Variance of $x + \xi(x)$, where $\xi(x) \sim \operatorname{Bin}(x, p)$.

Question

Let $y = x + \xi(x)$m where $\xi \sim \operatorname{Bin}(x,p)$. We want to calculate $\operatorname{Var}(y) = \mathbb{E}(\operatorname{Var}(x + \xi(x)|x)) + \operatorname{Var}(\mathbb{E}(x + \xi(x) | x))$.

The second term is easy-calculated: $\operatorname{Var}(\mathbb{E}(x + \xi(x) | x)) = \operatorname{Var}(x + xp) = (1+p)^2 \mathbb{E}(x)$.

But the first one is more complicated: $\mathbb{E}(\operatorname{Var}(x + \xi(x) | x)) = \mathbb{E}(x^2 + 2\mathbb{E}(x\xi(x)|x) + xpq + x^2 p^2 - (1 +p)^2 x^2))$.

The key problem is $\mathbb{E}(x\xi(x)|x)$. Does there any chance to calculate it fast?

Answer 1

The key observation is that the variance of $x + \xi(x) \mid x$ is the same as the variance of $\xi(x) \mid x$ because the variance is conditioned on $x$, which only affects the location, and not the variance.

So you would have $$\operatorname{Var}[x + \xi(x) \mid x] = xpq,$$ and its expectation is $pq \operatorname{E}[x]$.

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I should also point out that you have an error in your computation: it should be $$\operatorname{Var}[\operatorname{E}[x + \xi(x) \mid x]] = \operatorname{Var}[x + xp] = (1+p)^2 \color{red}{ \operatorname{Var}}[x].$$