1. ## about jointly continuous random variable

Suppose X and Y are jointly continuous with joint density

f(x,y)= (exp(-x/y)*exp(-y))/y for 0<x<oo and 0<y<oo

f(x,y)= 0 otherwise

Find P(X>1|Y=y).

I first find the density functions of X and Y respectively.
But I am in difficulty in integrating (exp(-x/y)*exp(-y))/y with respect to y. Can someoneone provides some hints?

Also P(X>1|Y=y)=P(X>1 and Y=y)/P(Y=y)
But for continuous random variable Y, P(Y=y)=0?

2. Originally Posted by ldpsong
Suppose X and Y are jointly continuous with joint density

f(x,y)= (exp(-x/y)*exp(-y))/y for 0<x<oo and 0<y<oo

f(x,y)= 0 otherwise

Find P(X>1|Y=y).

I first find the density functions of X and Y respectively.
But I am in difficulty in integrating (exp(-x/y)*exp(-y))/y with respect to y. Can someoneone provides some hints?

Also P(X>1|Y=y)=P(X>1 and Y=y)/P(Y=y)
But for continuous random variable Y, P(Y=y)=0?
Where is your single veriable distribution $\displaystyle f(y)$?

3. To obtain P(X>1|Y=y) you need the conditional density f(x|y) and you integrate x from 1 to infinity.

Both P(X>1 and Y=y) and P(Y=y) are zero.

$\displaystyle f(x|y)={f(x,y)\over f_Y(y)}$

4. Originally Posted by novice
Where is your single veriable distribution $\displaystyle f(y)$?
f(y)= S (exp(-x/y)*exp(-y))/y dx = exp(-y) <--------integration
f(x)= S (exp(-x/y)*exp(-y))/y dy =?....

I want to show that X and Y are independent,
so P(X>1|Y=y) = P(X>1)

5. SORRY, they are dependent by inspection.
The joint denisty does not factor.
This looks like a nasty integral.

6. there is another question :
Let X and Y be jointly continuous random variable with joint density function

f(x,y) = kxy for 0<=x<=2 and 0<y<x
f(x,y) =0 otherwise

There are 4 questions. I have solved the first two.

1, k=1/2
2,density function of X: f(x)=x^3/4 for 0<=x<=2
density function of Y: f(y)= y for 0<y<x

I have not solved the 3rd and 4th questions yet

3, Are X and Y independent?
4,Find P(X>3/2|Y=1)

It seems that the 2nd part is not correct.........

7. Originally Posted by matheagle
SORRY, they are dependent by inspection.
The joint denisty does not factor.
This looks like a nasty integral.
integrate exp(-x/y)/y with respect to x from 1 to infinity is exp(-1/y)
So P(X>1|Y=y)=exp(-1/y)

Is it correct?

8. any probability is between 0 and 1, inclusive.
So you need to find the correct density and integrate.

9. Originally Posted by ldpsong
there is another question :
Let X and Y be jointly continuous random variable with joint density function

f(x,y) = kxy for 0<=x<=2 and 0<y<x
f(x,y) =0 otherwise

There are 4 questions. I have solved the first two.

1, k=1/2
2,density function of X: f(x)=x^3/4 for 0<=x<=2
density function of Y: f(y)= y for 0<y<x

I have not solved the 3rd and 4th questions yet

3, Are X and Y independent?
4,Find P(X>3/2|Y=1)

It seems that the 2nd part is not correct.........

These two are dependent rvs since y<x.

To obtain k you need

$\displaystyle 1=k\int_0^2\int_0^x xy dydx$

AND the mariginal density of Y CANNOT have x in it.
It's a function of just y.
The mariginal of Y has support from 0 to 2.

10. Originally Posted by matheagle
These two are dependent rvs since y<x.

To obtain k you need

$\displaystyle 1=k\int_0^2\int_0^x xy dydx$

AND the mariginal density of Y CANNOT have x in it.
It's a function of just y.
The mariginal of Y has support from 0 to 2.
Do you means that this one
" density function of Y: f(y)= y for 0<y<x "
should be changed to this
"density function of Y: f(y)= y for 0<y<2 "

But I still found that density function cannot satisfy the condition ......
Can I change it to:
"density function of Y: f(y)= y for 0<y<2^(1/2) "