0
$\begingroup$

In one experiment we consider three events $A_1$, $A_2$ and $A_3$, with probabilities $p_1$,$p_2$ and $p_3$, such that $p_1+p_2+p_3=1 $

If we consider a sequence of $n$ independent trials of the experiment, the trinomial distribution arises when there are three possible outcomes. We can consider three r.v $X_i$ equal to the number of times that $A_i(i=1,2,3)$ occurs.

$X_1+X_2+X_3=n$.

$n_i$ is the value that the r.v $X_i$ takes.

$P(n_1,n_2,n_3)=\frac{n!}{n_1!n_2!n_3!}p_1^{n_1}p_2^{n_2}p_3^{n_3}$

My question arises when you find the marginal probability of $X_1$

$$P(n_1)=\sum_{n_2}P(n_1,n_2)=$$ $$\binom{n}{n_1}p_1^{n_1}\sum_{n_2=0}^{n-n_1}\binom{n-n_1}{n_2}p_2^{n_2}p_3^{n-n_1-n_2}=$$ $$\binom{n}{n_1}p_1^{n_1}(p_2+p_3)^{n-n_1}=\binom{n}{n_1}p_1^{n_1}(1-p_1)^{n-n_1} $$

In the proof above why $$\sum_{n_2=0}^{n-n_1}\binom{n-n_1}{n_2}p_2^{n_2}p_3^{n-n_1-n_2}$$equals $$(p_2+p_3)^{n-n_1} ? $$

Also in the sum $$\sum_{n_2=0}^{n-n_1}$$ the subindex $n_2$ starts at $0$. Is it right to consider the case that the event $A_2$ doesn't occur?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is an application of the binomial theorem,

$$ (x+y)^m=\sum_{k=0}^m\binom mkx^ky^{m-k}\;, $$

with $x=p_2$, $y=p_3$ and $m=n-n_1$.

I'm afraid I don't understand why you doubt that it's right to consider the case that the event $A_2$ doesn't occur.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .