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If we know $X\sim \operatorname{Pois}(\lambda)$, $Y\sim\operatorname{Pois}(\lambda_p)$:

When solving $E(X\mid Y)$, based on the law of iterated expectations, $E(X) = E(E(X\mid Y)) =\lambda$. And we know that $E(\lambda) = \lambda$, so can we just say that $E(X\mid Y)=\lambda$ ? But this answer is wrong I suppose.

Bonnaduck
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lemonnn29
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  • $\mathbb E[X\mid Y]$ is a random variable. We know that it has mean $\lambda$ -- this does not imply that it is equal to the constant $\lambda$, e.g. $X$ is a random variable that also has mean $\lambda$, but is not a constant. – jlammy Jan 01 '22 at 15:52
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    In general, there's not a huge amount we can say without knowing how $X$ and $Y$ depend on each other. If they were independent, then $\mathbb E[X\mid Y]=\mathbb E[X]=\lambda$, but this need not be true if $X$ and $Y$ depend on each other. – jlammy Jan 01 '22 at 15:57

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It is not possible in general to compute $E(X\mid Y)$ without any information on the dependence between $X$ and $Y$.

  • If $X$ is independent of $Y$, then indeed, $E(X\mid Y)=\lambda$.
  • If $\lambda>\lambda_p$ and $X=Y+Z$, where $Z$ is independent of $Y$ and has Poisson distribution with parameter $\lambda-\lambda_p$, then $E(X\mid Y)=Y+\lambda-\lambda_p$.
Davide Giraudo
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