Positive and Negative Parts of a Random Variable

Why Split a Random Variable Into Two Parts?

A signed random variable can take both positive and negative values. Its expectation must therefore account for contributions on both sides of zero. If the positive and negative contributions are each finite, their difference gives an ordinary finite expectation. If one contribution is infinite and the other finite, the signed expectation has an unambiguous infinite value. But if both contributions are infinite, subtracting them does not define a value. Separating the two contributions makes these cases precise.

This separation is useful even when a random variable is discrete: the positive values can contribute an infinite sum, as can the negative values. A formal expression that appears to balance the two does not by itself establish that the expectation exists. The key quantities are the expectations of the positive and negative parts, each of which is a nonnegative expectation and hence is well-defined, possibly with value \(+\infty\).

Positive and Negative Parts

Definition: Let \(X\) be a real-valued random variable. Its positive part and negative part are the random variables

X^+ = \max(X,0), \qquad X^- = \max(-X,0).

Both are nonnegative and measurable. At every outcome, \(X=X^+-X^-\) and \(|X|=X^++X^-\).

The name “negative part” can seem counterintuitive: \(X^-\) is not negative. It records the magnitude of the part of \(X\) below zero, and so is itself nonnegative. For example, if \(X=-4\) at an outcome, then \(X^+=0\) and \(X^-=4\). If \(X=4\), then \(X^+=4\) and \(X^-=0\). At \(X=0\), both parts are zero.

The formulas follow by considering whether \(X\) is nonnegative or negative. If \(X\geq 0\), then \(X^+=X\) and \(X^-=0\). If \(X<0\), then \(X^+=0\) and \(X^-=-X\). In either case, subtracting the parts gives \(X\), while adding them gives \(|X|\). Also, \(X^+X^-=0\) at every outcome: the two parts cannot both be positive.

Worked Example: Computing the Parts Outcome by Outcome

Suppose \(X\) takes the values \(-3,0,2\), with probabilities \(1/4,1/2,1/4\), respectively. Find the positive and negative parts and their expectations.

When \(X=-3\), \(X^+=0\) and \(X^-=3\). When \(X=0\), both parts are zero. When \(X=2\), \(X^+=2\) and \(X^-=0\). Therefore,

E[X^+] = \frac14(0)+\frac12(0)+\frac14(2)=\frac12, \qquad E[X^-] = \frac14(3)+\frac12(0)+\frac14(0)=\frac34.

Both quantities are finite, so the signed expectation is their difference:

E[X]=E[X^+]-E[X^-]=\frac12-\frac34=-\frac14.

Directly summing the original values gives \(\frac14(-3)+\frac12(0)+\frac14(2)=-\frac34+\frac12=-\frac14\), which agrees.

When Does the Signed Expectation Exist?

For a nonnegative random variable \(Y\), its expectation is defined as an extended nonnegative integral and may be finite or \(+\infty\). Thus \(E[X^+]\) and \(E[X^-]\) are each defined, with values in \([0,+\infty]\). The expectation of \(X\) is formed from these two quantities only when their difference is meaningful. The convention is that a finite quantity may be subtracted from \(+\infty\), but \(+\infty-(+\infty)\) is undefined.

Theorem: For a real-valued random variable \(X\), write \(a=E[X^+]\) and \(b=E[X^-]\). If \(a<\infty\) and \(b<\infty\), then \(E[X]=a-b\) is finite. If \(a=\infty\) and \(b<\infty\), then \(E[X]=+\infty\). If \(a<\infty\) and \(b=\infty\), then \(E[X]=-\infty\). If \(a=b=\infty\), the signed expectation \(E[X]\) is undefined. In particular, \(E[X]\) is a finite real number exactly when \(E[|X|]<\infty\).

Here an expectation of \(+\infty\) or \(-\infty\) is an extended expectation, not a finite mean. Some contexts reserve “the expectation exists” for a finite real number; others allow an extended value when one side is infinite and the other is finite. Stating the value of each part avoids ambiguity.

To justify the finite-expectation criterion, use the pointwise identity \(|X|=X^++X^-\). Additivity of the integral for nonnegative functions gives

$$ E[|X|]=E[X^+]+E[X^-], $$

where the sum is an extended nonnegative sum. This is finite exactly when both terms on the right are finite. The expectation of \(X\) is finite exactly in that case.

Proof of the Classification

The classification can be seen from the definition of the signed integral. Write \(X=X^+-X^-\), where both terms are nonnegative. If \(a\) and \(b\) are finite, the difference of their integrals is an ordinary real number, \(a-b\). If \(a=\infty\) and \(b<\infty\), the positive contribution is infinite while the negative contribution is finite, giving the extended value \(+\infty\). If \(a<\infty\) and \(b=\infty\), the corresponding value is \(-\infty\).

If \(a=b=\infty\), the proposed difference is \(\infty-\infty\), which is not defined. There is no signed integral obtained by subtracting these two infinite quantities. The pointwise identity \(X=X^+-X^-\) remains true; what fails is the attempt to integrate it as a difference of two infinite integrals. This proves the classification. \(\square\)

Worked Example: A Finite Expectation From Two Parts

Let \(Y\) take the value \(3\) with probability \(1/4\), the value \(-2\) with probability \(1/2\), and the value \(0\) with probability \(1/4\). Compute its positive and negative expectations and classify \(E[Y]\).

The positive part is \(3\) only when \(Y=3\), so \(E[Y^+]=\frac14(3)=\frac34\). The negative part is \(2\) only when \(Y=-2\), so \(E[Y^-]=\frac12(2)=1\). Both are finite; therefore,

E[Y]=E[Y^+]-E[Y^-]=\frac34-1=-\frac14.

Checking from the original distribution gives \(\frac14(3)+\frac12(-2)+\frac14(0)=\frac34-1=-\frac14\). The absolute expectation is \(E[|Y|]=\frac34+1=\frac74\), also finite.

Infinite Contributions: Discrete Examples

For a discrete random variable with possible values \(x_j\) and probabilities \(p_j\), the positive and negative expectations are the nonnegative sums

E[X^+]=\sum_j \max(x_j,0)p_j, \qquad E[X^-]=\sum_j \max(-x_j,0)p_j.

These sums can be examined separately, even when the usual signed sum is not valid. In particular, divergence of one sum does not prevent a conclusion if the other sum is finite.

Worked Example: A Positive Infinite Expectation

Define a discrete random variable \(X\) by

P(X=-1)=\frac12, \qquad P(X=n)=\frac{1}{2n(n+1)}, \quad n=1,2,\ldots.

These probabilities sum to one because

\frac12+\sum_{n=1}^{\infty}\frac{1}{2n(n+1)} =\frac12+\frac12\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right) =\frac12+\frac12=1.

The only negative value is \(-1\), so \(E[X^-]=\frac12\). Every positive support value contributes to \(X^+\), giving

E[X^+]=\sum_{n=1}^{\infty}n\frac{1}{2n(n+1)} =\frac12\sum_{n=1}^{\infty}\frac{1}{n+1} =\infty.

The last series diverges because it is a constant multiple of a tail of the harmonic series. Since \(E[X^+]=\infty\) and \(E[X^-]=1/2<\infty\), the classification gives \(E[X]=+\infty\).

Worked Example: Both Parts Have Infinite Expectation

Let \(c=\left(2\sum_{n=1}^{\infty}n^{-2}\right)^{-1}\), a positive finite constant, and define

P(X=n)=P(X=-n)=\frac{c}{n^2}, \quad n=1,2,\ldots.

The probabilities sum to \(2c\sum_{n=1}^{\infty}n^{-2}=1\), so this defines a probability distribution. The positive and negative parts have expectations

E[X^+]=\sum_{n=1}^{\infty}n\frac{c}{n^2} =c\sum_{n=1}^{\infty}\frac1n=\infty, \qquad E[X^-]=\sum_{n=1}^{\infty}n\frac{c}{n^2} =c\sum_{n=1}^{\infty}\frac1n=\infty.

Thus the signed expectation is undefined. The distribution assigns equal probabilities to \(n\) and \(-n\), but that symmetry does not make the expectation a finite value: each of the two nonnegative expectations is infinite.

Common Mistakes and Why the Distinction Matters

Calling \(X^-\) negative. It is nonnegative; it records the magnitude below zero. For \(X=-5\), \(X^-=5\), not \(-5\).
Subtracting infinite sums because the distribution is symmetric. Equal positive and negative contributions do not define a difference when both expectations are infinite. Symmetry alone does not establish a finite expectation.
Using a rearranged signed series without checking its parts. A series of positive and negative terms can be sensitive to how its terms are grouped. For a discrete expectation, first check the nonnegative sums defining \(E[X^+]\) and \(E[X^-]\).
Confusing infinite with undefined. If just one part has infinite expectation and the other is finite, the signed expectation has the corresponding infinite value. It is undefined when both parts have infinite expectation.

The positive-negative decomposition gives a reliable test before a signed expectation is used in a calculation. If \(E[|X|]<\infty\), the expectation is a finite number. If absolute integrability fails, inspecting \(E[X^+]\) and \(E[X^-]\) distinguishes a one-sided infinite expectation from an undefined one. This distinction matters whenever a calculation relies on a mean as an actual numerical quantity.

Key takeaway: Compute the two nonnegative quantities \(E[X^+]\) and \(E[X^-]\) separately. Their difference is a finite expectation when both are finite, an extended infinite value when only one is infinite, and undefined when both are infinite.

Check Your Understanding

For each question, use the positive and negative parts rather than relying on informal cancellation.

If \(X=-6\) at an outcome, what are \(X^+\) and \(X^-\) there? Verify that their difference is \(X\).
Suppose \(E[X^+]=4\) and \(E[X^-]=9\). Is \(E[X]\) finite, infinite, or undefined, and what is its value?
Suppose \(E[X^+]=\infty\) and \(E[X^-]=2\). Classify \(E[X]\).
Why does \(E[X^+]=E[X^-]=\infty\) not imply that \(E[X]=0\), even if the distribution is symmetric?
State the condition involving \(E[|X|]\) that is equivalent to \(X\) having a finite expectation.

Positive and Negative Parts of a Random Variable

What You'll Learn

Why Split a Random Variable Into Two Parts?