Consider $(X_1, X_2, X_3, X_4)$ is a multinomial distributed with parameters $n$ and p $= (p_1, p_2, p_3, p_4)$. What is the distribution of $X_1$ given $X_3+X_4 = m$?
From the conditional probability, we know $$
P(X_1 = x_1 | X_3 + X_4 = m) = \frac{P(X_1 = x_1, X_3 + X_4 = m)}{P(X_3 + X_4 = m)},
$$ then what we need to figure out are the denominator and numerator.
-
For the denominator, we can write it as (not sure)$$
P(X_3 + X_4 = m) = \frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(1-p_3-p_4)^{n-m} = \frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(p_1+p_2)^{n-m},
$$ since $\sum p_i = 1$. -
For the numerator, we can write it as (not sure)$$
P(X_1 = x_1, X_3 + X_4 = m) = P(X_1 = x_1, X_3 = m – x_4, X_4 = x_4, \sum_{k/\{1,3,4\}}X_k = n-m-x_1) = \frac{n!}{x_1!(m-x_4)!x_4!(n-m-x_1)!}p_1^{x_1}p_3^{m-x_4}p_4^{x_4}p_2^{n-m-x_1}.
$$
Plug 1. and 2. into the conditional probability and do some algebric calculation, it eventually yields to $$
P(X_1 = x_1 | X_3 + X_4 = m) = \frac{m!(n-m!)}{x_1!(m-x_4)!x_4!(n-m-x_1)!}(\frac{p_1}{p_2})^{x_1}(\frac{p_3}{p_3+p_4})^m(\frac{p_2}{p_1+p_2})^{n-m}(\frac{p_4}{p_3})^{x_4}.
$$ But this one seems it is not a multinomial distribution, so I am wondering whether the calculation is not correct or the conditional distribution of multinomial is not needed to be multinomial distribution?
Best Answer
For the denominator you have already found $$P(X_3 + X_4 = m) = \frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(p_1+p_2)^{n-m}$$ so the numerator would be $$P(X_1=x, X_3 + X_4 = m) = \frac{n!}{m!x!(n-m-x)!}(p_3+p_4)^{m}p_1^x p_2^{n-m-x}$$ making $$P(X_1=x \mid X_3 + X_4 = m) = \dfrac{\frac{n!}{m!x!(n-m-x)!}(p_3+p_4)^{m}p_1^x p_2^{n-m-x}}{\frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(p_1+p_2)^{n-m}} \\= \dfrac{{(n-m)!}p_1^x p_2^{n-m-x}}{{x!(n-m-x)!}(p_1+p_2)^{n-m}} \\= {n-m \choose x}\left(\frac{p_1}{p_1+p_2}\right)^x\left(1-\frac{p_1}{p_1+p_2}\right)^{n-m-x}$$ which is a binomial distribution with parameters $n-m$ and $\frac{p_1}{p_1+p_2}$.