Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Expected value ▷ Non-negativity

Theorem: If a random variable is strictly non-negative, its expected value is also non-negative, i.e.

$\label{eq:mean-nonneg} \mathrm{E}(X) \geq 0, \quad \text{if} \quad X \geq 0 \; .$

Proof:

1) If $X \geq 0$ is a discrete random variable, then, because the probability mass function is always non-negative, all the addends in

$\label{eq:mean-disc} \mathrm{E}(X) = \sum_{x \in \mathcal{X}} x \cdot f_X(x)$

are non-negative, thus the entire sum must be non-negative.

2) If $X \geq 0$ is a continuous random variable, then, because the probability density function is always non-negative, the integrand in

$\label{eq:mean-cont} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x$

is strictly non-negative, thus the term on the right-hand side is a Lebesgue integral, so that the result on the left-hand side must be non-negative.

Sources:

Metadata: ID: P52 | shortcut: mean-nonneg | author: JoramSoch | date: 2020-02-13, 20:14.