How do i solve $E(7+8X+X^2)$?

Question

I'm trying to solve the expected value of the random variable $E(7+8X+X^2)$. What I do know is that $E(X)=5$, and the variance is $Var(X)=2$.

Use the linearity of the mean, and that the variance is the mean of the square minus the square of the mean. — Sean Roberson, Sep 23 '16 at 15:19

operatorerror · Answer 1 · 2016-09-23T15:29:54.907

1

Using the fact that variance is defined as $\sigma^2=E(x^2)-[E(x)]^2$, expectation is linear and the expectation of a constant is the constant. $$E(7+8x+x^2)=E(7)+8E(x)+E(x^2)=7+8*5+E(x^2)=47+\sigma^2+[E(x)]^2=47+2+25=74$$

edited Sep 23 '16 at 15:29

answered Sep 23 '16 at 15:27

operatorerror

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I think you miscalculated there. It should be $E(x^2)=\sigma^2+(E(x))^2$ – Cahn Sep 23 '16 at 15:29
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@MarvinF. Oops fixing – operatorerror Sep 23 '16 at 15:29

score 0 · Accepted Answer · answered Sep 23 '16 at 16:07

We know that $\mathbb{E}[X]=5$ and $\text{Var}[X]=\mathbb{E}[(X-5)^2]=2$. Since $$ X^2+8X+7 = (X-5)^2+18 X-18 \tag{1}$$ by the additivity of the expected value it follows that $$ \mathbb{E}[X^2+8X+7]= 18\left(\mathbb{E}[X]-1\right)+\text{Var}[X] = 18\cdot 4+2=\color{red}{74}.\tag{2}$$

How do i solve $E(7+8X+X^2)$?

2 Answers2