Question

In Example 4f, we showed that the covariance of the multinomial random variables $N_i$and $N_j$is equal to $-m{P}_i{P}_j$by expressing $N_i$and $N_j$as the sum of indicator variables. We could also have obtained that result by using the formula $\mathrm{Var}\left(N_i+N_j\right)=\mathrm{Var}\left(N_i\right)+\mathrm{Var}\left(N_j\right)+2\mathrm{Cov}\left(N_i,N_j\right)$

(a) What is the distribution of $N_i+N_j$?

(b) Use the preceding identity to show that$\mathrm{Cov}\left(N_i,N_j\right)=-m{P}_i,P_j$

Short answer

a) The distribution of $N_i+N_j$is $P_i+P_j$.

b) It has been shown that the covariance of the multinomial random variables $N_i$and $N_j$is equal to $-m{P}_i{P}_j.$

Step-by-step solution

Given Information (Part a)

Using formula: $Var\left(N_i+N_j\right)=Var\left(N_i\right)+Var\left(N_j\right)+2Cov\left(N_i,N_j\right)$

The covariance of the multinomial random variables localid="1647410996201" $N_i+N_j=-m{P}_{i}Pj$

Sum of indicator variables$=N_i+N_j$

The distribution of$N_i+N_j=?$

Explanation (Part a) 

Find the distribution of $N_i+N_j$

Since the sum of the indicator variables $N_i$and $N_j$follows a Binomial with parameters $m$and$P_i+P_j$

Final Answer (Part a) 

Therefore, the distribution of$N_i+N_j$ is$P_i+P_j.$

Given Information (Part b) 

Using formula: $\mathrm{Var}\left(N_j+N_j\right)=\mathrm{Var}\left(N_i\right)+\mathrm{Var}\left(N_j\right)+2\mathrm{Cov}\left(N_i,N_j\right)$

The covariance of the multinomial random variables $N_i+N_j=-m{P}_i{P}_j$

Sum of indicator variables$=N_i+N_j$

Explanation (Part b)  

The total number of independent trails is $m$

The probability of success for the sum of the indicator variables is $P_i+P_j$

The probability of failure for the sum of the indicator variables is $1-(P_i+P_j)$

The formula for the variance of the binomial distribution is,

$V\left(X\right)=np(1-p)$

Explanation 

Find the variance for the sum of the indicator variables,

$\begin{array}{r}\mathrm{Var}\left(N_i+N_j\right)=m\left(P_i+P_j\right)\left(1-\left(P_i+P_j\right)\right)\end{array}$

$\begin{array}{r}=m\left(P_i+P_j\right)\left(1-P_i-P_j\right)\end{array}$

$\begin{array}{r}=\left(m{P}_i+m{P}_j\right)\left(1-P_i-P_j\right)\end{array}$

$\begin{array}{r}=m{P}_i-m{P}_i^2-m{P}_i{P}_j+m{P}_j-m{P}_i{P}_j-m{P}_j^2\end{array}$

role="math" localid="1647526313298" $=m{P}_i\left(1-P_i\right)+m{P}_j\left(1-P_j\right)-2m{P}_i{P}_j$

$=\mathrm{Var}\left(N_i\right)+\mathrm{Var}\left(N_j\right)-2\mathrm{Cov}\left(N_i,N_j\right)$

Therefore, the variance for the sum of the indicator variables is,

$\mathrm{Var}\left(N_i+N_j\right)=\mathrm{Var}\left(N_i\right)+\mathrm{Var}\left(N_j\right)-2\mathrm{Cov}\left(N_i,N_j\right)\dots \left(1\right)$

Final Answer 

From the equation (1), the covariance term is,

$\mathrm{Cov}\left(N_i,N_j\right)=-m{P}_i{P}_j$

So, the covariance of the multinomial random variables$N_i+N_j=P_i+P_j$