Actuarial Outpost Condtional Expectation
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#1
01-09-2009, 03:40 PM
 n00sh Member SOA Join Date: Jul 2008 Studying for MFE Posts: 118
Condtional Expectation

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.
#2
01-09-2009, 03:52 PM
 flachboard Member SOA Join Date: Dec 2008 Location: Holmen, WI Studying for Exam MLC College: UW Whitewater Favorite beer: Coffee Stout Posts: 74

Quote:
 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.
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#3
01-10-2009, 02:17 PM
 smhorne1 Join Date: Dec 2007 Location: Charlotte NC Studying for MFE Posts: 20

Quote:
 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.

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