Variance and Expectation Inequality
Which is larger, $E[X^2]$ or $(E[X])^2$? Prove your answer and state precisely when they are equal.
Hints
- Recall the definition of variance: $\text{Var}(X) = E[X^2] - (E[X])^2$. What do you know about the sign of variance?
- Alternatively, consider Jensen's inequality with the convex function $f(x) = x^2$.
- For the equality case: when is $\text{Var}(X) = 0$? A non-negative random variable equals zero in expectation only if it is zero almost surely.
Worked Solution
How to Think About It: This is one of the most fundamental inequalities in probability. The answer comes straight from the definition of variance: $\text{Var}(X) = E[X^2] - (E[X])^2$. Since variance is always non-negative, $E[X^2]$ must be at least as large as $(E[X])^2$. You can also see this as a special case of Jensen's inequality applied to the convex function $f(x) = x^2$.
Approach: Use the non-negativity of variance and characterize the equality case.
Formal Solution:
By definition of variance: $$\text{Var}(X) = E[X^2] - (E[X])^2$$
Since variance is non-negative for any random variable (it is the expected squared deviation from the mean): $$\text{Var}(X) \geq 0$$
Therefore: $$E[X^2] \geq (E[X])^2$$
Alternative proof via Jensen's inequality: The function $f(x) = x^2$ is convex (its second derivative is $2 > 0$). Jensen's inequality states that for any convex function $f$ and random variable $X$: $$E[f(X)] \geq f(E[X])$$
Applying this with $f(x) = x^2$: $$E[X^2] \geq (E[X])^2$$
Equality condition: $E[X^2] = (E[X])^2$ if and only if $\text{Var}(X) = 0$, which holds if and only if $X$ is constant almost surely -- meaning $P(X = c) = 1$ for some constant $c$.
To see why: $\text{Var}(X) = E[(X - \mu)^2] = 0$ requires $(X - \mu)^2 = 0$ almost surely (since a non-negative random variable has zero expectation only if it is zero a.s.), which means $X = \mu$ almost surely.
Answer: $E[X^2] \geq (E[X])^2$ always, with equality if and only if $X$ is constant almost surely.
Intuition
This inequality is the probabilistic version of "the square of the average is less than the average of the squares," and it captures a fundamental property of spread. Any time a random variable has uncertainty -- any non-trivial distribution -- the second moment exceeds the square of the first moment by exactly the variance. This shows up everywhere in quantitative finance: the expected value of a squared return (which drives P&L of gamma positions) always exceeds the square of the expected return. This is why owning convexity (being long gamma) has positive expected value in the presence of uncertainty -- you benefit from the gap between $E[X^2]$ and $(E[X])^2$. Jensen's inequality is the general version of this principle and is one of the most useful tools in a quant's toolkit.