$X_1$, $X_2$, $X_3$ 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..
a problem of trinomial distribution
2 Answers
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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0thank you, but how about the conditional probabilities of the trinomial distribution.I did not find clue in the wikipedia page. – 2012-07-30
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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0@perryzhu: I think you should unaccept my answer, and return to accepting the one by Raskalnikov, who answered the great bulk of your questions. – 2012-07-30