Cantelli's One-Sided Variance Bound

Why a One-Sided Bound Is Different

Knowing a random variable’s mean and variance limits how much probability can lie far from its mean. Chebyshev’s inequality uses this information to bound the probability of a deviation in either direction. Sometimes the question is one-sided: how likely is an unusually large value, or an unusually small one? Cantelli’s inequality gives a sharper bound for that question, without assuming a particular distribution or symmetry.

The improvement comes from using the direction of the event. A large positive deviation and a large negative deviation cannot both occur on the same outcome, and a one-sided argument can make more effective use of the known mean and variance. The result is also sharp: for every positive threshold and positive variance, some distribution attains the bound exactly.

Statement of Cantelli’s Inequality

Theorem (Cantelli’s inequality): Suppose $X$ has finite mean $\mu=E[X]$ and finite variance $\sigma^2=\operatorname{Var}(X)$. For every $t>0$, $$ P(X-\mu\ge t)\le \frac{\sigma^2}{\sigma^2+t^2}. $$ The corresponding lower-tail bound is $$ P(X-\mu\le -t)\le \frac{\sigma^2}{\sigma^2+t^2}. $$

The upper-tail form bounds the probability that $X$ is at least $t$ above its mean. The lower-tail form is the same statement applied to $-X$, whose mean is $-\mu$ and whose variance is still $\sigma^2$. If $\sigma^2=0$, then $X=\mu$ with probability one, so either event has probability zero when $t>0$; the displayed bounds remain valid.

Equivalently, if $a>\mu$, take $t=a-\mu$ to obtain $P(X\ge a)\le \sigma^2/(\sigma^2+(a-\mu)^2)$. For $a<\mu$, the lower-tail version gives $P(X\le a)\le \sigma^2/(\sigma^2+(\mu-a)^2)$. The threshold must be strictly separated from the mean for these forms; at $t=0$, the bound is merely the trivial upper bound of one.

Deriving the Bound

We prove the upper-tail result. Center the random variable by writing $Z=X-\mu$, so $E[Z]=0$ and $E[Z^2]=\sigma^2$. Fix a number $s\ge 0$. On the event $Z\ge t$, both sides of $Z+s\ge t+s>0$ are nonnegative, and therefore $(Z+s)^2\ge(t+s)^2$. The event indicator is consequently bounded pointwise by a nonnegative random variable:

\mathbf{1}_{\{Z\ge t\}}\le \frac{(Z+s)^2}{(t+s)^2}.

Taking expectations is valid because $Z$ has a finite second moment, and hence $(Z+s)^2$ is integrable. Expanding the square and using $E[Z]=0$ gives, for every $s\ge0$,

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

When $\sigma^2>0$, the right-hand side can be minimized over $s\ge0$. Define $g(s)=(\sigma^2+s^2)/(t+s)^2$. Differentiation yields $g'(s)=2(st-\sigma^2)/(t+s)^3$. Thus $g$ decreases for $s<\sigma^2/t$ and increases for $s>\sigma^2/t$; its minimum occurs at $s=\sigma^2/t$. Substitution gives the desired bound:

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

If $\sigma^2=0$, the centered variable is zero with probability one, so $P(Z\ge t)=0$ for $t>0$. This covers the remaining case and completes the proof. $\square$

Worked Example: Bounding an Upper-Tail Probability

A measurement $X$ has mean 40 and variance 25. What can be guaranteed about the probability that it is at least 50?

The event $X\ge 50$ is a deviation of at least $t=50-40=10$ above the mean. Cantelli’s inequality gives

P(X\ge 50)\le \frac{25}{25+10^2} =\frac{25}{125} =0.20.

Thus the probability is at most 0.20 on the basis of these two moments alone. The same calculation for a value at least 10 below the mean gives $P(X\le30)\le0.20$. These are separate one-sided statements; they do not say that the combined probability of either deviation is at most 0.20.

Comparison With Chebyshev’s Inequality

Chebyshev’s inequality states that, for a finite-variance random variable and $t>0$, $P(|X-\mu|\ge t)\le\sigma^2/t^2$. Since either one-sided event is contained in the two-sided event, this also gives a one-sided bound, usually reported with the probability capped at one. Cantelli uses the same information but directly targets a single tail.

\begin{aligned} \text{Cantelli:}\quad& P(X-\mu\ge t)\le \frac{\sigma^2}{\sigma^2+t^2},\\ \text{Chebyshev applied to one tail:}\quad& P(X-\mu\ge t)\le \min\left(1,\frac{\sigma^2}{t^2}\right). \end{aligned}

For positive variance and $t>0$, Cantelli’s bound is strictly smaller than the Chebyshev bound. If $t^2\ge\sigma^2$, compare the uncapped expressions: $\sigma^2/(\sigma^2+t^2)<\sigma^2/t^2$. If $t^2<\sigma^2$, the Chebyshev bound is one, while $\sigma^2/(\sigma^2+t^2)<1$. The improvement can matter especially when the threshold is not many standard deviations from the mean.

Worked Example: Comparing Bounds for a Uniform Distribution

Let $X$ be uniform on $[0,1]$. Its mean is $1/2$ and its variance is $1/12$. Compare bounds for $P(X\ge0.8)$, and check them against the actual probability.

The deviation above the mean is $t=0.8-0.5=0.3$. Cantelli gives

P(X\ge0.8)\le \frac{1/12}{1/12+(0.3)^2} =\frac{25}{52} \approx0.481.

Chebyshev’s one-sided use gives the bound $(1/12)/(0.3)^2=25/27\approx0.926$, which is weaker. Directly from the uniform distribution, $P(X\ge0.8)=1-0.8=0.2$. Both inequalities are valid, and Cantelli is tighter here, although it is not exact for this particular distribution.

Sharpness and Equality

Cantelli’s bound cannot be improved if all that is known is the mean and variance. To see why, fix $t>0$ and $\sigma^2>0$, and consider a variable $Z$ taking the value $t$ with probability $p=\sigma^2/(\sigma^2+t^2)$, and the value $-\sigma^2/t$ with probability $1-p=t^2/(\sigma^2+t^2)$. Its mean is zero:

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

Its variance is

E[Z^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.

Since the negative support point is below $t$, the event $Z\ge t$ consists exactly of the outcome $Z=t$. Its probability is $p=\sigma^2/(\sigma^2+t^2)$, which equals the Cantelli bound. Shifting $Z$ by any mean $\mu$ gives an equality example for a variable with mean $\mu$ and the same variance.

Worked Example: A Distribution Attaining the Bound

Construct a variable with mean 10 and variance 9 for which the Cantelli bound on $P(X\ge13)$ is exact.

Here $t=13-10=3$ and $\sigma^2=9$. The equality construction assigns probability $9/(9+3^2)=1/2$ to $X=10+3=13$, and the remaining probability $1/2$ to $X=10-9/3=7$. Its mean and variance check directly:

E[X]=\frac{13+7}{2}=10,\qquad \operatorname{Var}(X)=\frac{(13-10)^2+(7-10)^2}{2}=9.

The event $X\ge13$ occurs with probability $1/2$. Cantelli gives $9/(9+3^2)=1/2$, so equality holds. Chebyshev’s bound for the same one-sided event is $9/3^2=1$, which provides no improvement over the trivial probability bound.

Common Pitfall and Practical Use

A frequent mistake is to use Cantelli’s denominator for a two-sided event. The result bounds one tail, such as $P(X-\mu\ge t)$, not $P(|X-\mu|\ge t)$. For the latter event, Chebyshev’s inequality gives the relevant variance bound. Applying the one-sided inequality separately to both tails and adding gives the valid but generally non-sharp bound $P(|X-\mu|\ge t)\le2\sigma^2/(\sigma^2+t^2)$; it does not turn Cantelli’s one-tail formula into a two-sided formula.

Another error is to interpret an upper bound as an estimate of the actual probability. Cantelli is guaranteed by the mean and variance alone, but a specific distribution may have a much smaller tail, as the uniform example shows. When a model’s distribution is known, its tail probability may be calculated directly; the value of Cantelli is that it remains valid even when the distribution is otherwise unspecified.

Key takeaway: For a threshold $t>0$, a finite variance $\sigma^2$ implies that either specified one-sided deviation from the mean has probability at most $\sigma^2/(\sigma^2+t^2)$. The bound is distribution-free and sharp, and it improves on Chebyshev’s one-sided use.

Check Your Understanding

Work from the mean, variance, and event direction in each question.

A variable has mean 12 and variance 16. What upper bound does Cantelli give for $P(X\ge20)$?
State the lower-tail version of Cantelli’s inequality and explain why it follows from the upper-tail version.
For a deviation $t>0$ and positive variance, explain why Cantelli’s bound is strictly smaller than the Chebyshev bound used for one tail.
For mean zero, variance 4, and threshold $t=1$, describe the two-point distribution that attains Cantelli’s bound.
Which inequality directly bounds $P(|X-\mu|\ge t)$, and why is Cantelli’s formula not itself a two-sided bound?