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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 $i\neq j$ 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 $\mathbb{E}[N_i=i|N_j=j]=\mathbb{P}\dfrac{(N_i=i\cap N_j=j)}{\mathbb{P}(N_j=j)}$ but nothing came out of it, how should I proceed?

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

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

The terms in this fraction may be evaluated by noticing that $N_i\sim\mathcal{Binom}(n,p_i)$, $N_j\sim\mathcal{Binom}(n,p_j)$, and $(N_i\mid N_j{=}0)\sim\mathcal {Binom}(n, \tfrac{p_i}{1-p_j})$.

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

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$\newcommand\E{\mathbb{E}}\newcommand\P{\mathbb{P}}$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 $$ \E(N_i | N_j > 0) = \sum_{k=1}^n k \frac{1 - (1-p_j)^n}{\,\,\,\,\,1 - (1-p_j)^{n-k}} p_i^k (1-p_i)^k = \bigl[1 - (1-p_j)^n\bigr]\sum_{k=0}^n \frac{k p_i^k (1-p_i)^{n-k}}{1 - (1-p_j^{n-k})}. $$

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