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The binomial distribution is generalized by the multinomial distribution, which follows:

\begin{align} f(x_1,\ldots,x_k;n,p_1,\ldots,p_k) & {} = \Pr(X_1 = x_1\mbox{ and }\dots\mbox{ and }X_k = x_k) \\ \\ & {} = \begin{cases} { \displaystyle {n! \over x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}}, \quad & \mbox{when } \sum_{i=1}^k x_i=n \\ \\ 0 & \mbox{otherwise,} \end{cases} \end{align}

In particular, the "three"nomial distribution follows:

$${n! \over x_1! x_2!(n-x_1-x_2)!}p_1^{x_1}p_2^{x_2}p_3^{n-x_1-x_2}$$

I am not able to show why the marginal probability of this distribution, with respect to either $x_1$ or $x_2$ follows $b(n, p_1)$ or $b(n, p_2)$, respectively.

Please help!

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2 Answers 2

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The simplest way is "combinatorial/probabilistic." Recall that the "three-nomial" distribution measures the probability of $x_1$ Type 1 events, $x_2$ Type 2 events, and $n-x_1-x_2$ Type 3 events, when an experiment is repeated independently $n$ times, with probabilities of "success" respectively equal to $p_1$, $p_2$, and $p_3$, where $p_1+p_2+p_3=1$.

The probability of $x_1$ Type 1 events is therefore $$\binom{n}{x_1}p_1^{x_1}(p_2+p_3)^{n-x_1}.\tag{1}$$ It follows that the marginal distribution of $X_1$ is binomial.

If we really wish to sum, by the Binomial Theorem the probability (1) is equal to $$\binom{n}{x_1}p_1^{x_1}\sum_{x_2=0}^{n-x_1}\binom{n-x_1}{x_2}p_2^{x_2}p_3^{n-x_1-x_2}.$$ This is precisely the same as the result of summing over all (appropriate) $x_2$ the probability that $X_2=x_2$ and $X_3=n-x_1-x_2$. If we want to save writing down (1) until the very end, we can just write the sum argument backwards.

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To see it is relatively easy by the sum of indicators approach.

To verify/proof from the trinomial case:

$$\begin{align} \Pr\{X_1 = x_1\} &= \sum_{x_2=0}^{n-x_1} \Pr\{X_1 = x_1, X_2 = x_2\} \\ &= \sum_{x_2=0}^{n-x_1} \frac {n!} {x_1!x_2!(n-x_1-x_2)!} p_1^{x_1}p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2}\\ &= \frac {n!} {x_1!}p_1^{x_1} \sum_{x_2=0}^{n-x_1} \frac {1} {x_2!(n-x_1-x_2)!} p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2} \\ &= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} \sum_{x_2=0}^{n-x_1} \frac {(n - x_1)!} {x_2!(n-x_1-x_2)!} p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2} \\ &= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} [p_2 + (1 - p_1 - p_2)]^{n-x_1} \\ &= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} (1 - p_1)^{n-x_1}, x_1 = 0, 1, \ldots, n \end{align}$$

So the only key steps are: recognizing the support when $X_1 = x_1$ and recognizing the binomial theorem / expansion.

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