Cauchy–Schwarz Inequality for Random Variables¶

Theorem¶

For random variables $X$ and $Y$ with $\mathbb{E}[X^2]<\infty$ and $\mathbb{E}[Y^2]<\infty$ (which also makes $XY$ integrable),

If either second moment is zero, equality holds trivially.

For centred variables, the second moments are their Variance and Standard Deviation quantities: applying the inequality to $X-\mathbb{E}X$ and $Y-\mathbb{E}Y$ gives $|\operatorname{Cov}(X,Y)|\le\sigma_X\sigma_Y$${\sigma }_X{\sigma }_Y$.

Interpretation¶

The average product $\mathbb{E}[XY]$ of two variables with finite second moments can never exceed the product of their root-mean-square sizes $\sqrt{\mathbb{E}[X^2]}$ and $\sqrt{\mathbb{E}[Y^2]}$. Applied to centred variables, this bounds covariance and keeps correlation between $-1$ and $1$.

The figure draws $X$ and $Y$ as arrows of lengths $\sqrt{\mathbb{E}[X^2]}$ and $\sqrt{\mathbb{E}[Y^2]}$, with $\mathbb{E}[XY]$ in the role of the inner product. When $\mathbb{E}[X^2]>0$$0$, the shadow of $Y$ on $X$ has length $|\mathbb{E}[XY]|/\sqrt{\mathbb{E}[X^2]}$, and a shadow is never longer than the arrow itself; squaring gives the inequality. The inequality is also what makes $\cos\theta=\mathbb{E}[XY]/\bigl(\sqrt{\mathbb{E}[X^2]}\sqrt{\mathbb{E}[Y^2]}\bigr)$ a number in $[-1,1]$, so that $\theta$ is a genuine angle. Equality means the arrows lie on one line, the Key Result $Y=cX$ almost surely.

This is the Cauchy–Schwarz inequality from Math II in a new setting. If the sample space has $m$ equally likely outcomes, $\mathbb{E}[XY]$ is $\tfrac{1}{m}$ times the dot product of the vectors of values of $X$ and $Y$, so the two statements coincide. Taking $X-\mathbb{E}X$ and $Y-\mathbb{E}Y$ gives $|\operatorname{Cov}(X,Y)|\le\sigma_X\sigma_Y$${\sigma }_X{\sigma }_Y$, and when $\operatorname{Var}(X)>0$ equality holds only for an exact linear relation $Y=a+bX$ almost surely.

A tempting but false version is $\mathbb{E}[XY]\le\mathbb{E}X\cdot\mathbb{E}Y$$\mathbb{E}Y$. For $Y=X$ it would say $\mathbb{E}[X^2]\le(\mathbb{E}X)^2$$(\mathbb{E}X{)}^2$, which fails for every $X$ with positive variance; the correct bound uses second moments on the right.

Key Results¶

• $|\operatorname{Cov}(X,Y)|\le\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}$$\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}$.
• If both second moments are positive, equality holds exactly when $Y=cX$ almost surely for some real $c$.

Examples¶

• If $\mathbb{E}[X^2]=4$ and $\mathbb{E}[Y^2]=9$$9$, then $|\mathbb{E}[XY]|\le6$$6$.
• Taking $Y=1$ gives $(\mathbb{E}X)^2\le\mathbb{E}[X^2]$$\mathbb{E}[X^2]$; for a fair die this is $12.25\le91/6$$91/6$.

Prerequisites¶

• Variance and Standard Deviation

Related Concepts¶

• Covariance
• Correlation
• Cauchy-Schwarz Inequality

Sources¶

• Nicky van Foreest · Probability Distributions · Week 7

Symbols: infinity, sigma, square root, theta

Last updated: 20 October 2026 03:02