# a problem of trinomial distribution

$X_1$, $X_2$, $X_3$ are distributed according to the trinomial distribution with $n$($=X_1+X_2+X_3$) and $p_1$, $p_2$, $p_3$ ($p_1+p_2+p_3=1$). What is a correlation of $X_i$ and $X_j$? Is the conditional probability function $p(x_2,x_3|x_1)$ of $X_2$ and $X_3$ given that $X_1=x_1$ what special families of distributions? I think it need the probability function and moment generating function...but I am not sure..

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The probability distribution is

$$\mathbb{P}(X_1=k,X_2=l,X_3=m) = {{n}\choose{k,l,m}} p_1^k p_2^l p_3^m$$

with $k+l+m=n$ and $k,l,m \in \{0,\ldots,n\}$.

Multiply this by $x^k y^l z^m$ and sum over all possible values for the triple $(k,l,m)$ and you'll get the following generating function:

$$\Phi(x,y,z)=(p_1 x + p_2 y + p_3 z)^n \; .$$

By taking derivatives w.r.t the variables $x,y$ or $z$, you can easily find moments. For instance

$$\mathbb{E}[X_1 X_2] = \left.\frac{\partial^2\Phi}{\partial x \partial y}\right|_{(1,1,1)}=n(n-1)p_1p_2 \; .$$

Here's the wikipedia page for the multinomial distribution. Just put $k=3$ to find anything you need.

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I'm glad you were able to figure out what OP was talking about. I'm not sure I see how this answers questions about correlations, conditional probabilities, and whatever else it is that OP has in mind. – Gerry Myerson Jul 30 '12 at 5:26
I just gave the probability distribution and the generating function. From that he can find the covariance matrix. And conditional probabilities. I'm not sure I'm understanding his entire question either. I think he just wants to know all the properties of the trinomial distribution. – Raskolnikov Jul 30 '12 at 5:35
sorry, I think the definition of generating function is like φ(t)=E (e^tX), do you use this? – perry zhu Jul 30 '12 at 5:57
That's the moment generating function. I use the probability generating function. However they are identical, just make the substitution $x=e^{t}$ and likewise for the other variables. – Raskolnikov Jul 30 '12 at 6:01
thank you, but how about the conditional probabilities of the trinomial distribution.I did not find clue in the wikipedia page. – perry zhu Jul 30 '12 at 9:28

The conditional distribution of $(X_2,X_3)$ given that $X_1=x_1$:

We have possible outcomes $A_1$, $A_2$, $A_3$, where $\Pr(A_i)=p_i$. The experiment was repeated independently $n$ times. We want the probability that $A_2$ happened $x_2$ times, and $A_3$ happened $x_3$ times, given that $A_1$ happened $x_1$ times.

If we know that $X_1=x_1$, then $X_2+X_3=n-x_1$. The ratio of the conditional probabilities of $A_2$ and $A_3$ is unchanged at $p_2:p_3$. (We need to assume that $p_2+p_3\ne 0$, that is, that $p_1\ne 1$.)

But the conditional probabilities $p_2'$ and $p_3'$ of $A_2$ and $A_3$ must add up to $1$. Thus $p_2'=\frac{p_2}{p_2+p_3}$ and $p_3'=\frac{p_3}{p_2+p_3}$.

The conditional distribution of $(X_2,X_3)$ is therefore binomial (the case $2$ of the multinomial), and $$p(x_2,x_3|x_1)=\frac{(x_2+x_3)!}{x_2!x_3!}(p_2')^{x_2}(p_3')^{x_3},$$ where $x_2+x_3=n-x_1$.

If we know that $X_1=x_1$, then the value of $X_3$ is completely determined once we know $X_2$. So more simply we can say that $X_2$ has the ordinary binomial distribution, with probability of success $p_2'$, and number of trials $n-x_1$.

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ok I get it, thanks everyone. – perry zhu Jul 30 '12 at 15:23
@perryzhu: I think you should unaccept my answer, and return to accepting the one by Raskalnikov, who answered the great bulk of your questions. – André Nicolas Jul 30 '12 at 15:27
ok, thank you, I just realize that I can only accept one answer. – perry zhu Jul 30 '12 at 15:45