Condtional Expectation

[n00sh]

I know the thereotical formula for E[Y|X=x] in the continuous case, but when asked to find E[Y|X=1] in the continuous case, is the following formula correct? (Sorry for formatting but I don't know how to put into mathematical form on here)

E[Y|X=1]= int(a,b) y * f(1,y) dy / int(a,b) f(1,y) dy


int(a,b) is the integral with limits a to b where a and b are the limits of y given x = 1.

Thanks.

[flachboard]

Originally Posted by n00sh

I know the thereotical formula for E[Y|X=x] in the continuous case, but when asked to find E[Y|X=1] in the continuous case, is the following formula correct? (Sorry for formatting but I don't know how to put into mathematical form on here)

E[Y|X=1]= int(a,b) y * f(1,y) dy / int(a,b) f(1,y) dy


int(a,b) is the integral with limits a to b where a and b are the limits of y given x = 1.

Thanks.

E[Y|X=1]=int(a,b) y * f(1,y)dy/ fx(1)
In words, the integral from a to b of y times f of 1,y dy divided by the marginal distribution of x when x equals 1. I believe that's right.

[smhorne1]

Originally Posted by flachboard

E[Y|X=1]=int(a,b) y * f(1,y)dy/ fx(1)
In words, the integral from a to b of y times f of 1,y dy divided by the marginal distribution of x when x equals 1. I believe that's right.

This is correct.

Sometimes you can do the following - If the joint density doesn't depend on y, the f(y given x =1) is uniform, and you can find the E[y] using the continuous uniform formula = (a+b)/2. But you have to correctly define the interval in which y given x is uniform. I usually have to draw a picture to see.