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Find $ E[(x+2)^2] $ given that $ E[x] = 5, Var(X) = 2 $.

I'm not sure if I'm making this more complicated than it should be. However, this is what I did.

$$ E[(x+2)^2] = E[x^2+2x+4] = E[x^2]+E[2x]+4 = E[x^2]+2E[x]+4 = 25+10+4 = 39. $$ I'm unsure is if $E[x^2]$ is 25. And furthermore, if there's an easier method to solving this problem if $x+2$ was raised to a higher power

Jones
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3 Answers3

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Without any other information, it is not possible to compute the desired expectation. This is because $\operatorname{E}[X] = 5$ does not uniquely determine $\operatorname{E}[X^2]$. In fact, if $\operatorname{E}[X] = \mu$, we can write

$$\begin{align} \operatorname{E}[(X+2)^2] &= \operatorname{E}[(X - \mu + \mu + 2)^2] \\ &= \operatorname{E}[(X - \mu)^2 + 2(\mu + 2)(X - \mu) + (\mu + 2)^2] \\ &= \operatorname{E}[(X - \mu)^2] + 2(\mu + 2)\operatorname{E}[X - \mu] + (\mu+2)^2 \\ &= \operatorname{Var}[X] + 2(\mu + 2)(\mu - \mu) + (\mu + 2)^2 \\ &= \operatorname{Var}[X] + (\mu + 2)^2. \end{align}$$

This shows that the desired expectation equals the variance plus $(\mu + 2)^2 = 49$. So for instance, if $X$ is normally distributed with mean $5$, we can pick any nonnegative variance we please. All that we can say is that $\operatorname{E}[(X+2)^2] \ge 49$.

heropup
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Find $E[(x+2)^2]$ given that $E[x]=5$ ,$Var(X)=2$.

@Jones, observe that:

  • When you refer to rv you have to use always a Capital letter

  • your binomial expansion is wrong.

  • In the post you stated $\mathbb{V}(X)=2$, in your comment you stated it as 5. I assume correct what you stated in your post as it is the more recent edit.

  • remember that $\mathbb{E}(X^2)=\mathbb{V}(X)+\mathbb{E}^2(X)=2+5^2=27$

Given this, you get

$$\mathbb{E}[(X+2)^2]=\mathbb{E}(X^2)+4\mathbb{E}(X)+4=27+4\times 5+4=51$$

tommik
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No, $\displaystyle E\left[ X^{2}\right] \neq 25$.

But, you can probabily recal that: $\displaystyle Var( X) =E\left[ X^{2}\right] -( E[ X])^{2}$.

From here:

$\displaystyle E\left[ X^{2}\right] =Var( X) +( E[ X])^{2} =2+25=27$.

Now,

$\displaystyle \begin{array}{{>{\displaystyle}l}} E\left[( X+2)^{2}\right] =E\left[ x^{2} +4X+4\right] =E\left[ X^{2}\right] +E[ 4X] +E[ 4] =E\left[ X^{2}\right] +4E[ X] +4\\ =27+4\cdotp 5+4=51 \end{array}$

Iam Spano
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