Generalized Markov Bounds for Moments

From a First Moment to Higher Moments

Markov’s inequality turns a nonnegative random variable’s expectation into an upper bound on the probability that the variable exceeds a positive threshold. The same idea applies after transforming a real-valued random variable: raising its absolute value to a positive power preserves nonnegativity and makes large values count more heavily in the expectation. This gives a family of tail bounds indexed by the moment order.

The order $r$ need not be an integer. A finite first absolute moment gives a bound proportional to $1/t$; a finite second absolute moment gives one proportional to $1/t^2$. These are bounds, not formulas for the tail probability. Their usefulness depends both on the moment available and on the distribution of $X$.

Definition: For a real-valued random variable $X$ and $r>0$, the absolute moment of order $r$ is $E[|X|^r]$, allowing the value $+\infty$. It is finite when $E[|X|^r]<\infty$.

The Generalized Markov Bound

The event of interest is $\{|X|\ge t\}$, where $t>0$. On this event, $|X|^r$ is at least $t^r$. Off the event, the indicator of the event is zero. This comparison is the whole mechanism behind the bound.

Theorem: Let $X$ be a real-valued random variable, let $r>0$, and suppose $E[|X|^r]<\infty$. For every $t>0$, $$ P(|X|\ge t)\le \frac{E[|X|^r]}{t^r}. $$

Proof. The random variable $X$ is measurable, so $|X|^r$ is measurable and $\{|X|\ge t\}$ is an event. Pointwise,

\mathbf{1}_{\{|X|\ge t\}}\le \frac{|X|^r}{t^r}.

Indeed, on $\{|X|\ge t\}$, the right-hand side is at least $1$; on its complement, the left-hand side is zero and the right-hand side is nonnegative. Taking expectations preserves the inequality. The expectation of an event indicator is its probability, and the assumed finite moment makes the right-hand side finite. Therefore

P(|X|\ge t) =E[\mathbf{1}_{\{|X|\ge t\}}] \le \frac{E[|X|^r]}{t^r}.

This is Markov’s inequality applied to the nonnegative random variable $|X|^r$, with threshold $t^r$. $\square$

The denominator matters: $t$ is raised to the same power as the moment. The ratio is dimensionless: if $X$ is measured in some units, then both $E[|X|^r]$ and $t^r$ have those units to the $r$th power. When the displayed ratio exceeds $1$, the result is still correct but can be sharpened trivially by using the fact that a probability is at most $1$:

P(|X|\ge t)\le \min\left\{1,\frac{E[|X|^r]}{t^r}\right\}.

Working With Absolute Moments

Absolute values allow the same result to cover positive, negative, and mixed-sign random variables. No assumption that $X$ is nonnegative, symmetric, or centered is needed. The only moment condition in the theorem is that the chosen absolute moment is finite.

Worked Example: A Uniform Tail Bound

Let $X\sim\operatorname{Uniform}(-1,1)$. Find the bound from the second absolute moment for $P(|X|\ge 0.8)$, then calculate the actual probability for comparison.

The density is $1/2$ on $(-1,1)$, so

E[|X|^2] =\frac12\int_{-1}^{1}x^2\,dx =\frac12\left[\frac{x^3}{3}\right]_{-1}^{1} =\frac13.

Here $r=2$ and $t=0.8=4/5$. The generalized Markov bound is

P(|X|\ge 0.8) \le \frac{1/3}{(4/5)^2} =\frac{1}{3}\cdot\frac{25}{16} =\frac{25}{48} \approx 0.521.

The actual event consists of the intervals $[-1,-0.8]$ and $[0.8,1]$, whose combined length is $0.4$. Since the full interval has length $2$,

P(|X|\ge 0.8)=\frac{0.4}{2}=0.2.

The bound is valid but not exact: it allows a probability of about $0.521$, while the true probability is $0.2$.

Worked Example: An Exponential Tail Bound

Let $X$ have the exponential distribution with rate $1$, with density $e^{-x}$ for $x\ge 0$. Use its second moment to bound $P(|X|\ge 2)$, and compare the bound with the exact tail probability.

Because $X\ge 0$, $|X|=X$. Integration by parts gives

E[X^2]=\int_0^\infty x^2e^{-x}\,dx =\left[-x^2e^{-x}\right]_0^\infty +2\int_0^\infty xe^{-x}\,dx =2.

For the remaining integral, another integration by parts gives $\int_0^\infty xe^{-x}\,dx=1$; the boundary terms vanish because $xe^{-x}$ and $x^2e^{-x}$ tend to zero as $x\to\infty$. Applying the theorem with $r=2$ and $t=2$,

P(|X|\ge 2)\le \frac{2}{2^2}=\frac12.

The exact probability is

P(|X|\ge 2)=P(X\ge 2)=\int_2^\infty e^{-x}\,dx=e^{-2}\approx 0.1353.

Thus the bound $0.5$ exceeds the actual tail probability, as required. It gives a valid limit without needing to calculate the tail integral.

The theorem also applies to fractional moment orders. Such a moment may be available even when a higher-order moment is not, and it is still a valid input to the inequality.

Worked Example: A Fractional Absolute Moment

Again let $X\sim\operatorname{Uniform}(-1,1)$, but now use $r=1/2$ to bound $P(|X|\ge 1/2)$. This illustrates that the order in the theorem need not be a whole number.

Using the density,

E[|X|^{1/2}] =\frac12\int_{-1}^{1}|x|^{1/2}\,dx =\int_0^1 x^{1/2}\,dx =\left[\frac23x^{3/2}\right]_0^1 =\frac23.

Since $t^{1/2}=(1/2)^{1/2}$, the bound is

P(|X|\ge 1/2) \le \frac{2/3}{(1/2)^{1/2}} =\frac{2\sqrt{2}}{3} \approx 0.943.

The exact probability is $1/2$, because the two tail intervals together have length $1$ inside an interval of length $2$. The fractional-moment bound is valid, though not particularly sharp in this case.

Equality and the Role of the Distribution

A moment bound can be exact, but equality requires a special distribution. The pointwise comparison in the proof must be an equality with probability one; otherwise taking expectations leaves a positive gap.

Result: Under the theorem’s conditions, equality holds in $P(|X|\ge t)\le E[|X|^r]/t^r$ if and only if $|X|$ takes values only in $\{0,t\}$ with probability one.

Proof. Put $Y=|X|^r$, $a=t^r$, and $A=\{Y\ge a\}$. The difference between the two sides of the inequality is

\frac{E[Y]}{a}-P(A) =\frac{1}{a}\left(E[Y\mathbf{1}_{A^c}] +E[(Y-a)\mathbf{1}_A]\right).

Both terms inside the parentheses are expectations of nonnegative random variables. Their sum is zero exactly when both are zero. A nonnegative random variable has expectation zero only if it is zero with probability one. Consequently, $Y=0$ almost surely on $A^c$, and $Y=a$ almost surely on $A$. Since $Y=|X|^r$ and $a=t^r$, this says $|X|=0$ or $|X|=t$ almost surely. Conversely, if $|X|$ takes only those two values, the pointwise indicator inequality is an equality almost surely, so the probability bound is an equality. $\square$

Worked Example: Checking Equality in a Discrete Model

Suppose $X$ has probability $1/2$ at $0$, probability $1/4$ at $2$, and probability $1/4$ at $-2$. Use $r=1$ and $t=2$.

The absolute first moment is

$$ E[|X|] =(1/2)(0)+(1/4)(2)+(1/4)(2)=1. $$

The bound and the tail probability are

\frac{E[|X|]}{t}=\frac12, \qquad P(|X|\ge 2)=P(X=2)+P(X=-2)=\frac14+\frac14=\frac12.

Equality holds because $|X|$ takes only the values $0$ and $t=2$, exactly the condition in the result.

What Different Moment Orders Do—and Do Not—Guarantee

The theorem provides a bound for each positive order whose absolute moment is finite. It does not say that increasing $r$ must improve the bound at a fixed threshold. The numerator changes with $r$, as does the denominator, and their ratio depends on the distribution and on $t$.

Worked Example: Comparing Two Moment Bounds

Let $X$ equal $0$ with probability $0.90$, take the values $1$ and $-1$ with probability $0.045$ each, and take $10$ and $-10$ with probability $0.005$ each. At threshold $t=5$, the tail probability is

P(|X|\ge 5)=P(|X|=10)=0.01.

The first absolute moment is

$$ E[|X|]=(0.09)(1)+(0.01)(10)=0.19. $$

It gives the bound $0.19/5=0.038$. The second absolute moment is

$$ E[|X|^2]=(0.09)(1^2)+(0.01)(10^2)=0.09+1=1.09. $$

It gives the bound $1.09/5^2=1.09/25=0.0436$. Both bounds exceed the actual probability $0.01$, but in this example the second-moment bound is larger than the first-moment bound. A higher order is therefore not automatically a tighter bound at every threshold.

The inequality is informative when its right-hand side is meaningfully below $1$. If the chosen absolute moment is infinite, the theorem cannot give a finite bound using that moment. If the ratio is at least $1$, the result remains true but adds no useful restriction beyond the basic probability limit. The moment itself must be computed or justified for the particular distribution; the theorem does not supply it.

Common Mistake

A frequent error is to use a threshold $t$ in the denominator without raising it to the moment order. For example, after applying Markov’s inequality to $|X|^2$, writing $E[|X|^2]/t$ is incorrect. The threshold for the transformed variable is $t^2$, because $|X|\ge t$ implies $|X|^2\ge t^2$. The correct bound is $E[|X|^2]/t^2$.

A second error is to omit the absolute value when $X$ can be negative. The moment $E[X^r]$ may not even be defined for a non-integer $r$ when $X<0$, and signed odd moments can cancel positive and negative values. The generalized bound uses $E[|X|^r]$, which is nonnegative and measures the size of both tails.

Key takeaway: If $r>0$, $t>0$, and $E[|X|^r]$ is finite, then $P(|X|\ge t)\le E[|X|^r]/t^r$. Check the absolute value, the moment condition, and the matching power of the threshold before using the bound.

Check Your Understanding

Use the theorem and the reasoning in the examples to answer these questions.

State the assumptions needed to use the generalized Markov bound at a positive threshold.
If $E[|X|^3]=12$, what upper bound does the theorem give for $P(|X|\ge 2)$?
Why does the theorem use $E[|X|^r]$ rather than $E[X^r]$ for an arbitrary real-valued $X$?
For which values of $|X|$ does equality require the random variable to be concentrated?
Can a second-moment bound be larger than a first-moment bound at the same threshold? Explain briefly.