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

Question

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?

Answer 1

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

Answer 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.

Answer 3

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 $\mathbb{E}(X) = \mu$.