2

I need to find

  • $E[X^3\mid X+Y] $
  • $E[\max(X,Y)\mid \min(X,Y)]$

Knowing that $X$ and Y$$ are independent random variables with uniform distribution on the interval $[0,1]$.

My intuition tells me that the $E[\max(X,Y)\mid \min(X,Y)]= \frac{1+ \min(X,Y)}{2}$. That's because if we write that $\min(X,Y)=a$ then $\max(X,Y)$ has uniform distribution on the interval [a,1], so it's mean will be equal to $\frac{a+1}{2}$. Is there more formal way I can explain it?

Now I have no idea how to proceed with the $E[X^3\mid X+Y]$. I've tried to write down $X+Y=Z$ then I would have

$$\mathbb{E}(X^3\mid Z) = \int_0^\infty x^3 \frac{g_{(X^3, Z)}(x,z)}{g_{Z}(z)}\,dx$$

But I don't know how to find ${g_{(X^3, Z)}(x,z)}$ and I'm not sure that's the right way to approach it. Could you help me with it?

StubbornAtom
  • 17,052
ראָזעווע וואַלפיש
  • 399

1 Answers1

3

The way you have found $E\left[\max(X,Y)\mid \min(X,Y)\right]$ is quite formal. You can also do this by deriving the density of $\max(X,Y)$ conditioned on $\min(X,Y)$ and then find its mean.

For $E\left[X^3\mid X+Y\right]$, it suffices to find the density of $X$ conditional on $X+Y$.

Density of $(X,Y)$ is $$f_{X,Y}(x,y)=\mathbf1_{0<x,y<1}$$

Change variables $(x,y) \mapsto (x,z)$ with $z=x+y$, so that density of $(X,Z)$ where $Z=X+Y$ is

$$f_{X,Z}(x,z)=\mathbf1_{0<x<1,x<z<1+x}$$

From here, you can see that density of $Z$ is $$f_Z(z)=z\mathbf1_{0<z<1}+(2-z)\mathbf1_{1<z<2}$$

Conditional density of $X$ given $Z$ is therefore

\begin{align} f_{X\mid Z}(x\mid z)&=\frac{f_{X,Z}(x,z)}{f_Z(z)} \\&=\frac1z\mathbf1_{0<x<z<1}+\frac1{2-z}\mathbf1_{0<z-1<x<1} \end{align}

That is, $X$ given $Z=z$ has a uniform distribution on $(0,z)$ if $0<z<1$ and another uniform distribution on $(z-1,1)$ if $1<z<2$. In other words, the conditional distribution is uniform on $(\max(0,z-1),\min(1,z))$ for $0<z<2$.

The $n$th moment of a uniform distribution on $(a,b)$ is given by $\frac{b^{n+1}-a^{n+1}}{(n+1)(b-a)}$.

This means

\begin{align} E\left[X^3 \mid Z=z\right]&=\frac{(\min(1,z))^4-(\max(0,z-1))^4}{4(\min(1,z)-\max(0,z-1))} \\&=\begin{cases}\frac{z^3}4 &,\text{ if }0<z<1 \\ \frac{z}4 (z^2-2z+2) &, \text{ if } 1<z<2 \end{cases} \end{align}

You will get the same answer from the conditional density using $$E\left[X^3 \mid Z=z\right]=\int x^3 f_{X\mid Z}(x\mid z)\,dx$$

StubbornAtom
  • 17,052