# (Verification) Conditional pmf of Multinomial Distribution

Let $X=(X_1,X_2,...X_k)^t \thicksim Multi(n,(p_1,p_2,..p_k))^t, (1\le r\lt k)$.

Now I want to derive the conditional probability function of $(X_r+1,...X_k)^t$, $f_{X_{r+1},...X_k\mid X_1,...,X_r}(x_{r+1},...x_k \mid x_1,...x_r)$, where $X_1 = x_1,...X_r=x_r$.

Then $f_{X_{r+1},...X_k\mid X_1,...,X_r}(x_{r+1},...x_k \mid x_1,...x_r)= \dfrac{f_{1,...,k}}{f_{1,...,r}}\\=\dfrac{\begin{pmatrix} n\\x_1,...,x_k\end{pmatrix}p^{x_1}...p^{x_k}}{\begin{pmatrix} x_1+x_2...+x_r\\x_1,...,x_r\end{pmatrix}p^{x_1}...p^{x_r}}\\=\dfrac{n(n-1)...(x_1+x_2+...+x_r+1)}{x_{r+1}!...x_k!}p^{x_r+1}...p^{x_k}$

## 1 Answer

Leaving the algebra aside for a moment, what does the multinomial distribution mean? It is the probability of getting the results $1,\ldots, k$ some given numbers of times, in $n$ independent trials.

Suppose you know that

$$\forall_{1 \leq i \leq r} X_i = x_i.$$

Then you know the results for $\sum_{1 \leq i \leq r}x_i$ of the trials. There still remain $n - \sum_{1 \leq i \leq r}x_i$ for which you don't, and each such trial can only take results in $\{ r + 1, \ldots, k \}$. By independence, these results are independent from what you already know.

It follows that the distribution is

$$Multi \left(n - \sum_{1 \leq i \leq r}x_i, (p_{r + 1}, \ldots, p_k)\right).$$

• @Daschin Looks good. Jul 8, 2017 at 7:22
• @Daschin Meiguanxi. Zaijian. Jul 8, 2017 at 7:26
• I am korean but I could speak in Chinese well.. Bukeqi zaikljian. Jul 8, 2017 at 7:26
• @Daschin My apologies! 죄송합니다 Jul 8, 2017 at 7:31
• Jesus.. Some Computer scientists usually looks so good at natural language it might because of some kind of similarity between natural language and logical language.. without some semantical aspect. Jul 8, 2017 at 7:33