I know Expected value has the property of linearity $E(X+a)=E(X) + a$, but it also seems to hold for $E(X^3+a)$. But Variance also has the property $V(X+a)=V(X) + 0$, but it does not hold for $V(X^3+a)=V(X^3) + 0$.
I discovered this from a question that asked to find the expected value adn variance for $Y = X^3 + 2$ given a distribution for $X$.
$$\begin{array}{|r|r|r|r|r|}\hline x&-1&0&1&2\\\hline p(x)&0.1&0.4&0.3&0.2\\\hline\end{array}$$
The table shows the distribution for $X$, so I worked out $E(Y) = E(X^3 – 2) = 1.8 – 2 =-0.2$.
For variance I am thinking $V(Y) = V(X^3 – 2) = E(X^{3\times2}) – E(X^3)^2 – 0 =E(X^6)-0.2^2$ from which I get 11.16
Best Answer
From the definition of variance and the Linearity of Expectation:$$\begin{align}\mathsf V(X^3+a) &= \mathsf E((X^3+a)^2)-(\mathsf E(X^3+a))^2 \\ &= \mathsf E(X^6+2aX^3+a^2)-(\mathsf E(X^3)+a)^2\\ &=\bigl(\mathsf E(X^6)+2a\mathsf E(X^3)+a^2\bigr)-\bigl(\mathsf E(X^3)^2+2a\mathsf E(X^3)+a^2\bigr)\\&=\mathsf E(X^6)-\mathsf E(X^3)^2\\&=\mathsf V (X^3)\end{align}$$
$$\begin{array}{|r|r|r|r|r|}\hline x&-1&0&1&2\\\hline p(x)&0.1&0.4&0.3&0.2\\\hline\end{array}$$
$\mathsf E(X^3) = -1\cdot 0.1+1\cdot 0.3+0+2^3\cdot 0.2 = 1.8\\\mathsf E(X^6) = 1\cdot(0.1+0.3)+2^6\cdot0.2 = 13.2\\\mathsf {V}(X^3)= 13.2-1.8^2 = 9.96$