Conditional Distribution of a random variable in a Multinomial Distribution Given $X_1,X_2,X_3$ ~ $Multinomial(n, \theta_1, \theta_2, \theta_3)$ what is the conditional distribution of $X_2$ given that $X_1=x_1$?
My thoughts are:
P($X_2=x_2$ | $X_1=x_1$) = $n-x_1 \choose x_2$ $\theta_2^{x_2}$ ($1-\theta_2)^{n-x_1-x_2}$
Can someone tell me if this is correct?
 A: 
Disclaimer: This answer just received a downvote, nearly exactly three years after it was posted. Maybe one should consider this vote as a kind of birthday present? Anyway, the answer remains correct mathematically and it still addresses the question, naturally. Happy reading!

By definition, the parameters $(\theta_1,\theta_2,\theta_3)$ are nonnegative and sum to $1$, the support of the multinomial distribution is the set of the triples $(x_1,x_2,x_3)$ in $\mathbb N_0$ which sum to $n$, and, for every such triple $(x_1,x_2,x_3)$,
$$
P[X_1=x_1,X_2=x_2,X_3=x_3]={n\choose x_1,x_2,x_3}\theta_1^{x_1}\theta_2^{x_2}\theta_3^{x_3}.
$$
Thus, $x_3=n-x_1-x_2$ and, for every nonnegative $x_1$ and $x_2$ such that $x_1+x_2\leqslant n$,
$$
P[X_1=x_1,X_2=x_2]\propto\frac{\theta_2^{x_2}\theta_3^{n-x_2-x_1}}{x_1!(n-x_1-x_2)!},
$$
up to multiplicative factors independent of $x_2$. Considering
$$
k=n-x_1,\qquad p=\frac{\theta_2}{\theta_2+\theta_3},
$$
this yields
$$
P[X_2=x_2\mid X_1=x_1]\propto{k\choose x_1}p^{x_2}(1-p)^{k-x_2}.
$$
The sum of the RHS sum from $x_2=0$ to $x_2=k$ is $1$ hence the last $\propto$ is actually an equal sign.
For every integer $x_1$ between $0$ and $n$, the distribution of $X_2$ conditionally on $X_1=x_1$ is binomial with parameters $\left(n-x_1,\frac{\theta_2}{\theta_2+\theta_3}\right)$.
