Integrability Through Positive and Negative Parts

Why Split a Random Variable into Two Parts?

A real random variable may take both positive and negative values. Adding its positive contributions and subtracting its negative contributions is unambiguous when both totals are finite, and it can also give a well-defined infinite value when exactly one total is infinite. But if both totals are infinite, the proposed subtraction is $\infty-\infty$, which is undefined. The positive and negative parts make these cases precise.

Throughout, $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, so $\mathbb{P}(\Omega)=1$. A real random variable $X$ means an $\mathcal{F}$-measurable function $X:\Omega\to\mathbb{R}$, finite-valued at every point. Equalities of functions in the definitions below hold pointwise. For a nonnegative measurable function $Y:\Omega\to[0,\infty]$, write $\mathbb{E}[Y]=\int_\Omega Y\,d\mathbb{P}$, allowing the value $+\infty$.

Definition: For a real random variable $X$, its positive and negative parts are the nonnegative random variables $$ X^+(\omega)=\max\{X(\omega),0\},\qquad X^-(\omega)=\max\{-X(\omega),0\}. $$ They satisfy $X=X^+-X^-$ and $|X|=X^++X^-$ pointwise. The expectation of $X$ is a finite real number precisely when both $\mathbb{E}[X^+]$ and $\mathbb{E}[X^-]$ are finite; in that case $\mathbb{E}[X]=\mathbb{E}[X^+]-\mathbb{E}[X^-]$. If exactly one of these two nonnegative integrals is infinite, the extended expectation is respectively $+\infty$ or $-\infty$. If both are infinite, the expectation is undefined.

The expectation of a nonnegative measurable function is its Lebesgue integral, so its value is in $[0,\infty]$. We use the standard additivity of the nonnegative Lebesgue integral: for nonnegative measurable $U,V$, $\mathbb{E}[U+V]=\mathbb{E}[U]+\mathbb{E}[V]$, including infinite values. This result does not permit subtracting two infinite integrals. Because $\mathbb{P}$ is a probability measure, it is finite, but that does not force $\mathbb{E}[Y]$ to be finite for an unbounded nonnegative random variable $Y$.

Finite Expectation and Integrability

Proposition: For a real random variable $X$ on $(\Omega,\mathcal{F},\mathbb{P})$, the following are equivalent: (i) $\mathbb{E}[|X|]<\infty$; (ii) both $\mathbb{E}[X^+]$ and $\mathbb{E}[X^-]$ are finite. When these conditions hold, $X$ is called integrable, or $X\in L^1(\mathbb{P})$, and its expectation is the finite real number $\mathbb{E}[X]=\mathbb{E}[X^+]-\mathbb{E}[X^-]$.

Proof. The identities $X^+\ge0$, $X^-\ge0$, and $|X|=X^++X^-$ hold pointwise. Applying additivity of the nonnegative integral gives $\mathbb{E}[|X|]=\mathbb{E}[X^+]+\mathbb{E}[X^-]$. A sum of two nonnegative extended real numbers is finite if and only if each summand is finite. This proves the equivalence. If the two terms are finite, their difference is a well-defined finite real number, which is the definition of $\mathbb{E}[X]$ for an integrable random variable. $\square$

This criterion identifies exactly when the signed Lebesgue integral is finite: it is not enough that positive and negative values appear to cancel. Their magnitudes must have finite total integral. In particular, a real random variable with a finite expectation is integrable; an expectation equal to $+\infty$ or $-\infty$ is not a finite expectation.

Worked Example: A Finite Expectation from a Finite Law

Let $X$ take the values $-3,2,5$ with probabilities $1/4,1/2,1/4$, respectively. Compute its positive and negative parts and its expectation.

At the outcome where $X=-3$, $X^+=0$ and $X^-=3$. Where $X=2$, $X^+=2$ and $X^-=0$; where $X=5$, $X^+=5$ and $X^-=0$. Thus

\mathbb{E}[X^+] =0\cdot\frac14+2\cdot\frac12+5\cdot\frac14 =0+1+\frac54 =\frac94, \qquad \mathbb{E}[X^-] =3\cdot\frac14+0\cdot\frac12+0\cdot\frac14 =\frac34.

Both integrals are finite, so $X$ is integrable and $\mathbb{E}[X]=\frac94-\frac34=\frac64=\frac32$. Check: direct summation over the three values gives $(-3)\cdot\frac14+2\cdot\frac12+5\cdot\frac14=-\frac34+1+\frac54=\frac32$, the same result.

The Three Possible Outcomes for the Parts

For any real random variable, each of $\mathbb{E}[X^+]$ and $\mathbb{E}[X^-]$ lies in $[0,\infty]$. Consequently, there are four cases. If both are finite, the expectation is finite. If the positive-part integral is infinite and the negative-part integral is finite, the extended expectation is $+\infty$. If the positive-part integral is finite and the negative-part integral is infinite, it is $-\infty$. If both integrals are infinite, the expectation is undefined. The last case is not an infinite expectation: no subtraction of the two integrals is defined.

Classification: For a real random variable $X$, the extended expectation is $+\infty$ exactly when $\mathbb{E}[X^+]=\infty$ and $\mathbb{E}[X^-]<\infty$; it is $-\infty$ exactly when $\mathbb{E}[X^+]<\infty$ and $\mathbb{E}[X^-]=\infty$. It is finite exactly when both are finite, and undefined exactly when both are infinite.

Proof. By definition, $X^+$ and $X^-$ are nonnegative, so their integrals are nonnegative extended real numbers. If both integrals are finite, their difference is finite. If the first is infinite and the second finite, subtracting the finite value leaves $+\infty$; if the first is finite and the second infinite, the difference is $-\infty$. If both are infinite, the expression is $\infty-\infty$, which has no defined value. These cases exhaust all possibilities for the pair of nonnegative extended real numbers, proving the classification. $\square$

Worked Example: A Nonnegative Random Variable with Infinite Expectation

Take $\Omega=\mathbb{N}=\{1,2,\ldots\}$, $\mathcal{F}$ the set of all subsets of $\mathbb{N}$, and assign probability $\mathbb{P}(\{n\})=1/[n(n+1)]$. Let $X(n)=n$. Verify that this is a probability law, then determine the expectation.

For each $n\ge1$, $1/[n(n+1)]=1/n-1/(n+1)$. Hence the first $N$ probabilities sum to $\sum_{n=1}^N(1/n-1/(n+1))=1-1/(N+1)$, which tends to $1$. All masses are positive, so they define a probability measure. Since $X(n)=n>0$, the positive part is $X^+(n)=n$ and the negative part is $X^-(n)=0$. The partial sums for the positive-part integral are

\sum_{n=1}^N X^+(n)\mathbb{P}(\{n\}) =\sum_{n=1}^N\frac{n}{n(n+1)} =\sum_{n=1}^N\frac1{n+1}.

These sums diverge. Indeed, for each integer $j\ge1$, the terms $1/k$ with $2^j<k\le2^{j+1}$ include $2^j$ terms, each at least $1/2^{j+1}$; that block therefore sums to at least $1/2$. Infinitely many such blocks force the partial sums to grow without bound. Thus $\mathbb{E}[X^+]=\infty$, while $\mathbb{E}[X^-]=0$, and the classification gives $\mathbb{E}[X]=+\infty$. Check: the finite partial expectation is exactly $\sum_{n=1}^N1/(n+1)$, the same expression shown to diverge, and every term is nonnegative.

Worked Example: Symmetric Cancellation with Undefined Expectation

Let $\Omega=\{-1,1,-2,2,\ldots\}$, with all subsets measurable. Give each of $n$ and $-n$ probability $1/[2n(n+1)]$, for $n\ge1$, and let $X(\omega)=\omega$. Determine whether its expectation is defined.

For each $n$, the two masses at $n$ and $-n$ sum to $1/[n(n+1)]$. The telescoping calculation $\sum_{n=1}^N1/[n(n+1)]=1-1/(N+1)\to1$ shows that the specified masses form a probability measure. At $n>0$, $X^+(n)=n$ and $X^-(n)=0$; at $-n$, $X^+(-n)=0$ and $X^-(-n)=n$. Therefore, for either part, the partial sum over magnitudes $1,\ldots,N$ is

\sum_{n=1}^N n\frac{1}{2n(n+1)} =\frac12\sum_{n=1}^N\frac1{n+1}.

The same block argument as in the preceding example shows that these partial sums tend to $+\infty$. Thus both $\mathbb{E}[X^+]$ and $\mathbb{E}[X^-]$ are infinite, so $\mathbb{E}[X]$ is undefined. Check: if one pairs the terms at $n$ and $-n$ and truncates at $N$, the finite signed sum is $0$ for every $N$. But the separate positive and negative sums both diverge; the paired truncation does not turn $\infty-\infty$ into a defined expectation.

A Common Mistake: Cancelling Before Checking Integrability

A tempting but invalid calculation is to pair positive and negative contributions, obtain zero for every symmetric truncation, and conclude that the expectation is zero. The preceding example gives a concrete counterexample: every symmetric finite truncation sums to zero, but each of the positive and negative parts has infinite expectation. The correct procedure is to calculate $\mathbb{E}[X^+]$ and $\mathbb{E}[X^-]$ separately first. Subtract only when both are finite, or assign an extended value when exactly one is infinite.

For a random variable with a finite range, both parts have finite expectation because each is bounded and $\mathbb{P}(\Omega)=1$. For an unbounded random variable, finiteness must be checked rather than inferred from the fact that the underlying probability measure is finite. The positive-and-negative-part test is what distinguishes a finite expectation from an infinite or undefined one.

Check Your Understanding

For each question, identify the positive and negative parts before deciding what the expectation means.

Prove pointwise that $X=X^+-X^-$ and $|X|=X^++X^-$ for every real-valued random variable $X$.
If $\mathbb{E}[X^+]=7$ and $\mathbb{E}[X^-]=3$, is $X$ integrable, and what is $\mathbb{E}[X]$?
If $\mathbb{E}[X^+]=\infty$ and $\mathbb{E}[X^-]=4$, classify the expectation. Which condition prevents it from being undefined?
On $\mathbb{N}$ with $\mathbb{P}(\{n\})=1/[n(n+1)]$, let $Y(n)=-n$. Compute $\mathbb{E}[Y^+]$ and $\mathbb{E}[Y^-]$, and classify $\mathbb{E}[Y]$.
Why do symmetric finite truncations summing to zero not establish that a random variable with infinite positive and negative parts has expectation zero?

Integrability Through Positive and Negative Parts

What You'll Learn