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Let $X_i\sim \mathcal{Poisson}(\lambda)$, where $X_i$ come from a random sample of size $n$ (so they're independent and identically distributed). Let $T=I\lbrace X_1=0 \rbrace$ (indicator function); that is, $T\sim \mathcal{Bernoulli}(e^{-\lambda})$.

Now define $B=\sum_{i=1}^nX_i$. This implies $B\sim\mathcal{Poisson}(n\lambda)$. Using this information, what is the distribution of the conditional random variable:

$$T\mid_{B=b}$$

That is, what is the distribution of $T$ given that $B=b$?

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When you have $X_1, ..., X_n \sim \text{IID Pois}(\lambda)$ and you condition on their sum $B = \sum X_i$, the initial values now have a multinomial distribution. Letting $\boldsymbol{X} = (X_1, ..., X_n)$ denote the vector of these original count values, you have:

$$\boldsymbol{X} | B=b \sim \text{Mu} \Bigg( b, \Big( \frac{1}{n}, ..., \frac{1}{n} \Big) \Bigg).$$

From this distribution, the conditional values $X_1, ..., X_n$ have marginal binomial distributions $X_i | B= b \sim \text{Bin} (b, 1 / n)$. Hence, letting $T_i \equiv \mathbb{I}(X_i = 0)$ you therefore have:

$$\mathbb{P}(T_i = 1 | B = b) = \mathbb{P}(X_i = 0 | B = b) = \text{Bin} (0 | b, 1 / n) = \Big( \frac{n-1}{n} \Big)^b .$$

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  • $\begingroup$ So since, even conditionally, $T_i$ can only take values of 1 and 0, that essentially means that $T_i|_{S=s}\sim BERNOULLI\left( \left( \frac{n-1}{n}\right)^b \right)$, correct? $\endgroup$ – ereHsaWyhsipS Mar 3 '18 at 1:54
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One should not speak of a conditional random variable, but rather of the conditional distribution of a random variable. The conditional distribution of $T$ given $B$ is not the conditional distribution of a random variable called $\text{“ }T \text{ given } B\text{ ''}$; rather it is the conditional distribution given $B$, of the random variable $T.$ It is the distribution, not the random variable, that is conditional.

\begin{align} & \Pr(T=1\mid B=b) = \frac{\Pr(T=1\ \&\ B=b)}{\Pr(B=b)} = \frac{\Pr(X_1=0\ \&\ X_2+\cdots+X_n=b)}{\left(\dfrac{(n\lambda)^b e^{-n\lambda}}{b!}\right)} \\[10pt] = {} & \frac{\left( \dfrac{\lambda^0 e^{-\lambda}}{0!} \right)\left( \dfrac{((n-1)\lambda)^b e^{-(n-1)\lambda}}{b!} \right)}{\left(\dfrac{(n\lambda)^b e^{-n\lambda}}{b!}\right)} = \left( \frac{n-1} n \right)^b. \end{align}

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