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

• The denominator must be a sum over tuples $(y_1,\dots,y_k)$ that satisfy $y_i=x_i$ for $i=1,...,r$ and $y_1+\cdots+y_k=n$. Also the subscripts of the $p_i$ have gone lost. Jul 8 '17 at 7:33

Essential is that the conditional distribution is also multinomial.

To get hold of that try something "smaller". Suppose that there $n$ independent experiments with $3$ possible outcomes. $X_i$ denotes the number of outcomes $i$ and the probability that by an experiment we have outcome $i$ is $p_i$. This of course with $p_i\geq0$ and $p_1+p_2+p_3=1$.

Under the condition that $X_1=x_1$ we can just focus of the $n-x_1$ experiments where the outcomes $2,3$ can occur. Their probabilities to occur at such an experiment must now add up to $1$ but their ratio must stay the same, so and we come to $\frac{p_i}{1-p_1}$ for outcome $i=2,3$.

We end up with:$$P(X_2=x_2,X_3=x_3\mid X_1=x_1)=\binom{n-x_1}{x_2,x_3}\left(\frac{p_2}{1-p_1}\right)^{x_2}\left(\frac{p_3}{1-p_1}\right)^{x_3}$$

More generally we get:$$P(X_{r+1}=x_{r+1},\dots,X_k=x_k\mid X_1=x_1,\dots,X_r=x_r)=$$$$\binom{n-x_1-\cdots-x_r}{x_{r+1},\dots,x_k}\left(\frac{p_{r+1}}{1-p_1-\dots-p_r}\right)^{x_{r+1}}\cdot\cdot\cdot\left(\frac{p_k}{1-p_1-\cdots-p_r}\right)^{x_k}$$

• added some \cdots at the final result. Jul 8 '17 at 7:45
• Your idea looks easy and convenient but is that kind of reasoning like 3/1 case to n/r case directly permitted in mathematics? intuitively it looks perfectly understandable to me. like from n cases, x1+..xr had already happened, then among xr+1 to xk, we need to figure out which would happen with which weight, but the probability also has been modified into the confined way as it denoted.. Jul 8 '17 at 7:46
• Intuition is very important in math, but indeed sometimes the question rises: is this formal enough? In that sense there are often two steps in solving a problem. First let your intuition speak, then find a nice way to formalize it. I think that on e.g. an exam this will be accepted (and even more appreciated than wrestling with sums and fractions). Jul 8 '17 at 8:02
• this question might be off-topic, but last semester I had met two totally different stance of instructor one who argues that Mathematics is all about formal symbolic logic and other one insists that intuition is the very start of mathematics. Of course this two arguments could be well harmonized if one understands in a sense that "mathematics is a formal translation of our logical intuition.. but I am still convinced to the first argument since I had experienced more about mathematics that corrects my wrong and bad intuitional reasoning. Jul 8 '17 at 8:07