Conditional Expectation problem

*Consider n multinomial trials, where each trial independently results in outcome i with probability $\displaystyle p_i,\sum_{i=1}^{k}p_{i}=1$. With $\displaystyle X_i$ equal to the number of trials that result in outcome i, find $\displaystyle E[X_{1}|X_2>0].$*

I'm having a hard time understanding what the question is asking. Am I looking for the average number of trials that result in $\displaystyle X_1$ given that there's at least one trial that results in $\displaystyle X_2$?

EDIT:

After thinking about it, isn't the answer just

$\displaystyle 1-n\sum^{k}_{i=2}p_i$

I don't know, I don't think I'm getting the question.

Re: Conditional Expectation problem

Quote:

Originally Posted by

**downthesun01** *Consider n multinomial trials, where each trial independently results in outcome i with probability $\displaystyle p_i,\sum_{i=1}^{k}p_{i}=1$. With $\displaystyle X_i$ equal to the number of trials that result in outcome i, find $\displaystyle E[X_{1}|X_2>0].$*

I'm having a hard time understanding what the question is asking. Am I looking for the average number of trials that result in $\displaystyle X_1$ given that there's at least one trial that results in $\displaystyle X_2$?

To work out the conditional expectation you need the conditional probability:

EDIT:

After thinking about it, isn't the answer just

$\displaystyle 1-n\sum^{k}_{i=2}p_i$

I don't know, I don't think I'm getting the question.

$\displaystyle P(X_1=x_1|X_2>0)+P(X_1=x_1|X_2=0)=P(X_1=x_1$

So:

$\displaystyle P(X_1=x_1|X_2>0)=P(X_1=x_1)-P(X_1=x_1|X_2=0)$

Now the distribution of $\displaystyle X_1$ conditioned on $\displaystyle X_2=0$ is binomial with number of trials equal to $\displaystyle n$ and probability of a success being $\displaystyle p_1^*=p_1/(p_1+p_3+...+p_k)=p_1/(1-p_2)$, and the unconditional probability is binomial with number of trials $\displaystyle n$ and probability of a success $\displaystyle p_1$

Then:

$\displaystyle E(X_1|X_2>0)= \sum_{x_1=0}^n x_1 P(X_1=x_1|X_2>0)$

and we have enough information above to evaluate this.

CB