Mathematical Expectation

xuyuan · #1 $Old$ October 26th, 2009, 10:52 PM

Can someone explain how I can arrive at this solution? The book says to take the integral of g(X)f(x) evaluated from negative to positive infinity, but it gives no way to do it if F(x) is defined at more than a single place. The solution should be -11/6

If the probability density of X is given by

F(x) X/2 for 0<x<=1
1/2 for 1<x<=2
(3-x)/2 for 2<x<3
0 elsewhere

find the expected value of g(X)=X^2-5X+3

Chris L T521 · #2 $Old$ October 26th, 2009, 10:57 PM

Quote:

Originally Posted by xuyuan $View Post$

Can someone explain how I can arrive at this solution? The book says to take the integral of g(X)f(x) evaluated from negative to positive infinity, but it gives no way to do it if F(x) is defined at more than a single place. The solution should be -11/6

If the probability density of X is given by

F(x) X/2 for 0<x<=1
1/2 for 1<x<=2
(3-x)/2 for 2<x<3
0 elsewhere

find the expected value of g(X)=X^2-5X+3

$\int_{-\infty}^{\infty}g(x)f(x)\,dx=\int_0^1 \tfrac{1}{2}\left(x^3-5x^2+3x\right)\,dx+\int_1^2\tfrac{1}{2}\left(x^2-5x+3\right)\,dx$ $+\int_2^3\tfrac{1}{2}\left(-x^3+8x^2-18x+9\right)\,dx$

Do you see why this is the case?

xuyuan · #3 $Old$ October 26th, 2009, 10:59 PM

Quote:

Originally Posted by Chris L T521 $View Post$

$\int_{-\infty}^{\infty}g(x)f(x)\,dx=\int_0^1 \tfrac{1}{2}\left(x^3-5x^2+3x\right)\,dx+\int_1^2\tfrac{1}{2}\left(x^2-5x+3\right)\,dx$ $+\int_2^3\tfrac{1}{2}\left(-x^3+8x^2-18x+9\right)\,dx$

Do you see why this is the case?

Yes, thank you for the quick reply.

I believe the theorem is E(X) is the sum of the individual components. Correct? Kind of like just taking discrete probabilities such as 0.5*1/2+0.5*1/3=0.41667 so the same logic applies to the continuous RV Thanks again!