Random Variables X and Y follow a joint distribution:
f(x,y)=(6-x-y)/8 if 0<x<2, 2<y<4
and 0, else
Find P(1<Y<3|X=1)
I first got the marginal distribution for x and got (3-x)/4 (Integrated from 2 to 4)and then got the conditional dist and got (5-y)/4 and then integrated that from 1 to 3 and got 1.5, which means it's wrong. What am I doing wrong?
Thanks
To find P(1 < Y < 3|X = 1), you need to use the definition of conditional probability:
P(1 < Y < 3|X = 1) = P(1 < Y < 3, X = 1) / P(X = 1)
To find P(1 < Y < 3, X = 1), you integrate the joint distribution f(x,y) over the given region:
∫(1 < Y < 3, X = 1) f(x,y) dx dy
From the joint distribution, we have:
f(x,y) = (6 - x - y) / 8 if 0 < x < 2 and 2 < y < 4
f(x,y) = 0 otherwise
Plugging in X = 1 into the joint distribution:
f(1,y) = (6 - 1 - y) / 8 = (5 - y) / 8 if 2 < y < 4
f(1,y) = 0 otherwise
Now, we integrate f(1,y) over the given region:
∫(1 < Y < 3, X = 1) (5 - y) / 8 dy
Integrating from 1 to 3:
= (1/8) * ∫[1,3] (5 - y) dy
= (1/8) * [5y - (y^2/2)] evaluated from 1 to 3
= (1/8) * [(5*3 - (3^2/2)) - (5*1 - (1^2/2))]
= (1/8) * [(15 - (9/2)) - (5 - (1/2))]
= (1/8) * [(30/2 - 9/2) - (10/2 - 1/2)]
= (1/8) * [(21/2) - (9/2) - (9/2)]
= (1/8) * [21/2 - 18/2]
= (1/8) * 3/2
= 3/16
So, P(1 < Y < 3|X = 1) = 3/16.
You were correct in calculating the marginal distribution for X, but there seems to have been a mistake while calculating the conditional distribution or during integration.