Krdytkyu

I have a homework problem, which asks:

Let X = $(X_1, X_2, X_3, X_4, X_5) sim text{Mult}_5(n, p)$ with p = $(p_1, p_2, p_3, p_4, p_5)$.

(a) Find $E(X_1 | X_2) text{ and } Var(X_1 | X_2)$.

(b) Find $E(X_1 | X_2 + X_3)$.

So here is my approach to part (a):

To find $E(X_1|X_2 = x_2)$, I think of $X_2$ being set at a fixed value and remove it from the distribution. So first I have to normalize p, so say that:

$$ p_1^{'} = frac{p_1}{1-p_2}, p_3^{'} = frac{p_3}{1-p_2}, p_4^{'} = frac{p_4}{1-p_2}, p_5^{'} = frac{p_5}{1-p_2} $$
then let $p^{'} = (p_1^{'}, p_3^{'}, p_4^{'}, p_5^{'})$

So with $X_2$ fixed, I have a new distribution: $text{Mult}_4(n-X_2, p^{'})$. And to find the probability of $(X_1|X_2)$, I can treat it like any other marginal of a multinomial distribution, and say that: $(X_1|X_2 = x_2) sim text{Bin}(n-X_2, p_1^{'})$

I know that the expected value for a binomial distribution is $n cdot p$ and the variance is $n cdot p(1-p)$, so I can say that:

$$ E(X_1|X_2) = p_1^{'}(n-X_2)$$
$$ Var(X_1|X_2) = p_1^{'}(n-X_2)(1-p_1^{'}) $$

So I have two questions.

1. Is this approach OK?




2. If so, can I simply take the same steps for (b) but adding $X_2$ and $X_3$ together?