= 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 $x\geq0$; hence, for $x>y\geq0$,
$$
\boxed{\mathbb P(X>x\mid X>y)=e^{-\lambda(x-y)}.}
$$
Differentiating this conditional tail gives
$$
\boxed{f_{X\mid X>y}(x)=\begin{cases}\lambda e^{-\lambda(x-y)},&x>y,\\0,&x<y.\end{cases}}
$$
The 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:
$$
\boxed{\operatorname{Var}(X\mid X>y)=\frac1{\lambda^2}=\frac1{[\ln(\rho^{-1})]^2}.}
$$
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