Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness). Suppose $X$ is a continuous, nonnegative random variable with distribution function $F$ and probability density function $f$. If for $a>0,\ E(X|X>a)=a+E(X)$, find the distribution $F$ of $X$.

Hopefully there is a more elegant solution,
but let us say $\mu=\mathbb{E}[X]$, and start
with the definition of conditional expectation:
$$
\eqalign{
\mathbb{E}[X|X>a] &=& \int_0^{\infty}x\,\mathbb{P}[X|X>a]\,dx \\
\mu + a &=& \int_{a}^{\infty}x\,\frac{f(x)}{1-F(a)}\,dx \\
\left(\mu+a\right)\,\left(1-F(a)\right) &=& \int_{a}^{\infty}x\,f(x)\,dx \,.
}
$$
Differentiating with respect to $a$, we find
$$
\eqalign{
1-F(a)-\left(\mu+a\right)f(a) &=& -a\,f(a) \\\\
1-F(a)-\mu f(a) &=& 0 \\\\
F(a) + \mu F\,'(a) &=& 1
}
$$
which is an ordinary differential equation,
solvable by standard methods, e.g.,
by multiplying by the integrating factor:
$$
\eqalign{
F(x) + \mu F\,'(x) &=& 1
\qquad\text{for}\qquad x\ge0 \\\\
F\,e^{x/\mu} + \mu F\,'\,e^{x/\mu} &=& e^{x/\mu} \\\\
\left( \mu\,F\,e^{x/\mu} \right)' &=& e^{x/\mu} \\\\
\mu\,F(x)\,e^{x/\mu} &=& \int e^{x/\mu} dx = \mu \, e^{x/\mu} + c \\\\
\mu\,F(x) &=& \mu + c \, e^{-x/\mu}
}
$$
At $x=0$, since $X$ is continuous and nonnegative,
it must be the case that $F(0)=0$, from which it
follows that $c=\mu F(0)-\mu=-\mu$, giving us the CDF
$$
F(x) = 1 - e^{-x/\mu} = 1 - e^{-\lambda x}
$$
and the exponential density
$$
f(x) = \frac1\mu\,e^{-\mu x} = \lambda \, e^{-\lambda x}
$$
where the location parameter $\mu$ and (decay) rate parameter $\lambda$
are reciprocally related, i.e., $\lambda\mu=1$.
EDIT: There is indeed now a more elegant solution, thanks to Didier.
