Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness).

Question

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 $X$.

Answer 1

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.

Answer 2

About the necessary hypotheses (and in relation to a discussion somewhat buried in the comments to @bgins's answer), here is a solution which does not assume that the distribution of $X$ has a density, but only that $X$ is integrable and unbounded (otherwise, the identity in the post makes no sense).

A useful tool here is the complementary PDF $G$ of $X$, defined by $G(a)=\mathrm P(X\gt a)$. For every $a\geqslant0$, let $m=\mathrm E(X)$. The identity in the post is equivalent to $\mathrm E(X-a\mid X\gt a)=m$, which is itself equivalent to $\mathrm E((X-a)^+)=m\mathrm P(X\gt a)=mG(a)$. Note that $m\gt0$ by hypothesis. Now, for every $x$ and $a$, $$ (x-a)^+=\int_a^{+\infty}[x\gt z]\,\mathrm dz. $$ Integrating this with respect to the distribution of $X$ yields $$ \mathrm E((X-a)^+)=\int_a^{+\infty}\mathrm P(X\gt z)\,\mathrm dz, $$ hence, for every $a\gt0$, $$ mG(a)=\int_a^{+\infty}G(z)\,\mathrm dz. $$ This proves ${}^{(\ast)}$ that $G$ is infinitely differentiable on $(0,+\infty)$ and that $mG'(a)=-G(a)$ for every $a\gt0$. Since the derivative of the function $a\mapsto G(a)\mathrm e^{ma}$ is zero on $a\gt0$ and $G$ is continuous from the right on $(0,+\infty)$, one gets $G(a)=G(0)\mathrm e^{-ma}$ for every $a\geqslant0$.

Two cases arise: either $G(0)=1$, then the distribution of $X$ is exponential with parameter $1/m$; or $G(0)\lt1$, then the distribution of $X$ is a barycenter of a Dirac mass at $0$ and an exponential distribution. If the distribution of $X$ is continuous, the former case occurs.

${}^{(\ast)}$ By the usual seesaw technique: the RHS converges hence the RHS is a continuous function of $a$, hence the LHS is also a continuous function of $a$, hence the RHS integrates a continuous function of $a$, hence the RHS is a $C^1$ function of $a$, hence the LHS is also a $C^1$ function of $a$... and so on.

Answer 3

Yes, the exponential distribution is the ONLY continuous distribution satisfying the property, because the above differential equation has a unique solution. For the proof of this fact you can also have a look at the page on the exponential distribution at Statlect (the rate parameter and its interpretation - proof).