# If X and Y are independent and identically distributed with mean \mu and variance \sigma^{2}, find E[(X-Y)^{2}]

If X and Y are independent and identically distributed with mean $\mu$ and variance ${\sigma }^{2}$, find $E\left[\left(X-Y{\right)}^{2}\right]$
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Jaylen Fountain
Step 1
Consider the two normal random variables X and Y are independently and identically distributed.
$E\left(X\right)=E\left(Y\right)=\mu$
$V\left(X\right)=V\left(Y\right)={\sigma }^{2}$
Step 2
Consider,
$V\left(X\right)=E\left({X}^{2}\right)-\left[E\left(X\right){\right]}^{2}$
$E\left({X}^{2}\right)=V\left(X\right)-\left[E\left(X\right){\right]}^{2}$
$E\left({X}^{2}\right)={\sigma }^{2}-{\mu }^{2}$

Step 3
Now, the required quantity can be derived as follows:
$E\left(X-Y{\right)}^{2}=E\left({X}^{2}+{X}^{2}-2XY\right)$

$={\sigma }^{2}-{\mu }^{2}+{\sigma }^{2}-{\mu }^{2}-2{\mu }^{2}$
$=2{\sigma }^{2}-4{\mu }^{2}$