I am currently working through the solution for the following problem:
A firm will employ $N$ workers next year according to this year’s profit or loss $X$. What is the probability that exactly two workers will be employed? To be more specific, assume that $Λ := exp(X) ∼Gamma(2, 2)$ and, given $Λ = λ, N ∼ Poisson(λ)$
The solution states the answer is: $2\int_{0}^{\infty}\lambda^3e^{-3\lambda}=\frac{2}{3^4}\Gamma(4)=\frac{4}{27}$
But I believe the answer is $\frac{4}{9}$ since $\frac{2}{3^3}\int_{0}^{\infty}(3\lambda)^3e^{-3\lambda}=\frac{2}{3^3}\Gamma(4)=\frac{4}{9}$ Am I correct or is the solutions manual correct?
You want to compute $$I=\int_{0}^{\infty}\lambda^3e^{-3\lambda}\,\color{red} {d\lambda}$$ and you make $$3\lambda=x\implies \lambda=\frac x3\implies d\lambda=\frac {dx}3$$ which makes $$I=\frac{1}{81} \int_{0}^{\infty}x^3e^{-x} \,dx=\frac {\Gamma(4)}{81}=\frac{2}{27}$$
All your problem is that you did not write $\color{red} {d\lambda}$ and did not take it into account when you made the change of variable.