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

Question

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

Answer 1

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