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Let $(X,Y)$ be a random vector with uniform distribution at $0 \leq x \leq 1$, $x \leq y \leq x+h$ with $0<h<1$.

Find $E(X)$ and $E(XY)$.

What i did:

(1) Find densities:

$f_X(x) = \left\{ \begin{array}{lr} 1 & : 0 \leq x \leq 1 \\ 0 & : otherwise \end{array} \right.$

$f_Y(y) = \left\{ \begin{array}{lr} 1/h & : x \leq y \leq x+h \\ 0 & : otherwise \end{array} \right.$

(2) Find expected values

$E(X) = \displaystyle\int_0^1xf_X(x)dx = \displaystyle\int_0^1xdx =1/2$

$E(XY) = \displaystyle\int_{-\infty}^{+\infty}\displaystyle\int_{-\infty}^{+\infty}xf_{X,Y}(x,y)dxdy$, this mean i should find $f_{XY}$ or there's another way to do it?. Because here i don't know how can i get $f_{XY}$, my first attempt was consider $f_{XY}= f_{X}f_{Y}$ but it's only true if $X$ and $Y$ are independent.

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  • $\begingroup$ Probably, i can't get $f_{XY}$ with what i know, can i?. $\endgroup$ – Cure Jul 9 '13 at 7:12
  • $\begingroup$ $X$ and $Y$ certainly don't seem independent. On the other hand, the distribution is given - it's uniform on that parallelogram, presumably with respect to the usual measure on $\mathbb R^2$ - and so you should have enough information to solve the problem. I suggest writing $Z=Y-X$ and showing that $Z$ and $X$ are independent, then writing everything in terms of $X$ and $Z$. $\endgroup$ – Greg Martin Jul 9 '13 at 7:21
  • $\begingroup$ Sorry, I read badly the question. $f_Y$ is not the thing you wrote. The distribution of $f_{X,Y}$ is uniform in the set $A = \{(x,y): 0 \leq x \leq 1, x \leq y \leq x+h\}$ so $f_{X,Y}$ will be the indicator function of $A$ divided by its area. $\endgroup$ – Bunder Jul 9 '13 at 7:22
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Outline of Solution: We give two approaches, with the second significantly simpler than the first.

We have a constant density over a certain parallelogram. The corners are $(0,0)$, $(1,1)$, $(1,2)$, and $(0,1)$.

The density function of $X$ is right, it is uniform. The density function of $Y$ is not right. Even the shape is not right. The density function of $Y$ can only involve the variable $y$, you have to "integrate $x$ out. A look at the picture shows that $Y$ lives on the interval $[0,2]$, and that the densities near $y=0$ and $y=2$ are close to $0$. You can get the density of $Y$ either by integration or more easily by geometry. (First find the cdf. It will be a different formula for $y\lt 1$ than for $y\gt 1$).

To find the expectation of $XY$, we luckily don't need the distribution of $XY$. Note that our parallelogram has area $1$, so we just need to integrate $xy$ over the parallelogram.

Another way: Let $Z=Y-X$. Then $(X,Z)$ is uniform over the unit square. So to find the density of $Y$, you need the density of a sum of two uniforms. This may be a problem you have seen before.

To find the expectation of $XY$, we need the expectation of $X(X+Z)$. The expectation of $X^2$ is easy, you know the density of $X$. The expectation of $XZ$ is also easy. Add.

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