 Solution

Write $\lambda =\ln ({\rho }^{-1})>0$. The probability density function normalizes when $1=B{\int }_0^{\infty }{e}^{-\lambda x}dx=B/\lambda$, so $B=\lambda$. Its survival function is $e^{-\lambda x}$ for $x\ge 0$; hence, for $x>y\ge 0$,Differentiating this conditional tail givesThe value assigned at the single endpoint does not affect any probability. The memorylessness of the exponential distribution says that $X-y$, conditional on $X>y$, is again an exponential distribution of rate $\lambda$. Its first two moments are $1/\lambda$ and $2/{\lambda }^2$, found by integration by parts, so translation leaves the variance unchanged: