Linearity of expectation

For any r.v. $X$ and $Y$, the following are true:

• $\mathbb{E}(X+Y)=\mathbb{E}(X)+\mathbb{E}(Y)$
• $\mathbb{E}(cX)=c\cdot \mathbb{E}(X)$ for any constant $c$.

This ties back to us “doubling” r.v. $X$. Multiplying by two doesn’t change the probabilities, but the values that $X$ supports. This should intuitively also update $\mathbb{E}(X)$ when $c=2$.

Derived properties

Warning

These properties only hold for linear functions of $X$ and $Y$. For example, they would not hold for $\mathbb{E}(Z)$ where $Z=X^2$. In general, this is because

$$\mathbb{E}(X^2)\ge (\mathbb{E}(X){)}^2$$

so terms can’t just be “separated out” per se. Furthermore, for a nonlinear function $g$ it is not generally true that

$$\mathbb{E}(g(X))=g(\mathbb{E}(X))$$

For any constant $c$, we have $\mathbb{E}(X+c)=\mathbb{E}(X)+c$ for any constant $c$. This makes sense because we’re saying a r.v. $Y=c$ always.

If $X$ uses hours and we want to express as minutes, these are equivalent: $\mathbb{E}(60X)=60\cdot \mathbb{E}(X)$. This makes intuitive sense, we’re simply switching units.

Combining these properties, we also have $\mathbb{E}(aX+bY+c)=a\cdot \mathbb{E}(X)+b\cdot \mathbb{E}(Y)+c$.