Find E[X] and V[X]

Question

Let the random variable X satisfy $$E\left[(X −1)^2\right] = 10$$ and $$E\left[(X −2)^2\right] = 6.$$

No clue how to start this one... To find E[X] do I integrate both? Would appreciate some hints on how to start this one.

Kinda lost on how to start this one since I never encountered this problem before.

Answer 1

We have \begin{eqnarray} E[(X-1)^2]=10&\Rightarrow&E[X^2-2X+1]=10\nonumber\\ &\Rightarrow&E[X^2]-2E[X]+1=10\nonumber\\ &\Rightarrow&E[X^2]-2E[X]=9\nonumber\qquad(1) \end{eqnarray}

\begin{eqnarray} E[(X-2)^2]=6&\Rightarrow&E[X^2-4X+4]=6\nonumber\\ &\Rightarrow&E[X^2]-4E[X]+4=6\nonumber\\ &\Rightarrow&E[X^2]-4E[X]=2\nonumber\qquad(2) \end{eqnarray}

Subtract (2) from (1) to get $$E[X]=\frac{7}{2}$$

By (2) $$E[X^2]=16$$

Then $$Var(X)=E[X^2]-(E[X])^2=16-\frac{49}{4}$$

Answer 2

Hint: $E$ is a linear map. But to use this fact, you need the input to be a sum of random variables... can you turn a square into a sum somehow?