Solution
= Solution
The <law of the unconscious statistician> and the finite <variance> give
$$
\mathbb E(Y)=\int_0^\infty xg(x)\,dx
=\frac1\mu\int_0^\infty x^2f(x)\,dx
=\frac{\mathbb E(X^2)}\mu
=\boxed{\mu+\frac{\sigma^2}\mu}.
$$
Thus <size bias> increases the <mean> by $\sigma^2/\mu$, in accord with preferentially selecting larger values.