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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?

Did
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Faragó Dávid
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    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. – Did Oct 01 '16 at 09:06
  • But how do i show that $f_{X,Z}(x,z) = \lambda^2 e^{-\lambda z} 1_{0<x<z}$? – Faragó Dávid Oct 01 '16 at 09:19
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    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. – Did Oct 01 '16 at 12:19
  • http://math.stackexchange.com/q/473790 – Did Oct 01 '16 at 13:22

1 Answers1

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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}

Did
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Math1000
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  • 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. – Faragó Dávid Oct 01 '16 at 09:41
  • 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 $\frac1\lambda$.) – Math1000 Oct 01 '16 at 09:44
  • Yes, unfortunately that is exactly what I can't do. – Faragó Dávid Oct 01 '16 at 09:48
  • Recall that $$f_{X,Z} = f_{X\mid Z=z}\cdot f_Z $$ – Math1000 Oct 01 '16 at 10:11
  • 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. – Faragó Dávid Oct 01 '16 at 10:40
  • By symmetry - given $Z=z$, it is clear that $0<X<z$ and $0<Y<z$, and also that $X\mid Z=z$ and $Y\mid Z=z$ must have the same distribution. – Math1000 Oct 01 '16 at 11:03
  • Thanks a lot for your help. – Faragó Dávid Oct 01 '16 at 11:11
  • This reasoning does not allow to select the right conditional distribution since there are tons of other options, right? – Did Oct 01 '16 at 12:21
  • Much better now. +1. – Did Oct 01 '16 at 13:48
  • I suppose that fact does depend on $X$ and $Y$ being exponentially distributed...interesting. – Math1000 Oct 01 '16 at 13:53