 Solution

Write $F_X(x)=\mathbb{P}(X\le x)$ and $F_Y(x)=\mathbb{P}(Y\le x)$. For $0\le x\le \mu$, the factor $t/\mu$ is at most one throughout $0\le t\le x$, soFor $x>\mu$, compare the tails instead: $t/\mu \ge 1$ throughout $t\ge x$, givingRearranging provesThis is stochastic domination of $X$ by its size-biased distribution. The proof needs only the finite positive mean, not the finite variance assumed in (ii).