Question

Suppose X and Y have joint density f(x,y)=1 for 0 14 years ago

Answer 1

I'm confused on how to set up the limits for the double integral in this problem

Answer 2

So you're looking for the probability X*Y<=z, not X<=z and Y<=z, right?

Answer 3

Yes that's correct

Answer 4

I'll take a look at this in a minute.

Answer 5

Ok, so you're integrating over a square.

Answer 6

Yes exactly, a square with corners, (0,0) (0,1) (1,1) (1,0)

The distribution function F(x,y) = xy for 0≤x≤1, and 0≤y≤1

I just am confused where z comes into play for the double integral.

Answer 7

The idea is that for the outer integral you integrate over the entire region, so here from 0 to 1. The variable over which you integrate in the outer integral is fixed in the inner integral. So for the inner integral you integrate over those values for which the condition holds, here XY<=Z

Answer 8

Let's integrate over y in the outer integral, so we get: \[\int_0^1\int_a^b1dxdy\]

Answer 9

Now to find a and b.

Answer 10

Still with me so far?

Answer 11

Yes, I got that far, but I was thinking a = 0, and b = z/y? But this gives a ln(0) which is why I'm confused

Answer 12

That's what I had in mind too, I see the problem.

Answer 13

The problem is that z/y can become bigger than 1.

Answer 14

so you have to use min{1,z/y}?

Answer 15

That does work, but integrating that might be hard.

Answer 16

Yeah I wouldn't know how to integrate that

Answer 17

I might have an idea how to integrate that:

z/y<=1
y>=z

So you integrate 1 from 0 to z and integrate z/y from z to 1.

Answer 18

\[\int_0^1\text{min}(1,\frac{z}{y})dy=\int_0^z1dy+\int_z^1\frac{z}{y}dy\]

Answer 19

Is it that simple by just adding them together like that for the integral of a min{} ?

Answer 20

\[\int_0^1\text{min}(1,\frac{z}{y})dy=\int_0^z\text{min}(1,\frac{z}{y})dy+\int_z^1\text{min}(1,\frac{z}{y})dy\]

Answer 21

That's true more clearly, and on the interval [0,z] 1 is smaller than z/y, and on the interval [z,1] z/y is the smaller one.

Answer 22

Final answer is z-zln(z), which behaves properly: http://www.wolframalpha.com/input/?i=Plot [x-xln%28x%29%2C{x%2C0%2C1}]

Answer 23

Ok Thank you. I will work through this right now and see if I get the same conclusion

Answer 24

Perfect, Thanks for all your help

Answer 25

You're welcome.