Positivity and Expectation Bounds

Why Positivity Gives Bounds

An expectation averages the values a random variable can take, weighted by their probabilities. If the variable is nonnegative, values on a particular event cannot be offset by negative values elsewhere. This simple observation turns a lower bound on the variable over an event into a lower bound on its expectation.

For example, if \(X\) is nonnegative and \(X\) is at least \(a\) whenever \(X\) reaches a threshold \(a>0\), then the expectation must account for at least \(a\) on that part of the sample space. The resulting inequality is useful for relating the size of an expectation to how often large values occur. Its derivation rests on positivity and monotonicity, not on assumptions about a particular distribution.

Positivity and Monotonicity of Expectation

Definition: A random variable \(X\) is nonnegative almost surely if \(P(X\ge 0)=1\). For such an \(X\), its expectation \(E[X]\) is defined as the integral of \(X\) with respect to probability and may be a finite nonnegative number or \(+\infty\). If \(E[|X|]<\infty\), then \(X\) is integrable and its expectation is finite.

The phrase “almost surely” permits exceptions on events of probability zero; changing a random variable on such an event does not change its expectation. Positivity immediately implies \(E[X]\ge 0\). A more general fact is that pointwise ordering is preserved by expectation.

Theorem: If \(U\) and \(V\) are nonnegative random variables and \(U\le V\) almost surely, then \(E[U]\le E[V]\), with expectations allowed to be infinite. If \(U\) and \(V\) are integrable real-valued random variables and \(U\le V\) almost surely, the same inequality holds for their finite expectations.

Proof. In the nonnegative case, \(V-U\ge 0\) almost surely, so \(E[V-U]\ge 0\). Additivity of the integral for nonnegative random variables gives \(E[V]=E[U]+E[V-U]\), including when one of these expectations is infinite. Therefore \(E[U]\le E[V]\). In the integrable case, \(V-U\) is integrable and \(E[V]-E[U]=E[V-U]\ge 0\). \(\square\)

This order property is the key step in the bounds below: first establish an inequality between random variables, then take expectations. For an event \(A\), its indicator \(\mathbf{1}_A\) equals \(1\) on \(A\) and \(0\) outside \(A\). It is nonnegative, and its expectation is \(E[\mathbf{1}_A]=P(A)\), since it takes the value \(1\) with probability \(P(A)\) and \(0\) otherwise.

A General Bound From an Event

The threshold bound is one instance of a more general statement. Suppose \(X\ge 0\) almost surely, \(A\) is an event, and \(a>0\) is a constant such that \(X\ge a\) almost surely on \(A\). On \(A\), \(a\mathbf{1}_A=a\), while outside \(A\), \(a\mathbf{1}_A=0\le X\). Thus \(a\mathbf{1}_A\le X\) almost surely.

Theorem: If \(X\ge 0\) almost surely and \(X\ge a\) almost surely on an event \(A\), where \(a>0\), then

E[X]\ge aP(A).

This holds in the extended sense if \(E[X]=+\infty\).

Proof. The assumptions give \(a\mathbf{1}_A\le X\) almost surely. Both sides are nonnegative random variables, so monotonicity gives \(E[a\mathbf{1}_A]\le E[X]\). Since \(a\) is a constant and \(E[\mathbf{1}_A]=P(A)\), \(E[a\mathbf{1}_A]=aP(A)\). Combining these statements proves the result. \(\square\)

Take \(A=\{X\ge a\}\). On this event \(X\ge a\), so the theorem gives the threshold bound.

Key bound: For any nonnegative random variable \(X\) and any \(a>0\),

E[X]\ge aP(X\ge a).

No density, mass function, or independence assumption is required.

Worked Examples

Worked Example: A Discrete Distribution

Let \(X\) take the values \(0,2,5\) with probabilities \(0.5,0.3,0.2\), respectively. Check the threshold bound at \(a=2\).

First, the probabilities sum to \(0.5+0.3+0.2=1\), and \(X\ge 0\). Its expectation is

$$ E[X]=0(0.5)+2(0.3)+5(0.2)=0+0.6+1=1.6. $$

The event \(\{X\ge 2\}\) occurs when \(X=2\) or \(X=5\), so

P(X\ge 2)=0.3+0.2=0.5, \qquad 2P(X\ge 2)=2(0.5)=1.

Consequently, \(E[X]=1.6\ge 1=2P(X\ge 2)\), as required. The expectation includes the contribution \(0.6\) from the value \(2\) and \(1\) from the value \(5\); the threshold bound counts at least \(2\) for each outcome in the threshold event, giving a total lower bound of \(1\).

Worked Example: A Uniform Distribution

Let \(X\sim \operatorname{Uniform}(0,4)\), with density \(f(x)=1/4\) for \(0\le x\le4\), and zero otherwise. Set \(a=3\).

The mean is

E[X]=\int_0^4 x\frac14\,dx =\frac14\left[\frac{x^2}{2}\right]_0^4 =\frac14(8)=2.

The threshold probability is the probability of falling in the interval from \(3\) to \(4\):

P(X\ge3)=\int_3^4\frac14\,dx =\frac14(4-3)=\frac14.

Thus \(aP(X\ge a)=3(1/4)=3/4\), and the bound checks as \(2\ge3/4\). The inequality need not be an equality: values of \(X\) above \(3\) contribute more than \(3\) to the expectation, and values below \(3\) contribute additional nonnegative amounts.

Worked Example: An Exponential Distribution

Let \(X\) have an exponential distribution with rate \(2\), so its density is \(f(x)=2e^{-2x}\) for \(x\ge0\). Use the threshold \(a=1\).

Integration by parts gives the mean. Taking \(u=x\) and \(dv=2e^{-2x}\,dx\), so \(du=dx\) and \(v=-e^{-2x}\), yields

E[X]=\int_0^\infty 2xe^{-2x}\,dx =\left[-xe^{-2x}\right]_0^\infty+\int_0^\infty e^{-2x}\,dx =0+\frac12 =\frac12.

The boundary term at infinity is zero because \(xe^{-2x}\to0\). The probability above the threshold is

P(X\ge1)=\int_1^\infty 2e^{-2x}\,dx =\left[-e^{-2x}\right]_1^\infty =e^{-2}.

The lower bound is \(1\cdot e^{-2}\), which is less than \(1/2\) because \(e^2>2\). Therefore \(E[X]=1/2\ge e^{-2}=aP(X\ge a)\), in agreement with the general result.

Worked Example: When the Bound Is Exact

Let \(X=10\) with probability \(0.2\) and \(X=0\) with probability \(0.8\). For \(a=10\), the event \(\{X\ge10\}\) has probability \(0.2\), and

E[X]=10(0.2)+0(0.8)=2, \qquad 10P(X\ge10)=10(0.2)=2.

Here equality holds. On the threshold event, \(X\) is exactly \(10\); off that event, \(X\) is exactly zero. Hence the pointwise comparison \(X\ge10\mathbf{1}_{\{X\ge10\}}\) is an equality for every possible value of \(X\).

Interpreting the Bound Carefully

The general event version is useful whenever a variable is known to be large on some event, even if that event is not described by a threshold of the variable itself. For instance, if \(X\ge 4\) on an event \(A\) with probability \(0.3\), then \(E[X]\ge4(0.3)=1.2\), provided \(X\ge0\) almost surely. This conclusion uses only the lower bound on \(A\); the distribution of \(X\) elsewhere does not need to be specified.

The condition \(X\ge0\) matters. If negative values are possible outside \(A\), they can offset the positive contribution from \(A\), so the same argument does not give \(E[X]\ge aP(A)\) for a signed variable without additional assumptions. Likewise, the constant \(a\) must be positive for the threshold statement as given. A zero threshold gives only the trivial inequality \(E[X]\ge0\); a negative threshold is not the positive lower bound considered here.

Common Mistake

Ignoring the nonnegativity condition. The incorrect reasoning is: “On \(\{X\ge a\}\), \(X\ge a\), so \(E[X]\ge aP(X\ge a)\), even if \(X\) can be negative elsewhere.” The pointwise comparison \(X\ge a\mathbf{1}_{\{X\ge a\}}\) may fail outside the threshold event when \(X<0\). The correction is to require \(X\ge0\) almost surely, as in the theorem, or to establish a different bound that accounts for negative values.
Replacing the event probability with the threshold value. The contribution guaranteed by the event is \(a\) multiplied by the probability of that event, namely \(aP(X\ge a)\), not simply \(a\).
Assuming equality is automatic. The comparison can be strict because \(X\) may exceed \(a\) on the event or be positive off the event. The exponential and uniform examples show that a valid bound need not be tight.

Key takeaway: If \(X\ge0\) almost surely and \(X\ge a\) on an event \(A\), then the expectation is at least \(a\) times the probability of that event. The proof is the pointwise inequality \(a\mathbf{1}_A\le X\), followed by monotonicity of expectation.

Check Your Understanding

Use positivity and monotonicity to answer each question.

If \(X\ge0\) almost surely and \(X\ge5\) on an event \(A\) with \(P(A)=0.12\), what lower bound follows for \(E[X]\)?
For a nonnegative \(X\), explain why \(E[\mathbf{1}_{\{X\ge a\}}]=P(X\ge a)\).
Let \(X\) equal \(0\) with probability \(0.7\) and \(4\) with probability \(0.3\). Check the threshold bound for \(a=4\).
Why does the proof not apply unchanged if \(X\) can be negative outside the event \(A\)?
Give a condition on the values of \(X\) under which the event bound \(E[X]\ge aP(A)\) is an equality.