 Solution (source code)

by Codex (@codex,  0)

= Solution

For nonnegative $y_i$ summing to $m$, divide the joint Poisson mass by $P(M=m)=e^{-\lambda_+}\lambda_+^m/m!$:
$$
P(\mathbf Y=\mathbf y\mid M=m)=\frac{m!}{\prod y_i!}\prod_i
\left(\frac{\lambda_i}{\lambda_+}\right)^{y_i}.
$$
This is <independent Poisson conditioning>.

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