$\mathbb E[X^2]$ vs $\mathbb E[X]^2$ in Statistics

Question

In a statistics problem solution it stated: $$ \mathbb E[\bar{X}]^2 = \operatorname{Var}(\bar{X})+ (\mathbb E[\bar{X}])^2 $$

I remember that

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

Are we saying that $ \mathbb E[{X}]^2 = \mathbb E[X^2] $ ?

Or is it because we have $\bar{X}$ instead of ${X}$?

What's the difference between $ \mathbb E[{X}]^2$ and $\mathbb E[X^2]$ ?

Answer 1

$\mathbb E[X]^2$ and $\mathbb E[X^2]$ are not the same thing. The first one is the square of the expected value. The second one is the expected value of the square of the random variable $X$.

For example, if $X$ is the outcome of a fair die numbered from $1$ to $6$, then the expected value of a single roll is $$\mathbb {E}[X] = \sum_{x=1}^6 x \Pr[X = x] = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + \cdots + 6 \cdot \frac{1}{6} = \frac{7}{2}.$$ The square of this value is $$\mathbb {E}[X]^2 = \frac{49}{4}.$$ However, the square of the set of outcomes is $$X^2 \in \{1, 4, 9, 16, 25, 36\},$$ hence the expected value of the square is $$\mathbb{E}[X^2] = \sum_{x=1}^6 x^{\color{red}{2}} \Pr[X = x] = 1 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + \cdots + 36 \cdot \frac{1}{6} = \frac{91}{6}.$$ And this is obviously not the same as $\mathbb{E}[X]^2$.

The first equation you wrote is not technically correct. The left hand side should be written $\mathbb E[\bar X^2]$. It's possible that $\bar X^2$ might be confused as $$\frac{1}{n} \sum_{i=1}^n X_i^2,$$ that is to say, the mean of the squared sample, but this in my view is incorrect: I would instead write $\overline{X^2}$ to denote this quantity. This is because the horizontal bar notation in $\bar X^2$ goes over the $X$ first, then it is squared; whereas in $\overline{X^2}$, the $X$ is squared first, then the bar goes over the entire expression.

Answer 2

Let $\xi = \overline{X}$. We know that $Var[\xi]=E[\xi^2]-(E[\xi])^2$. Now you can add to both sides of this equality $(E[\xi])^2$ and get

$Var[\xi] + (E[\xi])^2 =E[\xi^2]-(E[\xi])^2 + (E[\xi])^2 \Rightarrow Var[\xi] + (E[\xi])^2 = E[\xi^2]$.

So now you just have different means to say the one thing, $E\xi^2 := E[\xi^2]$ and $E[\xi]^2 := E[\xi^2]$. It is just the notation which means "we take random variable take it's square and only after that count it's mathematical expectation". It differs from $(E[\xi])^2$ where we "take random variable, count it's expectation and only after it we square number we got."