0
$\begingroup$

I have the following homework assignment:

Let $X,Y$ be two independent $\operatorname{Exp}(\lambda)$ r.v. and $Z := X + Y$. Show that for any non-negative measurable $h$ we have $\mathbb{E}(h(X)|Z) = \frac{1}{Z} \int_0^Z h(t) dt$.

So far I've done this:

We know that $Z \sim \Gamma(2, \lambda)$, so $$ f_Z(z) = \frac{\lambda^2}{\Gamma(2)} z^{2-1} e^{-\lambda z} = \frac{\lambda^2}{1!} z^1 e^{-\lambda z} = z \cdot \lambda^2 \cdot e^{-\lambda z}. $$ Furthermore \begin{align*} F_{X,Z}(x,z) &= \mathbb{P}(X \leq x, Z \leq z) \\ &= \mathbb{P}(X \leq x, X + Y \leq z) \\ &= \mathbb{P}(0 \leq x, Y \leq z) \\ &= 1 \cdot \mathbb{P}(Y \leq z), \quad \text{if } x \geq 0 \\ &= 1 - e^{-\lambda z}, \quad \text{if } x \geq 0, \end{align*} derivating, the p.d.f. is $f_{X,Z}(x,z) = \lambda e^{-\lambda z}$, $x \geq 0$.

Thus the expected value is $$ \mathbb{E}(h(X)|Z) = \int\limits_\Omega h(x) \frac{f(x,z)}{f_Z(z)} dx = \int\limits_0^Z h(x) \frac{\lambda \cdot e^{-\lambda Z}}{Z \cdot \lambda^2 \cdot e^{-\lambda Z}} dx = \frac{1}{Z} \int\limits_0^Z h(x) \frac{1}{\lambda} dx. $$ As you can see, this is almost correct, with the exception of $\frac{1}{\lambda}$ in the integral. My question would be, how do I get rid of that?

$\endgroup$
4
  • 1
    $\begingroup$ Not quite sure how you computed $f_{X,Z}$ (the step $\mathbb{P}(X \leq x, X + Y \leq z) = \mathbb{P}(0 \leq x, Y \leq z)$ being highly suspect) but anyway, $$f_{X,Z}(x,z)=\lambda^2e^{-\lambda z}\mathbf 1_{0<x<z}$$ and the rest follows. $\endgroup$ – Did Oct 1 '16 at 9:06
  • $\begingroup$ But how do i show that $f_{X,Z}(x,z) = \lambda^2 e^{-\lambda z} 1_{0<x<z}$? $\endgroup$ – Faragó Dávid Oct 1 '16 at 9:19
  • 1
    $\begingroup$ Always surprising to see people turn to conditional densities and then recovering the joint density, although the direct route is automatic... Here you know that $(X,Y)$ has PDF $f_{X,Y}$ and you are asking for the PDF $f_{X,Z}$ of $(X,Z)=(X,X+Y)$. Well, there is a theorem for this, based on the general change of variable formula involving the Jacobian of the map $(x,y)\to(x,z)$. Here $(x,z)=(x,x+y)$ hence the Jacobian determinant is $1$ and $$f_{X,Z}(x,z)=f_{X,Y}(x,z-x)$$ end of story. $\endgroup$ – Did Oct 1 '16 at 12:19
  • $\begingroup$ math.stackexchange.com/q/473790 $\endgroup$ – Did Oct 1 '16 at 13:22
3
$\begingroup$

As noted in the comments, the joint density of $(X,Z)$ is \begin{align} f_{X,Z}(x,z) &=f_{X,Y}(x,z-x)\\ &= \left(\lambda e^{-\lambda x}\mathsf 1_{(0,\infty)}(x)\right) \left(\lambda e^{-\lambda (z-x)}\mathsf 1_{(0,\infty)}(z-x)\right)\\ &= \lambda^2 e^{-\lambda z}\mathsf 1_{(0,z)}(x). \end{align} Hence the density of $Z$ is $$f_Z(z)=\int_\mathbb R f_{X,Z}(x,z)\,\mathsf dx=\lambda^2 e^{-\lambda z}\mathsf 1_{(0,\infty)}(z)\int_0^z\,\mathsf dx=\lambda^2 ze^{-\lambda z}\mathsf 1_{(0,\infty)}(z).$$ For $z>0$ we compute \begin{align} f_{X\mid Z=z}(x) &= \frac{f_{X,Z}(x,z)}{f_Z(z)}\\ &= \frac{\lambda^2 e^{-\lambda z}}{\lambda^2 z e^{-\lambda z}}\mathsf 1_{(0,z)}(x)\\ &= \frac1z \mathsf 1_{(0,z)}(x). \end{align}

It follows that for every nonnegative measurable $h$, \begin{align} \mathbb E[h(X)\mid Z] &= \int_0^\infty h(x)f_{X\mid Z}(x\mid Z)\,\mathsf dx\\ &= \int_0^\infty h(x)\frac1Z\mathsf 1_{(0,Z)}(x)\,\mathsf dx\\ &= \frac1Z\int_0^Z h(x)\,\mathsf dx. \end{align}

$\endgroup$
10
  • $\begingroup$ While this is indeed a solution of my homework (and I'll probably turn it in this way, for it is rather simple), my question was how to get rid of the $\frac{1}{\lambda}$ in my work. That's why I won't accept this as an answer. $\endgroup$ – Faragó Dávid Oct 1 '16 at 9:41
  • $\begingroup$ Given the conditional density $f_{X\mid Z=z}$ and the marginal density $f_Z$, can you not compute the joint density $f_{X,Z}$? (That is where you're missing a factor of $\frac1\lambda$.) $\endgroup$ – Math1000 Oct 1 '16 at 9:44
  • $\begingroup$ Yes, unfortunately that is exactly what I can't do. $\endgroup$ – Faragó Dávid Oct 1 '16 at 9:48
  • $\begingroup$ Recall that $$f_{X,Z} = f_{X\mid Z=z}\cdot f_Z $$ $\endgroup$ – Math1000 Oct 1 '16 at 10:11
  • $\begingroup$ One last question: Can you tell me how do I know that $X$ given $Z=z$ is uniform on $(0,z)$? I remember having learnt that a few years ago, but can't recall it. $\endgroup$ – Faragó Dávid Oct 1 '16 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.