Cantelli's One-Sided Chebyshev Inequality

Why a One-Sided Bound Is Useful

Knowing only a random variable’s mean and variance does not determine its tail probabilities. It does, however, constrain how much probability can lie far from the mean. Chebyshev’s inequality controls deviations in either direction; when the question concerns only an upper or lower tail, Cantelli’s inequality gives a sharper variance-only bound.

This distinction matters in settings where an unusually large value is the concern—for example, a measurement exceeding a safety threshold. The bound requires no normality or other distributional form. Its strength comes from using only the available mean and variance information while accounting for the direction of the event.

Definition: Let $X$ be a random variable with finite mean $\mu=E[X]$ and finite variance $\sigma^2=\operatorname{Var}(X)$. For any $t>0$, the event $X-\mu\ge t$ is an upper-tail deviation of at least $t$. Its probability is bounded by Cantelli’s inequality.

Statement and Derivation

Theorem (Cantelli’s inequality): If $E[X]=\mu$ and $\operatorname{Var}(X)=\sigma^2<\infty$, then for every $t>0$, $$ P(X-\mu\ge t)\le \frac{\sigma^2}{\sigma^2+t^2}. $$

The proof uses Markov’s inequality: if $W$ is nonnegative with finite expectation and $a>0$, then $P(W\ge a)\le E[W]/a$. Write $Y=X-\mu$, so $E[Y]=0$ and $E[Y^2]=\sigma^2$. For any $c\ge0$, on the event $Y\ge t$, we have $Y+c\ge t+c>0$, and therefore $(Y+c)^2\ge(t+c)^2$. Applying Markov’s inequality to the nonnegative random variable $(Y+c)^2$ gives the following bound for every such $c$:

P(Y\ge t) \le \frac{E[(Y+c)^2]}{(t+c)^2} = \frac{\sigma^2+c^2}{(t+c)^2}.

The equality in the numerator follows by expanding the square: $E[(Y+c)^2]=E[Y^2]+2cE[Y]+c^2=\sigma^2+c^2$. We now choose $c$ to make the bound as small as possible. For $\sigma^2>0$, define $q(c)=(\sigma^2+c^2)/(t+c)^2$ for $c\ge0$. Differentiating,

q'(c)=\frac{2(ct-\sigma^2)}{(t+c)^3}.

Since $t>0$, the derivative is negative when $c<\sigma^2/t$ and positive when $c>\sigma^2/t$. Thus the minimum occurs at $c=\sigma^2/t$. Substitution yields

q\left(\frac{\sigma^2}{t}\right) = \frac{\sigma^2+\sigma^4/t^2}{(t+\sigma^2/t)^2} = \frac{\sigma^2(t^2+\sigma^2)/t^2}{(t^2+\sigma^2)^2/t^2} = \frac{\sigma^2}{\sigma^2+t^2}.

If $\sigma^2=0$, then $E[(X-\mu)^2]=0$, so $X=\mu$ almost surely and $P(X-\mu\ge t)=0$. The stated inequality holds in that case as well. This proves Cantelli’s inequality for all finite $\sigma^2\ge0$ and $t>0$. $\square$

The optimizing shift $c=\sigma^2/t$ is not an extra assumption about $X$. It is a choice made in the proof using the known variance and the selected threshold. A larger threshold reduces the bound; a larger variance increases it. If $t$ and $\sigma$ have the same units as $X$, the ratio $\sigma^2/(\sigma^2+t^2)$ is dimensionless, as a probability bound must be.

Worked Applications

Worked Example: An Upper-Tail Bound for a Measurement

A clinical measurement has mean $50$ units and variance $16$ square units. Without assuming any particular distribution, bound the probability that one measurement is at least $10$ units above its mean.

Here $\mu=50$, $\sigma^2=16$, and the deviation is $t=10$. The event is $X-\mu\ge10$, or equivalently $X\ge60$. Cantelli’s inequality gives

P(X\ge60) \le \frac{16}{16+10^2} =\frac{16}{116} =\frac{4}{29} \approx 0.138.

Thus, based on the mean and variance alone, the probability is at most about $13.8\%$. This is a bound, not an estimate of the actual probability: the distribution might place much less probability above $60$.

Worked Example: A One-Sided Bound for a Sample Mean

Suppose $X_1,\ldots,X_{25}$ are independent observations with common mean $120$ and common variance $100$. Find a variance-only upper bound for the probability that their sample mean is at least $4$ above $120$.

Let $\bar X=25^{-1}\sum_{i=1}^{25}X_i$. Independence and the common finite variance imply

E[\bar X]=120, \qquad \operatorname{Var}(\bar X) =\frac{1}{25^2}\sum_{i=1}^{25}\operatorname{Var}(X_i) =\frac{25(100)}{625} =4.

The requested event is $\bar X-E[\bar X]\ge4$. Applying Cantelli’s inequality to $\bar X$, with variance $4$ and $t=4$, gives

P(\bar X\ge124) \le \frac{4}{4+4^2} =\frac{4}{20} =0.20.

The independence assumption is used to calculate the variance of the average. Once that variance is known, Cantelli’s inequality itself does not require the observations to have any particular distributional shape.

Worked Example: A Lower-Tail Question

A variable $X$ has mean $80$ and variance $25$. Bound the probability that $X\le72$.

Cantelli’s stated form is an upper-tail bound. Apply it instead to $Z=-X$. Then $E[Z]=-80$, $\operatorname{Var}(Z)=25$, and $X\le72$ is equivalent to $Z-(-80)\ge8$. Therefore,

P(X\le72) =P(Z-E[Z]\ge8) \le\frac{25}{25+8^2} =\frac{25}{89} \approx0.281.

The sign change converts a lower-tail event into an upper-tail event without changing the variance. The result is a bound of about $28.1\%$, not a claim that this probability equals $28.1\%$.

Sharpness: The Bound Can Be Attained

Cantelli’s inequality is sharp: with only a mean and variance specified, one cannot generally replace its right-hand side by a smaller universal bound. The following two-point distribution attains equality for any chosen $t>0$ and $\sigma^2>0$.

Set $Y=X-\mu$, and assign probability $p=\sigma^2/(\sigma^2+t^2)$ to $Y=t$, and probability $1-p=t^2/(\sigma^2+t^2)$ to $Y=-\sigma^2/t$. Its mean is

E[Y] = \frac{\sigma^2}{\sigma^2+t^2}t + \frac{t^2}{\sigma^2+t^2}\left(-\frac{\sigma^2}{t}\right) =0.

Its second moment, and hence its variance because its mean is zero, is

E[Y^2] = \frac{\sigma^2t^2}{\sigma^2+t^2} + \frac{t^2\sigma^4/t^2}{\sigma^2+t^2} = \frac{\sigma^2t^2+\sigma^4}{\sigma^2+t^2} =\sigma^2.

The upper-tail event $Y\ge t$ occurs exactly when $Y=t$, so its probability is $p=\sigma^2/(\sigma^2+t^2)$, exactly the Cantelli bound. For a numerical instance, take $\mu=50$, $\sigma^2=36$, and $t=6$. Then $X$ equals $56$ or $44$, each with probability $1/2$. Its mean is $50$, its variance is $36$, and $P(X-50\ge6)=1/2=36/(36+36)$.

Interpreting the Bound and Avoiding Common Errors

Cantelli’s inequality is useful when the distribution is unknown or when a conservative guarantee is needed from moment information alone. It can be applied to a single random variable or to a statistic—such as a sample mean—once that statistic’s mean and finite variance have been established. It may be conservative for a particular distribution, but the equality example shows why no uniformly stronger bound based only on those two moments is possible.

Do not confuse a bound with a probability calculation. The inequality says the probability cannot exceed the displayed number; it does not say that the probability equals that number.
Do not omit the direction of the event. The theorem bounds $P(X-\mu\ge t)$. For $P(X-\mu\le-t)$, apply the result to $-X$.
Do not use a nonpositive deviation threshold in this formula. The statement requires $t>0$. For $t=0$, the event includes $X=\mu$, and the formula does not provide a useful version of the stated bound.
Do not substitute the standard deviation for the variance. The numerator is $\sigma^2$, and the denominator is $\sigma^2+t^2$; both quantities are squared units.

For comparison, the two-sided Chebyshev bound controls $P(|X-\mu|\ge t)$, which includes both tails. Cantelli targets just one direction. Its denominator includes $t^2$ in addition to the variance, producing a bound that can be smaller for an upper-tail question. Neither inequality identifies the true tail probability without more information about the distribution.

Key takeaway: For any finite-variance $X$ and $t>0$, $P(X-E[X]\ge t)\le\operatorname{Var}(X)/(\operatorname{Var}(X)+t^2)$. The result is distribution-free, applies to a single tail, and is sharp.

Check Your Understanding

Use the hypotheses and direction of Cantelli’s inequality carefully in each question.

A variable has mean $30$ and variance $9$. What upper bound does Cantelli’s inequality give for the probability it is at least $6$ above its mean?
Why must the threshold $t$ be positive in the stated theorem?
If $X$ has mean $\mu$ and variance $\sigma^2$, which variable would you apply Cantelli’s inequality to in order to bound $P(X-\mu\le-t)$?
For the two-point construction in the sharpness argument, verify that the probability assigned to $Y=t$ equals the Cantelli bound.
A sample mean is based on independent observations with common variance $v$. What is its variance for a sample of size $n$, and what additional quantity is needed to apply Cantelli’s inequality to a specified deviation?