 Solution

Repeated integration by parts gives, for every nonnegative integer $j$,With $A={\lambda }^2$, the first two moments are $\mathbb{E}[X]={\lambda }^2(2!/{\lambda }^3)=2/\lambda$ and $\mathbb{E}[X^2]={\lambda }^2(3!/{\lambda }^4)=6/{\lambda }^2$. Subtracting the square of the mean givesIf $X$ carries units, $\lambda$ has inverse units of $X$, so the mean and standard deviation have the units of $X$ and the variance has their square.