# Multivariable Probability Distribution

1. Apr 26, 2012

### mathmajor23

f(x,y) = ke^(-x-y), 0<x<∞ , 0<y<∞ , and x<y

Find k to make this a PDF.

So I set up: ∫(from 0 to ∞)∫(from 0 to ∞) [ke^(-x-y)]dxdy. Is this integral right?

I also need to find the marginals of X and Y but my k has to be right first.

2. Apr 26, 2012

### Dustinsfl

I am not familiar with multi-variable, but if it is like single, you would need to set the integral = 1 and you could find your k.

3. Apr 26, 2012

### mathmajor23

Yes, it's the same as single variable. I'm just unsure on my bounds of my integrals.

4. Apr 26, 2012

### Dustinsfl

[0,Infinity) seem reasonable.

5. Apr 26, 2012

### Ray Vickson

Your integration is over the two-dimensional set {(x,y}: 0 ≤ x < ∞, 0 ≤ y < ∞}, but you told us that f is defined only for x < y.

RGV

6. Apr 27, 2012

### mathmajor23

Oops! So it should be ∫from 0 to ∞ ∫ from 0 to x [f(x,y)dxdy] ?

7. Apr 27, 2012

### Ray Vickson

Yes, or the other way if you prefer (y from 0 to ∞ and x from 0 to y).

RGV.

8. Apr 27, 2012

### mathmajor23

What's the integral of the inner part evaluate to? I tried using integration by substitution but it's not working.

9. Apr 27, 2012

### Ray Vickson

Are you saying you don't know how to do the integral $\int_0^x e^{-y} \, dy?$ I hope you are joking.

RGV

10. Apr 28, 2012

### mathmajor23

I got that k=2, and checked it by seeing it integrated to 1.

For the Marginal Distributions of X and Y, I got :
f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x)
f(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y)

Then, for the conditionals, I got:
f(x|y) = e^(-x) / [1-e^(-y)] I[0,infinity) (x)
f(y|x) = e^(-y) / [1-e^(-x)] I[0,infinity) (y)

Can someone check those? Thanks!

11. Apr 29, 2012

### Ray Vickson

Your marginal for y is correct but your marginal for x is incorrect. Do NOT use the same f for three different functions. You already used f for f(x,y), so call the marginals something different, like g(x) and h(y), or fX(x) and fY(y).

RGV

Last edited: Apr 29, 2012
12. Apr 29, 2012

### mathmajor23

Was my conditional of f(x|y) correct?

Hmm is this the MD of x then?

Integral from 0 to infinity [2e^(-x-y)dy]
= -2(e^(-x-y)] from y=0 to y=infinity
=2[1+e^(x)] I [0,infinity) (x)

Last edited: Apr 29, 2012
13. Apr 29, 2012

### Ray Vickson

There is no point in answering questions about conditional densities until you get both marginal densities correct. The integral above is wrong.

RGV

14. Apr 29, 2012

### mathmajor23

Which part of f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x) is wrong?

15. Apr 29, 2012

### Ray Vickson

All of it. You seem to have claimed
$$\int_x^{\infty} e^{-x-y}\, dy = e^{-x} - e^{-2x}.$$
but this is absolutely not correct.

RGV

16. Apr 29, 2012

### mathmajor23

Let's start over.

I originally had for the marginal of X:
∫ from 0 to x of [2e^(-x-y)]dy
=-2[e^(-x-y)], evaluated from y=0 to y=x
= 2[e^(-x)-e^(-2x)] I[0,∞) (x)

To check this, I integrated this and it came out to 1, so I'm not sure why this is wrong.

17. Apr 29, 2012

### Ray Vickson

It is wrong because you integrated y from 0 to x instead of from x to infinity.

RGV

18. Apr 29, 2012

### mathmajor23

Ok, hopefully here's the final check:

fx(x) =
∫ from x to infinity [2e^(-x-y)]dy
=-2[e^(-x-y)] from y=x to y=∞
= 2e^(-2x) I[0,∞) (x)

fy(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y)

f(x|y) = 2e^(-x-y) / 2[e^(-y)-e^(-2y)]
= [e^(-x)] / (1-e^(-y)) I[0,∞) (x)

f(y|x) = 2e^(-x-y) / 2e^(-2x)
= e^(-y) / e^(-x) I[0,∞) (y)

19. Apr 30, 2012

### Ray Vickson

These look OK now, except you should get rid of those annoying I[0,∞)(x), etc., symbols. Just say in words that x and y are >= 0.

RGV

20. Apr 30, 2012

### mathmajor23

Thanks a bunch!

I, too, hate the annoying indicator functions, but my professor seems the need to "require" us to use them.