Intuitive meaning of $E(X^2)$ and $E(X+a)$?

Question

I understand conceptually that $E(X)$ is the average expected value of random variable $X$ over multiple trials over a long period of time i.e. the mean. Similarly, I understand for conditionals that $E(X|Y)$ is the average value of $X$ for all cases where $Y$ has already happened.

However, what is the intuitive meaning behind expressions like $E(X^2)$, and other expectations besides $E(X)$? For instance, the meaning of $E(X+a)$, where $a$ is a constant? I can't seem to grasp the concept of those expressions.

Answer 1

$X^2$ or $X + a$ are both random variables, pretty much the same as $X$. In the first case, you are to compute the mean value of $X^2$ and in the second one, you should compute the mean of $X + a$.

A little example:

Let $X$ be a random variable that takes value $-2$ with probability $\dfrac{1}{2}$, $1$ with probability $\dfrac{1}{3}$ and $3$ with probability $\dfrac{1}{6}$. Hence, $$E(X) = -2\cdot \dfrac{1}{2} + 1 \cdot \dfrac{1}{3} + 3 \cdot \dfrac{1}{6} = -\dfrac{1}{6}$$

Then $X^2$ is $(-2)^2$ with probability $\dfrac{1}{2}$, $1^2$ with probability $\dfrac{1}{3}$ and $3^2$ with probability $\dfrac{1}{6}$.

$$E(X^2) = 4\cdot \dfrac{1}{2} + 1 \cdot \dfrac{1}{3} + 9 \cdot \dfrac{1}{6} = \dfrac{23}{6}$$

As regards to, say, $X+2$, we can merely write down $$E(x+2) = E(x) + 2 =\dfrac{11}{6}$$

Answer 2

• $E(X+a):$

is a mean value of some new random variable, exactly the $X+a$, which has just shifted ALL values of $X$ by $a$. Why is it happening? Just remember what is a formal definition of some r.v. $X$, it's a function that takes values in $\Omega$ and map them into some subset of $\mathbb{R}$, i.e.: $$X:\Omega \to \mathbb{R}.$$

So we just move all that real values by $a$. And hence get a new r.v. with the mean $E(X+a)=E(X)+a.$

• $E(X^2)$

Is similarly a mean of r.v. $X^2$. And how we get that random variable from $X$? We just square all of it ($X$) values, but the probability of each squared value will be the same as in $X$ without squaring. But in can be that $X$, for example, takes values $b$ and $-b$ with probability $p(b)$ and $p(-b)$, then in $X^2$ we will have $b^2$ and $b^2$ (which is the same), so the total probability of $b^2$ in $X^2$ will be $p(b)+p(-b).$