Calculating Variance for Y=3x^2+3x+3

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To calculate the variance of Y=3x^2+3x+3, the initial approach incorrectly assumes that the covariance between X^2 and X is zero. The correct variance calculation should account for this covariance, as it is not generally true that Cov(X^2, X) equals zero. The variance expression Var(Y) = 9Var(x^2) + 9Var(x) is flawed without this consideration. The discussion highlights the importance of recognizing covariance when calculating variances in polynomial expressions. Understanding these assumptions is crucial for accurate statistical analysis.
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Homework Statement


Find the variance of Y=3x^2+3x+3

The Attempt at a Solution


Let Y = 3x^2 +3x +3

Var(Y) = Var(3x^2 +3x +3) = 9Var(x^2) +9Var(x) = 9 [E[X^4] - E[X^2]^2 +E[X^2] - E[X]^2]
This is wrong.
 
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hellokitten said:
Var(3x^2 +3z +1) = 9Var(x^2) +9Var(x)
I guess the z should be X, or 9var(x) should be 9Var(Z). Either way, what assumption does that step require?
 
haruspex said:
I guess the z should be X, or 9var(x) should be 9Var(Z). Either way, what assumption does that step require?

Oh. It assumes no covariance.
 
Last edited:
hellokitten said:

Homework Statement


Find the variance of Y=3x^2+3x+3

The Attempt at a Solution


Let Y = 3x^2 +3x +3

Var(Y) = Var(3x^2 +3x +3) = 9Var(x^2) +9Var(x) = 9 [E[X^4] - E[X^2]^2 +E[X^2] - E[X]^2]
This is wrong.

You have assumed that ##\text{Cov}(X^2,X) = 0,## which is not generally true.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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