# how to justify $\operatorname{E} [X^2] = \sigma^2 + \mu^2$

I am learning this video, which gives a justification of the formula about Sample variance.

at this time point, the teach is giving this formula

$$\operatorname{E} [X^2] = \sigma^2 + \mu^2$$

where does this formula come from?

• Just write the definition of $\sigma^2$ out. – lulu May 11 '19 at 11:55

## 3 Answers

this post has justified

$$\operatorname{Var} (X) = \operatorname{E} [X^{2}] - [\operatorname{E} (X)]^{2}$$

which is the same one on your video, call it equation_1

the term on the left of that equation above $$\operatorname{Var} (X) = \sigma^2$$

the second term on the right of equation above $$[\operatorname{E} (X)]^{2} = \mu^2$$

the one as follow is the exactly same as equation_1.

$$\sigma^2 = \operatorname{E} [X^{2}] - \mu^2$$ Add $$\mu^2$$ to both sides $$\sigma^2 + \mu^2 = \operatorname{E} [X^2]$$

$$\sigma^{2}=E(X-\mu)^{2}=E(X^{2}-2\mu X+\mu^{2})=EX^{2}-2 \mu^{2}+\mu^{2}=EX^{2}-\mu^{2}$$. Add $$\mu^{2}$$ to both sides.

You already have $$\mathbb{V}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$$ so that $$\mathbb{E}(X^2) =\mathbb{V}(X) + (\mathbb{E}(X))^2$$. But on the other hand $$\mathbb{V}(X) = \sigma^2$$ and $$\mathbb{E}(X) = \mu$$.