Find mean $\mu$ and variance $\sigma^2$ given $E[(X-1)^2] = 10$ and $E[(X-2)^2] = 6$

Question

How do I find $\mu$ and $\sigma^2$ given $E[(X-1)^2] = 10$ and $E[(X-2)^2] = 6$? I totally have no clue. Any idea/hint would be good. :)

Answer 1

Recall that $\mu={\rm E}[X]$ and that $\sigma^2={\rm E}[X^2]-{\rm E}[X]^2$. Now, simply expand $(X-1)^2$ and $(X-2)^2$ and use linearity of the expecation after which you'll obtain two equations in two unknowns (the two unknowns being ${\rm E}[X]$ and ${\rm E}[X^2]$). Solve this.

Answer 2

$E(X-1)^2=EX^2-2EX+1=10$

$E(X-2)^2=EX^2-4EX+4=6$

Subtract the equations to get

$2EX-3=4$ $EX=3.5$ Then EX^2=10+2 \times 3.5-1=16$

$Var(X)=16-3.5^2$

Answer 3

Alternative:

In general for a constant $c$ we have $\mathbb{E}\left(X-c\right)=\mathbb{E}X-c$ and $\text{Var}\left(X-c\right)=\text{Var}X$.

Making use of that we find the folowing equations that are not difficult to solve:

$$10=\mathbb{E}\left(X-1\right)^{2}=\text{Var}\left(X-1\right)+\left(\mathbb{E}\left(X-1\right)\right)^{2}=\sigma^{2}+\left(\mu-1\right)^{2}$$

$$6=\mathbb{E}\left(X-2\right)^{2}=\text{Var}\left(X-2\right)+\left(\mathbb{E}\left(X-2\right)\right)^{2}=\sigma^{2}+\left(\mu-2\right)^{2}$$