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