Csdrhrt

Consider $n$ independent trials, each of which results in one of the outcomes ${1, ..., k}$, with respective probabilities $p_1, p_2, ...,p_k$ where those probabilites sum to $1$. Let $N_i$ denote the number of trials that result in outcome $i$ where $i = 1, ..., k$. For $ineq j$ find $mathbb{E}[N_i|N_j>0]$.

I tried to write it as a double sum on $i$ and $j$, and expanding the conditional probability as $mathbb{E}[N_i=i|N_j=j]=mathbb{P}dfrac{(N_i=icap N_j=j)}{mathbb{P}(N_j=j)}$ but nothing came out of it, how should I proceed?

Hint: use the Law of Total Expectation:$$mathsf E(N_i)=mathsf E(N_imid N_j{=}0)~mathsf P(N_j{=}0)+mathsf E(N_imid N_j{>}0)~mathsf P(N_j{>}0)$$

$$therefore mathsf E(N_imid N_j>0)=dfrac{mathsf E(N_i)-mathsf E(N_imid N_j{=}0)~mathsf P(N_j{=}0)}{mathsf P(N_j{>}0)}$$

The terms in this fraction may be evaluated by noticing that $N_isimmathcal{Binom}(n,p_i)$, $N_jsimmathcal{Binom}(n,p_j)$, and $(N_imid N_j{=}0)simmathcal {Binom}(n, tfrac{p_i}{1-p_j})$.

[When given that none of the trials are outcome $j$ the conditional probability that a particular trial is outcome $i$ is $p_i/(1-p_j)$]

$newcommandE{mathbb{E}}newcommandP{mathbb{P}}$Using Bayes' rule we can reduce$$
P(N_i = k | N_j > 0) = frac{P(N_j > 0 | N_i = k)P(N_i = k)}{P(N_j > 0)} = frac{p_i^k(1-p_i)^{n-k}}{1 - (1-p_j)^n}P(N_j > 0 | N_i = k).
$$
But when $N_i = k$, there are $n-k$ more independent trials to possibly affect $N_j$. As a result
$$
P(N_j > 0 | N=k) = 1 - (1-p_j)^{n-k},
$$
giving
$$
P(N_i = k | N_j > 0) = frac{1 - (1-p_j)^n}{,,,,,1 - (1-p_j)^{n-k}} p_i^k (1-p_i)^{n-k}.
$$
Thus
$$
E(N_i | N_j > 0) = sum_{k=1}^n k frac{1 - (1-p_j)^n}{,,,,,1 - (1-p_j)^{n-k}} p_i^k (1-p_i)^k = bigl[1 - (1-p_j)^nbigr]sum_{k=0}^n frac{k p_i^k (1-p_i)^{n-k}}{1 - (1-p_j^{n-k})}.
$$