What's the distribution of $X$?

Question

Let X be a continuous, positive random variable. We also have that $E(X−a|X≥a)$ does not depend on $a$. What's the distribution of $X$?

I think the expected value is $E(X−a|X≥a)=\int (x-a) f(x-a) dx$, and it's derivate w.r.t a is $\int x f(x-a) dx + \int(x-a)f(x-a) dx$, and it must be 0, so we have that $-\int x f(x-a) dx = \int(x-a)f(x-a) dx$, but I can do nothing with it. How should I do it?

Answer 1

BY definition of expected value in terms of the distribution function, $$ \frac{\int_a^\infty(x-a)f(x)\, dx}{\int_a^\infty f(x)\, dx} $$ is independent of $a$; say its value is the constant $k$. Then $$ \int_a^\infty(x-a)f(x)\, dx = k\int_a^\infty f(x)\, dx $$ $$ \int_a^\infty xf(x)\, dx = (k+a)\int_a^\infty f(x)\, dx $$ $$ \frac{d}{da}\int_a^\infty xf(x)\, dx = \frac{d}{da}\left[(k+a)\int_a^\infty f(x)\, dx\right] $$ $$ -af(a) = \frac{d}{da}\left[(k+a)\right]\int_a^\infty f(x)\, dx+(k+a) \frac{d}{da}\left[\int_a^\infty f(x)\, dx\right] $$ $$ -af(a) = \int_a^\infty f(x)\, dx - (k+a) f(a) $$ $$ \int_a^\infty f(x)\, dx = kf(a) $$ $$ \frac{d}{da}\int_a^\infty f(x)\, dx = k \frac{df(a)}{da} $$ $$ \frac{df(a)}{da} = -\frac1k f(a) $$ $$f(a) = \alpha e^{-a/k}$$ which is an exponential distribution.

Answer 2

Hint: can you show that $X$ is memorylessness, i.e. satisfies $\mathbb{P}(X>t+s \ | \ X>t)=\mathbb{P}(X>s)$?

Since you didn't manage to show this yourself, a proof goes as follows: we know that we have a continuous, positive, random variable, and we are given that $\mathbb{E}(X-a\ | \ X\geq a)$ does not depend on $a\geq0$. I assume $a\geq0$ because otherwise the question does not make sense. Then we have $\mathbb{E}(X-a\ | \ X\geq a)=\mathbb{E}(X-0\ | \ X\geq 0)=\mathbb{E}(X)$ since $X$ is positive. Thus $$\int_{a}^{\infty}(x-a)\frac{f(x)}{\mathbb{P}(X\geq a)}\mathrm{d}x=\int_{0}^{\infty}xf(x)\mathrm{d}x,$$ and then by the change of variables $x\mapsto x+a$, $$\int_{0}^{\infty}xf(x)\mathrm{d}x=\int_{0}^{\infty}x\frac{f(x+a)}{\mathbb{P}(X\geq a)}\mathrm{d}x.$$ Then, because this equality holds for any $a\geq0$, we conclude $f(x)=\frac{f(x+a)}{\mathbb{P}(X\geq a)}$. Now we take the integral over the region $[b,\infty]$ for a $b\geq0$, and we see that $$\mathbb{P}(X> b)=\mathbb{P}(X\geq b)=\frac{\mathbb{P}(X\geq a+b)}{\mathbb{P}(X\geq a)}=\frac{\mathbb{P}(X> a+b)}{\mathbb{P}(X> a)}=\mathbb{P}(X>a+b\ | \ X>a),$$ as desired.