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. Can someone tell me what. I'm am I doing wrong?
To find P(1<Y<3 | X=1), you need to use the conditional probability formula:
P(1<Y<3 | X=1) = P(1<Y<3, X=1) / P(X=1).
Let's calculate each term separately:
1. P(1<Y<3, X=1):
We can determine the range of possible values for Y when X=1 by looking at the joint distribution:
f(x,y) = (6-x-y)/8 if 0<x<2, 2<y<4, and 0, else
Since X=1, the range for Y is still 2<y<4.
So, the joint distribution becomes:
f(1, y) = (6-1-y)/8 = (5-y)/8 if 2<y<4, and 0, else
Integrating f(1, y) over the range 1<Y<3, we get:
Integral[((5-y)/8), {y, 2, 3}] = Integral[(5-y)/8, {y, 2, 3}] = [5y/8 - y^2/16] evaluated from y=2 to y=3
= [(15/8) - (9/16)] - [(10/8) - (4/16)]
= 2 - (13/8)
= (16 - 13)/8
= 3/8.
Therefore, P(1<Y<3, X=1) = 3/8.
2. P(X=1):
To find P(X=1), we can integrate the marginal distribution for X:
f(x) = Integral[((6-x-y)/8), {y, 2, 4}] evaluated from x=1 to x=1
= Integral[(6-x-y)/8, {y, 2, 4}]
= [(6-x)y/8 - y^2/16] evaluated from y=2 to y=4
= (6-x)/8 * (4-2) - [4^2/16 - 2^2/16]
= (6-x)/8 * 2 - (16/16 - 4/16)
= (6-x)/4 - (12/16)
= (6-x)/4 - 3/4
= (6-x-3)/4
= (3-x)/4.
Substituting x=1, we get:
P(X=1) = (3-1)/4 = 2/4 = 1/2.
Finally, putting it all together:
P(1<Y<3 | X=1) = P(1<Y<3, X=1) / P(X=1)
= (3/8) / (1/2)
= (3/8) * (2/1)
= 6/8
= 3/4.
Therefore, the correct value for P(1<Y<3 | X=1) is 3/4.
To find P(1<Y<3|X=1), you need to calculate the conditional probability of Y being between 1 and 3 given that X is equal to 1.
You correctly found the marginal distribution of X, which is (3-x)/4 for 2 ≤ x ≤ 4. However, the value of X is given as 1, so the marginal distribution for X being 1 is (3-1)/4 = 1/2.
Next, you need to find the conditional distribution of Y given X=1. To do this, you need to calculate the joint probability density function (PDF) of X and Y when X=1 and then normalize it.
f(y|x=1) = f(x=1,y) / f(x=1)
To find f(x=1,y), substitute X=1 into the joint distribution:
f(1,y) = (6-1-y)/8 = (5-y)/8 (for 2 ≤ y ≤ 4)
To find f(x=1), integrate f(1,y) over the range of y:
f(1) = ∫[(5-y)/8] dy (from 2 to 4)
Calculate the integral:
f(1) = [(5y - y^2)/8] (from 2 to 4) = 1/2
Now calculate the conditional distribution:
f(y|x=1) = f(1,y) / f(1) = [(5-y)/8] / (1/2) = (5-y)/4
Finally, integrate the conditional distribution over the range of y from 1 to 3 to find P(1<Y<3|X=1):
P(1<Y<3|X=1) = ∫[(5-y)/4] dy (from 1 to 3)
Calculate the integral:
P(1<Y<3|X=1) = [(5y - (y^2)/2)/4] (from 1 to 3) = (9/8)
Therefore, P(1<Y<3|X=1) = 9/8.