I have three independent Poisson variables: B, C and D, whose parameters $\lambda_B$, $\lambda_C$ and $\lambda_D$ are unknown.
I sample once the variable: $$ A_1 \equiv 0.9\cdot B + 0.1\cdot C $$ and I get the result: $ A_1 = a_1 $.
$A_1$ is the sum of two Poissonian distributions and therefore it is Poissonian as well. On the basis of this I make an estimate of $ \lambda_{A_1} $: $ \hat\lambda_{A_1} = a_1 $ and the variance of the estimator $ \hat\lambda_{A_1} $ is $ \sigma_{\hat\lambda_{A_1}}^2 = a_1 $.
Thereafter, I sample once the variable: $$ A_2 \equiv 0.9\cdot B + 0.1\cdot D $$ and I get the result: $ A_2 = a_2 $.
Now in principle I should estimate: $ \hat\lambda_{A_2} = a_2 $ and $ \sigma_{\hat\lambda_{A_2}}^2 = a_2 $ as before.
But isn't it possible to get an estimator $ \hat\lambda_{A_2} $ with a smaller variance than $ a_2 $, exploiting the result of the first sampling of $ A_1 $, seen that $ A_1 $ and $ A_2 $ have in common the addend $ 0.9\cdot B $ ? Maybe applying some Bayes's theory ?
My goal is to estimate $A_2$ with a small estimator's variance. I am not interested in B, C and D.