# Math Help - stuck

1. ## stuck

Let X and Y have the joint probability distribution given by

p(x,y) = exp(-(x+y)) for x>0 and y>0

what is P(X+Y<3)?

i know that i would need to use double integrals for dx and dy but i do now know what i can put as the range of x and y such that the condition is fulfilled.

for what values of Y is the conditional density p(x l y) defined?

thanks!

2. Use the conditions you have.

$X>0, Y>0$

$X+Y>3$

$=> 3-X

Now what are your boundaries for $X?$

Also, $p(x|y) = \frac{p(x,y)}{p(y)}$

What can you infer about the bounds here?

Let me know if you have questions.
Anonymous

3. are my boundaries 0<X<3 and x1<Y<3

im still quite confused actually..

and also, how do i solve the other questions? thanks!!

4. Originally Posted by alexandrabel90
are my boundaries 0<X<3 and x1<Y<3

im still quite confused actually..

and also, how do i solve the other questions? thanks!!
I remember doing this stuff in probability also. The bounds were always the cause of confusion. No worries though! After working with bounds on a variety of different problems it will become more intuitive.

Basically what I did was use the facts you were given:

$X>0, Y>0$ and $X+Y>3$

Then, I wrote $Y$ in terms of $X,$ $Y> 3-X$ and we know $Y<3,$ since if $Y>3$ our condition is automatically not satisfied. So we obtain:

$3-X

For $Y$ we integrate over integrate over $(3-X,3).$

The first integral is usually straightforward. $X$ can take on values from $(0,3).$ So you have:

$\int_0^3 \int_{3-x}^3 f_Y (y)f_X(x)dydx$

Hey, you erased your original question?????

5. Originally Posted by alexandrabel90
for what values of Y is the conditional density p(x l y) defined?

and for any Y>0, how do i find the conditional density function of X given that Y=y?

thanks!
We have that,

$p(x|y) = \frac{p(x,y)}{p(y)}$

So you are given $p(x,y)$ already. Now we just need to find the marginal density $p(y)$ in order to determine the conditional density. Then we can use the stipulations on $p(x,y)$ and $p(y)$ to determine for what values of $x$ and $y$ the conditional density is defined.

The marginal density is defined:

$p(y) = \int_{0}^{\infty} e^{-(x+y)}dx
$

Go ahead and calculate that and report back if you run into problems.
Anonymous

6. p(y) =exp(-x)

what does it mean that the conditional densities is defined? does it mean that p(y) cannot be zero?

7. how do you know that x can take the values from 0 to 3? from your workings, i can understand the range of y but im still confused how you derived the range of x..

opps. i just reedited the questions(:

thanks for helping

8. Generally the first integral covers all possible values. Clearly, X can't < 0, be cause that is given. Also, X cannot be >3 because then, there is no way for X+Y to be <3.

There may be exceptions, but this is the rule not the exception:

The first integral covers all possible values of one of the variables, in this case X, and the second integral covers the restricted bounds of the second variable. So,
(1)Choose a variable to restrict. Use the methods shown above to determine your bounds for that variable.
(2)Use the first integral to cover all possible values of your unrestricted variable.

9. Yes the conditional density cannot be undefined. Don't get to caught up on the bounds. For now, calculate your conditional density and show me what you get.

10. Thanks for the explanation..

by the way, p(y) =exp(-y)

and since p(y) cant be zero or the conditional desities wont be defined, does it mean that the y cannot go to infinity?

11. im wondering, shouldnt the denominator be the marginal desitie of X instead of Y?

if it is the marginal desity of X then it is exp(-x)
and if it is the marginal desity of Y then it is exp(-y)

12. Nice observation, but that's a rather controversial statement. Can anything go to $\infty?$

Best of luck.
Anonymous

13. not that i can think off..

so does that mean that it will always be defined since the denominator will hardly ever go to infinity?

The answer is $\int_0^3\int_0^{3-x}e^{-x}e^{-y}dy dx$