lwfrgin

2021-01-08

Since $Px=-1=Px=1=\frac{1}{2}$ , calculate the expected value and variance of the X random variable.
$E\left[{\left(X+1\right)}^{2}\right]=$?

Neelam Wainwright

Step 1
Solution:
From the given information,
$P\left\{x=-1t\right\}=\frac{1}{2}$
$P\left\{x=1\right\}=\frac{1}{2}$
Step 2
Then, the expected value of the X random variable is
$E\left(X\right)=\sum xP\left(x\right)$
$=\left(-1\right)×\frac{1}{2}+\left(1×\frac{1}{2}\right)=0$
Thus, the expected value of the X random variable is 0.
Step 3
$E\left({X}^{2}\right)=\sum {x}^{2}P\left(x\right)$
$=\left({\left(-1\right)}^{2}×\frac{1}{2}\right)+\left({1}^{2}×\frac{1}{2}\right)$
$=\frac{1}{2}+\frac{1}{2}=1$
Then, the variance of the X random variable is
$V\left(X\right)=E\left({X}^{2}\right)-{\left\{E\left(X\right)\right\}}^{2}$
$=1-{0}^{2}=1$
Thus, the variance of the X random variable is 1.
Step 4
$E\left[{\left(X+1\right)}^{2}\right]=E\left[{X}^{2}+2X+1\right]$
$=E\left[{X}^{2}\right]+2E\left[X\right]+E\left[1\right]$
$=1+2\left(0\right)+1=2$

Jeffrey Jordon