# Marginal Multinomial PMF of a subset of the random vector $X_1, ..., X_m$

Given that $(X_1, X_2, ..., X_m)$ have multinomial distribution $(n, p_1, ..., p_m)$, I want to find the marginal PMF of $(X_1, X_2, ..., X_k)$, where $k < m$. I have

$$P(X_1 = x_1, ..., X_k = x_k) = \sum_{x_{k+1}}...\sum_{x_{m}} P(X_1 = x_1, ..., X_m = x_m)$$

But I am stuck at this point, the summation seems too complicated. Is there a way to simplify this marginal pmf?

Let $y_k=x_1+x_2+\dots +x_k$, $q_k=1-p_1-\dots-p_k$, and $$\mathcal{A}_k=\left\{(x_{k+1},\dots,x_m) : \substack{x_{k+1},\dots,x_m\ge 0 \\ x_{k+1}+\dots+x_m=n-y_k}\right\}.$$
$$\mathbb{P}\{X_1 = x_1, ..., X_k = x_k\} = \sum_{\mathcal{A}_k} \frac{n!}{x_1!\dots x_m!}p_{1}^{x_{1}}\cdots p_{m}^{x_{m}} \\ =\frac{n!}{x_1!\dots x_k!(n-y_k)!}p_{1}^{x_{1}}\cdots p_{k}^{x_{k}} \times\sum_{\mathcal{A}_k} \frac{(n-y_k)!}{x_{k+1}!\dots x_m!}p_{k+1}^{x_{k+1}}\cdots p_{m}^{x_{m}} \\ =\frac{n!}{x_1!\dots x_k!(n-y_k)!}p_{1}^{x_{1}}\cdots p_{k}^{x_{k}} \times q_k^{n-y_k}.$$
• If it were multinomial, the terms $(n-y_k)!$ and $q_k ^{n-y_k}$ wouldn't be there Nov 19, 2016 at 19:07
• Actually it is multinomial? Because $\mathbb{P} \{X_1 = x_1, ..., X_k = x_k\} = \mathbb{P} \{X_1 = x_1, ..., X_k = x_k, Y = n - y_k\}$ Nov 19, 2016 at 19:42
• @ZoeyA It is multinomial with $k+1$ options... Nov 20, 2016 at 4:38