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?
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.........
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) "