Suppose thatthe joint density of X and Y is given by
(x,y)=e-(x/y).e(-y) ,0<x<�‡,0<y<�‡ and is o otherwise. Find P(X>1 / Y=y)
To find P(X>1 | Y=y), we need to calculate the conditional probability of X being greater than 1 given Y=y. Let's use the definition of conditional probability:
P(X>1 | Y=y) = P(X>1 and Y=y) / P(Y=y)
To find the numerator P(X>1 and Y=y), we integrate the joint density function over the region where X>1 and y=y:
∫∫[e^(-x/y) * e^(-y)] dx dy, where x > 1 and 0 < y < ∞.
Since the joint density function is given by e^(-x/y) * e^(-y), we can integrate with respect to x first:
∫[e^(-x/y) * e^(-y)] dx = ∫e^(-x/y) dx = -y * e^(-x/y) |x=1 to ∞
Evaluating this integral, we get:
[-y * e^(-x/y)] |x=1 to ∞ = - y * (e^(-∞/y) - e^(-1/y))
Since e^(-∞) approaches 0, the first term in the above expression becomes 0, so we have:
[-y * e^(-x/y)] |x=1 to ∞ = y * e^(-1/y)
Now, let's find the denominator P(Y=y):
P(Y=y) = ∫∫[e^(-x/y) * e^(-y)] dx dy, where 0 < x < ∞ and 0 < y < ∞.
To calculate this integral, we integrate first with respect to x:
∫[e^(-x/y) * e^(-y)] dx = ∫e^(-x/y) dx = -y * e^(-x/y) |x=0 to ∞
Evaluating this integral, we get:
[-y * e^(-x/y)] |x=0 to ∞ = - y * (e^(-∞/y) - e^(0/y))
Again, since e^(-∞) approaches 0, the first term in the above expression becomes 0, so we have:
[-y * e^(-x/y)] |x=0 to ∞ = y
Now we can calculate the conditional probability:
P(X>1 | Y=y) = P(X>1 and Y=y) / P(Y=y)
= (y * e^(-1/y)) / y
= e^(-1/y)
Therefore, the conditional probability P(X>1 | Y=y) is e^(-1/y).