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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?

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1 Answer 1

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Let $y_k=x_1+x_2+\dots +x_k$, $q_k=1-p_1-\dots-p_k$, and $$ \mathcal{A}_k=\left\{(x_{k+1},\dots,x_m) : \substack{x_{k+1},\dots,x_m\ge 0 \\ x_{k+1}+\dots+x_m=n-y_k}\right\}. $$

Then

$$ \mathbb{P}\{X_1 = x_1, ..., X_k = x_k\} = \sum_{\mathcal{A}_k} \frac{n!}{x_1!\dots x_m!}p_{1}^{x_{1}}\cdots p_{m}^{x_{m}} \\ =\frac{n!}{x_1!\dots x_k!(n-y_k)!}p_{1}^{x_{1}}\cdots p_{k}^{x_{k}} \times\sum_{\mathcal{A}_k} \frac{(n-y_k)!}{x_{k+1}!\dots x_m!}p_{k+1}^{x_{k+1}}\cdots p_{m}^{x_{m}} \\ =\frac{n!}{x_1!\dots x_k!(n-y_k)!}p_{1}^{x_{1}}\cdots p_{k}^{x_{k}} \times q_k^{n-y_k}. $$

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  • $\begingroup$ So this marginal is not multinomial? $\endgroup$
    – Zoey A
    Nov 18, 2016 at 5:32
  • $\begingroup$ @ZoeyA How did u conclude that? $\endgroup$
    – d.k.o.
    Nov 18, 2016 at 6:03
  • $\begingroup$ If it were multinomial, the terms $(n-y_k)!$ and $ q_k ^{n-y_k}$ wouldn't be there $\endgroup$
    – Zoey A
    Nov 19, 2016 at 19:07
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    $\begingroup$ Actually it is multinomial? Because $\mathbb{P} \{X_1 = x_1, ..., X_k = x_k\} = \mathbb{P} \{X_1 = x_1, ..., X_k = x_k, Y = n - y_k\}$ $\endgroup$
    – Zoey A
    Nov 19, 2016 at 19:42
  • $\begingroup$ @ZoeyA It is multinomial with $k+1$ options... $\endgroup$
    – d.k.o.
    Nov 20, 2016 at 4:38

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