Paley-Zygmund Lower Bounds from Two Moments

Why Lower Bounds Need More Than Markov’s Inequality

Markov’s inequality uses a nonnegative random variable’s mean to bound the probability of a large value from above. A different question is whether the mean and second moment can guarantee that the variable exceeds some fraction of its mean with at least a certain probability. The Paley–Zygmund inequality answers this with a lower bound. It is useful when an upper-tail estimate alone does not describe how much probability must remain away from zero.

The bound depends on the ratio of the squared mean to the second moment. If most of the second moment is not excessively large relative to the mean, the variable cannot put nearly all its probability close to zero while still retaining that mean. The inequality makes this idea precise, though its guaranteed probability can be conservative.

Definition: Let $X$ be a random variable with $E[X^2]<\infty$ and positive mean $\mu=E[X]>0$. For $0\leq\theta<1$, the Paley–Zygmund lower bound is $$ P(X>\theta\mu)\geq \frac{(1-\theta)^2\mu^2}{E[X^2]}. $$ The usual statement assumes $X\geq0$ almost surely. In fact, nonnegativity is sufficient but not necessary: the displayed inequality also holds for any square-integrable $X$ with positive mean when $0\leq\theta<1$.

The second moment in the denominator is $E[X^2]$, not $\operatorname{Var}(X)$. Since $E[X^2]=\operatorname{Var}(X)+\mu^2$, the bound can also be written $$ P(X>\theta\mu)\geq \frac{(1-\theta)^2}{1+\operatorname{Var}(X)/\mu^2}. $$ This form highlights that the lower bound is large when variability is small relative to the squared mean.

Proof of the Paley–Zygmund Inequality

The argument isolates the part of the mean contributed by values above the threshold. Write $A=\{X>\theta\mu\}$. On the complementary event, $X\leq\theta\mu$, so the contribution to the mean there cannot exceed $\theta\mu$. Cauchy–Schwarz then connects the remaining contribution to the probability of $A$.

Theorem (Paley–Zygmund inequality): If $E[X^2]<\infty$, $E[X]=\mu>0$, and $0\leq\theta<1$, then $$ P(X>\theta\mu)\geq \frac{(1-\theta)^2\mu^2}{E[X^2]}. $$ In particular, the conclusion holds if $X\geq0$ almost surely.

Proof. Square integrability implies integrability: by Cauchy–Schwarz, $E[|X|]\leq\sqrt{E[X^2]}<\infty$, so the expectations below are defined. Let $A=\{X>\theta\mu\}$. Because $\theta\geq0$ and $\mu>0$, the threshold $\theta\mu$ is nonnegative. On $A^c$, $X\leq\theta\mu$, and therefore

E[X\mathbf{1}_{A^c}] \leq \theta\mu P(A^c) \leq \theta\mu.

This inequality does not require $X$ to be nonnegative: it follows by integrating the pointwise upper bound $X\leq\theta\mu$ on $A^c$. Decomposing the expectation across $A$ and $A^c$ gives

E[X\mathbf{1}_A] =\mu-E[X\mathbf{1}_{A^c}] \geq (1-\theta)\mu.

In particular, this expectation is positive. Applying Cauchy–Schwarz to $X$ and $\mathbf{1}_A$ yields

\big(E[X\mathbf{1}_A]\big)^2 \leq E[X^2]E[\mathbf{1}_A^2] =E[X^2]P(A).

Combining the last two inequalities and dividing by $E[X^2]>0$ proves the result:

P(A)\geq \frac{\big(E[X\mathbf{1}_A]\big)^2}{E[X^2]} \geq \frac{(1-\theta)^2\mu^2}{E[X^2]}.

Thus $P(X>\theta\mu)$ satisfies the claimed lower bound. $\square$

A Fixed-Threshold Version

The fraction-of-the-mean form can be expressed using a threshold $t$. The restriction $t\geq0$ matters: the proof bounds the contribution to the mean on $\{X\leq t\}$ by $t$, which gives the needed upper bound only when $t$ is nonnegative.

Corollary: If $E[X^2]<\infty$, $E[X]=\mu>0$, and $0\leq t<\mu$, then $$ P(X>t)\geq\frac{(\mu-t)^2}{E[X^2]}. $$

To prove the corollary, take $A=\{X>t\}$. On $A^c$, $X\leq t$, so $E[X\mathbf{1}_{A^c}]\leq tP(A^c)\leq t$. Hence $E[X\mathbf{1}_A]\geq\mu-t>0$. Cauchy–Schwarz gives $\big(E[X\mathbf{1}_A]\big)^2\leq E[X^2]P(A)$. Combining these inequalities gives the stated bound.

Worked Applications

Worked Example: A Bernoulli Random Variable

Let $X\sim\operatorname{Bernoulli}(p)$, where $0<p\leq1$. Then $X$ is nonnegative, $E[X]=p$, and $E[X^2]=p$ because $X^2=X$. For any $0\leq\theta<1$, the threshold $\theta p$ is less than 1. The event $X>\theta p$ occurs exactly when $X=1$, so its probability is $p$. Paley–Zygmund gives

P(X>\theta p) \geq \frac{(1-\theta)^2p^2}{p} =(1-\theta)^2p.

This is valid but not sharp: the exact probability $p$ is at least $(1-\theta)^2p$, since $0\leq1-\theta\leq1$. For instance, at $\theta=1/2$, the inequality guarantees at least $p/4$, while the exact probability is $p$.

Worked Example: An Exponential Random Variable

Let $X$ have an exponential distribution with rate 2. Its density is $f(x)=2e^{-2x}$ for $x\geq0$, and its first two moments are $E[X]=1/2$ and $E[X^2]=2/2^2=1/2$. At $\theta=1/2$, the threshold is $\theta E[X]=1/4$. The bound is

P(X>1/4) \geq \frac{(1-1/2)^2(1/2)^2}{1/2} =\frac18.

For comparison, the exact probability from the exponential survival function is $P(X>1/4)=e^{-2(1/4)}=e^{-1/2}$, approximately $0.6065$. The guaranteed lower bound $0.125$ is smaller, as a lower bound need not equal the actual tail probability.

Worked Example: A Binomial Count

Let $S\sim\operatorname{Binomial}(10,0.2)$, the number of successes in ten independent trials with success probability $0.2$. Its mean is $E[S]=10(0.2)=2$, and its variance is $10(0.2)(0.8)=1.6$. Thus $E[S^2]=1.6+2^2=5.6$. Taking $\theta=1/2$ gives the threshold 1, so

P(S>1)\geq \frac{(1-1/2)^2(2)^2}{5.6} =\frac{1}{5.6} \approx 0.1786.

Because $S$ is integer-valued, $S>1$ means $S\geq2$. The exact probability is

P(S\geq2) =1-P(S=0)-P(S=1) =1-(0.8)^{10}-10(0.2)(0.8)^9 \approx 0.6242.

The exact tail exceeds the guarantee, and both calculations use the strict event $S>1$, not $S\geq1$.

Worked Example: The Inequality Can Hold Without Nonnegativity

Suppose $X=-1$ with probability $1/4$ and $X=3$ with probability $3/4$. Then

E[X]=(-1)(1/4)+3(3/4)=2,\qquad E[X^2]=1(1/4)+9(3/4)=7.

For $\theta=1/2$, the threshold is 1. The event $X>1$ occurs exactly when $X=3$, so its probability is $3/4$. The inequality guarantees

P(X>1)\geq\frac{(1-1/2)^2(2)^2}{7} =\frac17.

Indeed, $3/4\geq1/7$. This example illustrates the more general theorem: a variable may take negative values and still satisfy the bound when its mean is positive and the threshold fraction lies in the stated range.

Common Pitfalls and Interpretation

A frequent mistake is to say that Paley–Zygmund applies to a variable that can take negative values only after shifting it to make it nonnegative. Nonnegativity is a standard sufficient hypothesis, but it is not necessary for this inequality. With $E[X]>0$ and $0\leq\theta<1$, the proof remains valid because the event complement satisfies $X\leq\theta E[X]$; the upper bound on its contribution follows even when some values of $X$ are negative.

A different mistake is to use the fixed-threshold bound for $t<0$. For example, if $X=1$ almost surely and $t=-1$, then $P(X>t)=1$, but the proposed expression $(E[X]-t)^2/E[X^2]$ equals $4$. A probability cannot be at least 4. The argument for the threshold form requires $t\geq0$. If $X\geq0$ and $t<0$, then $P(X>t)=1$, but the displayed fixed-threshold formula is not thereby justified.

Finally, the bound can be weak when $E[X^2]$ is large compared with $E[X]^2$. In the variance form, this occurs when $\operatorname{Var}(X)/E[X]^2$ is large. The inequality then guarantees only a small probability above the selected fraction of the mean. Its role is a rigorous guarantee from two moments, not a claim that the tail probability is accurately approximated.

Key takeaway: For a square-integrable random variable with positive mean, the probability of exceeding a nonnegative fraction $\theta<1$ of that mean is at least $(1-\theta)^2E[X]^2/E[X^2]$. The standard nonnegative-variable assumption is sufficient, and the fixed-threshold version requires $0\leq t<E[X]$.

Check Your Understanding

Work through these questions using the conditions and proof above.

State the moment and threshold assumptions in the Paley–Zygmund inequality.
If $E[X]=3$, $E[X^2]=15$, and $\theta=1/3$, what lower bound does the inequality give for $P(X>1)$?
Why does the proof allow $X$ to take negative values when $E[X]>0$ and $0\leq\theta<1$?
For $X\geq0$, explain why the fixed-threshold result as stated does not cover $t=-2$.
How does the lower bound change when $\operatorname{Var}(X)/E[X]^2$ increases while $\theta$ stays fixed?