When an Expected Value Exists

When Does an Expected Value Exist?

An expected value is a probability-weighted average, but not every random variable has a finite average. For a finite set of outcomes, the calculation is always a finite sum. With infinitely many possible values or a continuous distribution, however, the defining sum or integral may diverge. A signed random variable presents an additional risk: positive and negative terms can appear to cancel even when the expectation is not defined.

The decisive test is absolute integrability. The expected value of \(X\) is finite when the expected absolute value \(E[|X|]\) is finite. This criterion is aligned with the SOA VEE Mathematical Statistics syllabus and applies whether \(X\) is discrete or continuous.

Definition: A random variable \(X\) has a finite expected value if \(E[|X|]<\infty\). For a discrete random variable with possible values \(x_j\) and probabilities \(p_j\), this requires \(\sum_j |x_j|p_j<\infty\). For a continuous random variable with density \(f_X\), it requires \(\int_{-\infty}^{\infty}|x|f_X(x)\,dx<\infty\). When this condition holds, \(E[X]\) is a finite real number.

Positive and Negative Parts

For a signed random variable, define its positive and negative parts by \(X^+=\max(X,0)\) and \(X^-=\max(-X,0)\). Both are nonnegative, and \(X=X^+-X^-\), while \(|X|=X^++X^-\). Their expectations are nonnegative sums or integrals, so they do not involve cancellation.

If both \(E[X^+]\) and \(E[X^-]\) are finite, their difference gives a finite expected value. If exactly one is infinite, the expectation can be described as \(+\infty\) or \(-\infty\), respectively, but it is not finite. If both are infinite, the difference is undefined: subtracting infinity from infinity does not produce an expected value.

E[|X|]=E[X^+]+E[X^-], \qquad E[X]=E[X^+]-E[X^-]\quad\text{when both terms are finite.}

Key result: A finite expected value exists exactly when both the positive and negative contributions are finite, equivalently when \(E[|X|]<\infty\). A symmetric distribution or a symmetric-looking formula does not, by itself, guarantee that the expectation exists.

Why Absolute Integrability Is the Test

The criterion follows directly from the positive-negative decomposition. For a discrete variable, the absolute-value sum is the sum of the positive and negative contributions:

\sum_j |x_j|p_j = \sum_j x_j^+p_j+\sum_j x_j^-p_j.

For a continuous variable, the same identity holds inside the density integral:

\int |x|f_X(x)\,dx = \int x^+f_X(x)\,dx+\int x^-f_X(x)\,dx.

If either right-hand contribution is infinite, the two parts cannot both be combined into a finite difference. In particular, if both diverge, rearranging or pairing terms cannot repair the problem. Absolute convergence is what makes the signed sum or integral well-defined independently of the order of summation or the way the domain is divided.

For a nonnegative random variable, \(X^-=0\), so there is no cancellation issue. Its expectation is finite if the defining sum or integral converges, and is \(+\infty\) if it diverges. The same convergence test can be used for \(E[|X|]\) when \(X\) can be negative.

Worked Examples

Worked Example: A Discrete Distribution With Infinite Mean

A hypothetical claim-count variable \(N\) takes positive integer values with probability \(P(N=n)=1/[n(n+1)]\), for \(n=1,2,\ldots\). Determine whether \(E[N]\) is finite.

Method: first verify the probability model, then test the expected-value series. The probabilities are positive. Using \(1/[n(n+1)]=1/n-1/(n+1)\), their sum is

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

Thus they define a probability distribution. Since \(N\geq 1\), \(|N|=N\), and

E[|N|]=E[N] =\sum_{n=1}^{\infty}n\frac{1}{n(n+1)} =\sum_{n=1}^{\infty}\frac{1}{n+1}.

The partial sum through \(n=M\) is \(\sum_{n=1}^{M}1/(n+1)\), which grows without bound as \(M\) increases; it is a tail of the harmonic series. Therefore \(E[|N|]=\infty\), and \(E[N]=+\infty\). Check: the probabilities sum to one, but the expected-value series does not converge. The common mistake this method avoids is assuming that a valid probability distribution must have a finite mean.

Worked Example: Symmetric Outcomes With No Expected Value

A hypothetical signed fluctuation \(X\) takes values \(+n\) and \(-n\), for every positive integer \(n\), with \(P(X=n)=P(X=-n)=1/[2n(n+1)]\). Determine whether \(E[X]\) exists.

Method: evaluate the positive and negative parts separately. For each \(n\), the combined probability assigned to \(+n\) and \(-n\) is \(1/[n(n+1)]\). Summing over \(n\) gives 1 by the telescoping calculation in the preceding example, so this is a valid distribution.

The positive-part expectation is

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

By the same calculation, the negative-part expectation is

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

Consequently, \(E[|X|]=E[X^+]+E[X^-]=\infty\), and the formal difference \(E[X^+]-E[X^-]\) is \(\infty-\infty\), which is undefined. Pairing \(+n\) with \(-n\) gives zero for each pair, but that rearrangement does not define the expectation. Check: both parts diverge, so the correct conclusion is “undefined,” not zero. The common mistake this method avoids is treating symmetry or paired cancellation as proof that a mean exists.

Worked Example: A Continuous Variable With a Finite Improper Expectation

Let \(U\) have the uniform distribution on \((0,1)\), with density \(f_U(u)=1\) there. Define \(Y=1/\sqrt{U}\). Determine whether \(E[Y]\) is finite and calculate it if it is.

Method: use the density of \(U\) to form the expectation of the function \(1/\sqrt{U}\), then check the improper integral. Since \(Y\) is positive, \(E[|Y|]=E[Y]\). The integral is improper at zero:

E[Y] =\int_0^1 \frac{1}{\sqrt{u}}\,du =\lim_{\varepsilon\downarrow 0}\int_{\varepsilon}^1 u^{-1/2}\,du =\lim_{\varepsilon\downarrow 0}\left[2\sqrt{u}\right]_{\varepsilon}^1 =2.

The limit is finite, so \(E[|Y|]=2<\infty\), and \(E[Y]=2\). Although \(Y\) becomes unbounded as \(U\) approaches zero, this alone does not make its expectation infinite; the rate of growth and the probability density together determine whether the integral converges. Check: the lower-end limit contributes \(2\sqrt{\varepsilon}\), which tends to zero. The common mistake this method avoids is declaring an expectation infinite merely because the integrand is unbounded at an endpoint.

Worked Example: A Continuous Distribution With Infinite Mean

A hypothetical severity variable \(X\) has density \(f_X(x)=1/x^2\) for \(x\geq 1\), and zero otherwise. Determine whether its mean is finite.

Method: verify the density and test the nonnegative expectation integral. The density integrates to one:

\int_1^\infty \frac{1}{x^2}\,dx =\lim_{b\to\infty}\left[-\frac1x\right]_1^b =1.

Because \(X\geq1\), its absolute value is \(X\). Its expected value is

E[|X|]=E[X] =\int_1^\infty x\frac{1}{x^2}\,dx =\lim_{b\to\infty}\int_1^b\frac1x\,dx =\lim_{b\to\infty}\ln b =\infty.

Therefore \(X\) has infinite expectation, not a finite mean. Check: the density is valid, but the integral defining the mean diverges logarithmically. The common mistake this method avoids is confusing a density that integrates to one with a density whose first moment must also be finite.

A Reliable Convergence Check

When a defining sum or integral has infinitely many terms or an unbounded domain, apply the test to the absolute value before trying to simplify a signed expression. For a discrete variable, examine \(\sum_j |x_j|p_j\). For a continuous variable, examine \(\int |x|f_X(x)\,dx\), including any improper endpoints or infinite tails. A finite result establishes a finite expected value. A divergent result means the expectation is not finite; for a nonnegative variable it is \(+\infty\), while for a signed variable the positive and negative parts must be checked separately to determine whether an extended value is meaningful or the expectation is undefined.

Write the model and the expectation.
Identify the possible values and probabilities, or the density and its support, before evaluating a sum or integral.

Test absolute integrability.
Calculate or assess \(E[|X|]\), using an improper limit where needed. Do not rely on cancellation in a signed expression.

State the right conclusion.
If \(E[|X|]\) is finite, \(E[X]\) is finite. If \(X\geq0\) and the integral or sum diverges, \(E[X]=+\infty\). If \(X\) is signed, compare \(E[X^+]\) and \(E[X^-]\); two infinite parts make the expectation undefined.

This distinction matters because later calculations may use an expected value as a finite quantity. Such use is justified only after the relevant expectation is known to exist. A divergent sum cannot be made finite by writing a formal algebraic expression for it, and a symmetric cancellation is not a substitute for absolute convergence.

Check Your Understanding

For each problem, check the absolute-value sum or integral before deciding what the expectation is.

A nonnegative integer-valued variable \(N\) has \(P(N=n)=1/[n(n+1)]\) for \(n=1,2,\ldots\). Is \(E[N]\) finite? (Answer: No; \(E[N]=+\infty\).)
A signed variable \(Z\) has \(P(Z=n)=P(Z=-n)=1/[2n(n+1)]\) for each positive integer \(n\). What are \(E[Z^+]\) and \(E[Z^-]\), and is \(E[Z]\) defined? (Answer: Both are infinite; \(E[Z]\) is undefined.)
Let \(U\) be uniform on \((0,1)\), and set \(W=U^{-2/3}\). Evaluate \(E[W]\) and state whether it is finite. (Answer: \(E[W]=3\), finite.)
A continuous variable \(X\) has density \(f_X(x)=3/x^4\) for \(x\geq1\). For \(T=X^3\), determine whether \(E[T]\) is finite. (Answer: \(E[T]=+\infty\).)
A positive integer-valued variable \(N\) has \(P(N=n)=2^{-n}\) for \(n=1,2,\ldots\). Let \(W=(-1)^n n\) when \(N=n\). Determine whether \(E[W]\) exists and calculate it. (Answer: It exists, and \(E[W]=-2/9\).)

What You'll Learn