Question

Consider Example 4f, which is concerned with the multinomial distribution. Use conditional expectation to compute E[NiNj], and then use this to verify the formula for Cov(Ni, Nj) given in Example 4f.

Short answer

Therefore it has been computed that $E\left[N_i{N}_j\right]=m(m-1)p_i{p}_j$by using conditional expectation

Step-by-step solution

Given information

Use conditional expectation to compute E[NiNj],

Solution

Compute the $E\left[N_i{N}_j\right]$using conditional expectation.

So,

$E\left[N_i{N}_j\mid N_i\right]=N_{i}E\left[N_j\mid N_i\right]$

$=N_i\left(m-N_i\right)\frac{p_j}{1-p_i}$

Now, each of the $m-N_i$ trails no resulting in outcome i will independently result inj with probability $p_j\left(1-p_i\right)$.

Solution

Therefore,

$E\left[N_i{N}_j\right]=\frac{p_j}{1-p_i}\left[mE\left[N_i\right]-E\left[N_i^2\right]\right]$

$=\frac{p_j}{1-p_i}\left[m^2{p}_i-m^2{p}_i^2-m{p}_i\left(1-p_i\right)\right]$

$=\frac{p_j}{1-p_i}\left[m(m-1)p_i\left(1-p_i\right)\right]$

$=m(m-1)p_i{p}_j$

And,

$\mathrm{Cov}\left(N_i,N_j\right)=m(m-1)p_i{p}_j-m^2{p}_i{p}_j$

$=m^2{p}_i{p}_j-m{p}_i{p}_j-m^2{p}_i{p}_j$

$=-m{p}_i{p}_j$

Final answer 

Therefore it has been computed that $E\left[N_i{N}_j\right]=m(m-1)p_i{p}_j$by using conditional expectation