= Solution
First $\mu>0$. If a nonnegative <random variable> had zero <expected value>, it would equal zero almost surely, which is impossible for a distribution having a <probability density function> on $[0,\infty)$: a singleton has probability zero. Thus the given division by $\mu$ is legitimate.
Because $x\geq0$ and $f(x)\geq0$, $g(x)=xf(x)/\mu\geq0$. Moreover
$$
\int_0^\infty g(x)\,dx=\frac1\mu\int_0^\infty xf(x)\,dx=\frac\mu\mu=1.
$$
Extend $g$ by zero for negative $x$. \b[It is a <probability density function>], defining the <size-biased distribution> of $X$: larger observations receive weight proportional to their value.
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