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

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. – Ami Tavory Jul 8 '17 at 7:22
• @Daschin Meiguanxi. Zaijian. – Ami Tavory Jul 8 '17 at 7:26
• I am korean but I could speak in Chinese well.. Bukeqi zaikljian. – Daschin Jul 8 '17 at 7:26
• @Daschin My apologies! 죄송합니다 – Ami Tavory Jul 8 '17 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. – Daschin Jul 8 '17 at 7:33