Variance of a function g(x)=(2x+1)^2 where x is random variable of values 3 6 9 with probabilities 1/6 1/2 and 1/3 respectively

Question

Solving for $Var[g(x)]$ by computing $E[g(x)^2]-(E[g(x)])^2$ gives the answer:

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While solving for variance by breaking down the g(x) into individual values gives $Var[g(x)] = Var[(2x+1)^2]$

$Var[(4x^2)+(4x)+1] = Var[4x^2]+Var[4x]+0 = 16(E[x^4]-(E[x^2])^2 + E[x^2] - E[x])^2)$

$E[x^4] = 2745/4 , E[x^2] = 46.5 , E[x] = 6.5$

$Var[g(x)] = 11048$

Answer 1

We have

\begin{align} Var(4X^2+4X+1) &= 16Var(X^2+X)\\ &= 16 Cov(X^2+X, X^2+X) \\ &= 16[Var(X^2)+Var(X) + 2Cov(X^2, X)] \end{align}

\begin{align}Cov(X^2, X) &= E[X^3] - E[X^2]E[X] \\ &=\left(3^3 \cdot \frac16 + 6^3\cdot\frac12 + 9^3\cdot \frac13\right)-(46.5)(6.5) \\ &=53.25\end{align}

$$11048+32(53.25)=12752$$