=\\=\sum{\frac{\sum{mP\left\{ X=m,N=n \right\}}}{P\left\{ N=n \right\}}P\left\{ N=n \right\}}=\\=\sum{\sum{mP\left\{ X=m,N=n \right\}}}=EX\\b:\\X|N=n\sim Bin\left( n,\theta \right) \\E\left( X|N \right) =N\theta \\i:\\P\left( N=k \right) =C_{M}^{k}\theta ^k\left( 1-\theta \right) ^{M-k}\sim Bin\left( M,\theta \right) \\EX=EE\left( X|N \right) =E\left( N\theta \right) =\theta EN=\theta \cdot M\theta =M\theta ^2\\P\left( X=x \right) =\sum{P\left( X=x|N=n \right) P\left( N=n \right)}=\sum_{n\geqslant x}{C_{n}^{x}\theta ^x\left( 1-\theta \right) ^{n-x}C_{M}^{n}\theta ^n\left( 1-\theta \right) ^{M-n}}\\ii:\\P\left( N=k \right) =\theta \left( 1-\theta \right) ^k\sim Geom\left( \theta \right) \\EX=EE\left( X|N \right) =E\left( N\theta \right) =\theta EN=\theta \cdot \frac{1-\theta}{\theta}=1-\theta \\P\left( X=x \right) =\sum{P\left( X=x|N=n \right) P\left( N=n \right)}=\sum{C_{n}^{x}\theta ^x\left( 1-\theta \right) ^{n-x}\theta \left( 1-\theta \right) ^n}=\\=\frac{\theta ^x}{\left( 1-\theta \right) ^x}\sum_{n\geqslant x}{C_{n}^{x}}\left( 1-2\theta +\theta ^2 \right) ^n\\iii:\\P\left( N=k \right) =\frac{\theta e^{-\theta k}}{k!}\sim Poiss\left( \theta \right) \\EX=EE\left( X|N \right) =E\left( N\theta \right) =\theta EN=\theta \cdot \theta =\theta ^2\\P\left( X=x \right) =\sum{P\left( X=x|N=n \right) P\left( N=n \right)}=\sum_{n\geqslant x}{C_{n}^{x}\theta ^x\left( 1-\theta \right) ^{n-x}\frac{\theta e^{-\theta n}}{n!}}\\iv:\\P\left( N=k \right) =\theta \left( 1-\theta \right) ^{k-1}\sim Geom_1\left( \theta \right) \\EX=EE\left( X|N \right) =E\left( N\theta \right) =\theta EN=\theta \cdot \frac{1}{\theta}=1\\P\left( X=x \right) =\sum{P\left( X=x|N=n \right) P\left( N=n \right)}=\sum{C_{n}^{x}\theta ^x\left( 1-\theta \right) ^{n-x}\theta \left( 1-\theta \right) ^{n-1}}=\\=\frac{\theta ^x}{\left( 1-\theta \right) ^{x+1}}\sum_{n\geqslant x}{C_{n}^{x}}\left( 1-2\theta +\theta ^2 \right) ^n$