Question

If $X$and Y are independent and identically distributed with mean $\mu$and variance $\sigma 2$, find

$E\left[(X-Y{)}^2\right]$

Short answer

The value of $E\left[(X-Y{)}^2\right]$is

$E\left[(X-Y{)}^2\right]=2{\sigma }^2$

Step-by-step solution

Step 1:Given Information

$X$and$Y$are autonomous and identically allocated with mean$\mu$and variance ${\sigma }^2$.

Step 2:Explanation

$E\left[X\right]=\mathrm{E}\left[Y\right]=\mu$

$\mathrm{Var}\left(X\right)=\mathrm{Var}\left(Y\right)={\sigma }^2$

$\mathrm{Var}\left(X\right)=E\left[X^2\right]-\left(E\right[X]{)}^2$

$E\left[X^2\right]=\mathrm{Var}\left(X\right)+\left(E\right[X]{)}^2$

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

$X$ and $Y$ are independent,

$E\left[(X-Y{)}^2\right]=E\left[X^2-2XY+Y^2\right]$

$=E\left[X^2\right]-2E\left[X\right]E\left[Y\right]+E\left[Y^2\right]$

$=\left({\sigma }^2+{\mu }^2\right)-2\mu \mu +\left({\sigma }^2+{\mu }^2\right)$

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

$=2{\sigma }^2$

Step 3:Final Answer

$E\left[(X-Y{)}^2\right]=2{\sigma }^2$