Consider a multinomial distribution with $r$ different outcomes, where the $i$th outcome having the probability $p_i$, $i$=1,…,$r$, $\sum_{i=1}^r p_i = 1$. Denote $X_i$ be the number of times the $i$th type outcome occurs, $i$=1,…,r.
I find it difficult to construct the conditional probability $P(X_i=x_i, i=1,…,r-1\mid X_r=j)$, for $0 \le j \le n$.
Here is my work:
I first use the definition of conditional probability.
$$P(X_i=x_i\mid X_r=j)= \frac{P(X_i=x_i \cap X_r=j)}{P(X_r=j)}$$
Now, for the numerator, I use the multinomial distribution, which gives
$$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!} p_i^{x_i} p_r^j$$
For the denominator, I write
$$P(X_r=j) =\frac{n!}{j! (n-j)!} p_r^j (1-p_r)^{n-j}$$
Combining the above, I then conclude
$$P(X_i=x_i\mid X_r=j)= \frac{(n-j)! p_i^{x_i}}{x_i! (1-p_r)^{n-j}}$$
I am completely new in learning probability theory and I welcome any comments and ideas towards my work and the clarity of the question. Thank you.
Best Answer
Actually, using the multinomial distribution, one gets $$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!\color{red}{(n-x_i-j)!}} p_i^{x_i} p_r^j\color{red}{(1-p_i-p_r)^{n-x_i-j}}.$$ Then one should get that the conditional distribution of $X_i$ conditionally on $[X_r=j]$ is binomial $(k,q)$, for some well chosen parameters $k$ and $q$.
Note also that you mention the joint conditional distribution of $(X_i)_{i\ne r}$ conditionally on $[X_r=j]$. True, this can be computed following the same idea, but this question is not addressed simply by computing $P(X_i=x_i \cap X_r=j)$.