Finding Expected Value of fy(Y) = 3(1-y)^2

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SUMMARY

The expected value of the function fy(Y) = 3(1-y)^2 is calculated using the integral of y multiplied by fy(Y) over the interval from 0 to 1. The correct formula is indeed y * fy(Y), leading to the integral of y * 3(1-y)^2. The final result of this calculation is 1/4, confirming the book's answer. A common mistake noted in the discussion is neglecting the y term in the integral, which can lead to incorrect results.

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  • Understanding of probability density functions
  • Knowledge of integral calculus
  • Familiarity with the concept of expected value
  • Experience with definite integrals
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semidevil
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ok, so to find the expected value of fy(Y) = 3(1-y)^2 for 0 <= 1 <= 1

I thought the formula is y * fy(Y) which is the intgeral from 0 to 1 of y* 3(1-y)^2. right?

the book says the answer is 1/4...but I get a whole number answer...
 
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Hi,

I did the integral of y*f(y) from 0 to 1 and got the answer 1/4...double check your calculation...This is what I did:

Int[3*(1-y)^2] = 3*Int[y^3-2y^2+y] from 0 to 1
 
oops, I forgot the y in the 1st integral expression
 

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