3
$\begingroup$

Let the joint probability distribution of X and Y be $f(x,y)=Te^{-x-y}$ for $0<y<x<\infty$

Find cov(X,Y)

My approach is to first find T. I did this by evaluating T from:

$\int_0^\infty\int_0^xTe^{-x-y}dydx=1$

I got T=2. Is this correct?

Next, I want to use $Cov(X,Y)=E(XY) - E(X)E(Y)$

However, I'm having a problem in getting the marginal distribution of x. I was doing $\int_0^\infty2e^{-x-y}dy$ but this leads to $2e^{-x}$ which is wrong because it is not a valid pdf.

$\endgroup$
  • 4
    $\begingroup$ In the marginal, the integral should be from 0 to $x$ not $0$ to $\infty$. $\endgroup$ – Greenparker Jul 3 '17 at 9:02
3
$\begingroup$

A good way to check one's results is to use a completely different analysis. I will offer two: statistical reasoning and simulation.

Statistical reasoning

The density is proportional to the product $e^{-x}e^{-y}$, which is instantly recognizable as the product of Exponential densities, and the only additional restriction is $y \lt x$. That shows us $(y,x)$ have the same distribution as the order statistics of two independent exponential variables. That is, letting $Z_1$ and $Z_2$ be those variables, set $Z_{(1)} = \min(Z_1,Z_2)$ and $Z_{(2)}=\max(Z_1,Z_2)$. The vector $(Z_{(1)}, Z_{(2)})$ has the same distribution as $(y,x)$.

Let the expectation of an Exponential variable be $\mu=1$. Then because $Z_1+Z_2 = Z_{(1)}+Z_{(2)}$ and $Z_{(1)}Z_{(2)}=Z_1Z_2$,

$$E(x+y) = E(Z_{(1)}+Z_{(2)}) = E(Z_1 + Z_2) = E(Z_1)+E(Z_2) = 2\mu=2\tag{1}$$

and

$$E(xy) = E(Z_{(1)}Z_{(2)}) = E(Z_1Z_2) = E(Z_1)E(Z_2) = \mu^2=1\tag{2}.$$

The Exponential distribution is memoryless: this means that $x-y$, conditional on $x \gt y$, also has an Exponential distribution. Thus

$$E(x)-E(y) = E(x-y) = \mu = 1\tag{3}.$$

The simultaneous linear equations $(1)$ and $(3)$ have the unique solution

$$E(y)=1/2,\ E(x) = 3/2.\tag{4}$$

Plug $(2)$ and $(4)$ into a standard formula for the covariance:

$$\operatorname{Cov}(x,y) = E(xy) - E(x)E(y) = 1 - \frac{3}{2}\frac{1}{2}=\frac{1}{4}.$$

Simulation

It's fast and easy to draw realizations $(x,y)$ from this distribution. The most obvious way--although it's a little inefficient--is to draw positive $(x,y)$ from the distribution with density $e^{-x-y}$ and discard any for which $y \ge x$. Since this density factors into $e^{-x}e^{-y}$, it suffices to draw an independent pair from the univariate distribution with density $e^{-x},\ x\gt 0$: the Exponential. The sample moments for a large set of realizations ought to approximate those of the underlying distribution. Here is R code to produce and analyze approximately a million such realizations; it takes one second to run.

n <- 2e6
set.seed(17)
a <- matrix(rexp(2*n), ncol=2)
a <- a[a[,2] < a[,1], ]
x <- a[,1]
y <- a[,2]
signif(c(E.xy=mean(x*y), E.x=mean(x), E.y=mean(y), Cov=cov(x,y)), 3)

The output is

E.xy  E.x  E.y  Cov 
1.00 1.50 0.50 0.25 

The values agree exactly with the preceding solution to the three significant figures shown.

$\endgroup$
  • $\begingroup$ Should I get the same answer by evaluating $E(xy)=\int_0^\infty\int_0^x2xye^{-x-y}dydx$ then subtracting by (1.5)(0.5)? Why are X and Y considered independent when Y is restricted to be less than X? $\endgroup$ – user164144 Jul 4 '17 at 22:03
  • $\begingroup$ Yea, I got it wrong. I evaluated the above expectation again and got 1. Thanks! $\endgroup$ – user164144 Jul 4 '17 at 22:15
  • $\begingroup$ I don't know what you mean by "X" and "Y". If they are the same as "x" and "y", then you're right: they're not independent. The point is that $Z_1$ and $Z_2$ are independent. That makes them easier to work with. $\endgroup$ – whuber Jul 5 '17 at 12:45
0
$\begingroup$

Thanks to @Greenparker for pointing out where I got it wrong.

So I solved the following to get to the answer:

$f(x)=\int_0^x2e^{-x-y}dy=2e^{-x}-2e^{-2x}$

$E(x)=\int_0^\infty x(2e^{-x}-2e^{-2x})dx=1.5$

$f(y)=\int_y^\infty 2e^{-x-y}dy=2e^{-2y}$

$E(y)=\int_0^\infty 2e^{-2y}=0.5$

$E(xy)=\int_0^\infty\int_0^x2xye^{-x-y}dydx=1$

So, $COV(xy)=1-(1.5)(0.5)=\frac{1}{4}$

$\endgroup$
  • 1
    $\begingroup$ Since $X$ takes on positive values only, warning bells ought to go off in your mind when you find that $E[X]$ -- the average value of $X$ -- is a negative number.... $\endgroup$ – Dilip Sarwate Jul 3 '17 at 22:45
  • $\begingroup$ Sorry about that. Yes, that should be $1.5$. I made a mistake in evaluating the integral. $\endgroup$ – user164144 Jul 3 '17 at 22:55

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.