Expectation from Densities and the LOTUS Formula

Expectation by Integrating Against a Density

When a random variable \(X\) has a density, the expectation of a function \(g(X)\) can often be calculated directly from that density. There is no need first to derive the probability distribution of the new random variable \(g(X)\). This direct calculation is called the Law of the Unconscious Statistician, or LOTUS. Its practical simplicity depends on a precise condition: the integral must be interpreted correctly, and for a finite signed expectation it must be absolutely integrable.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A real-valued random variable \(X\) has density \(f\) with respect to Lebesgue measure if \(f(x)\ge 0\), \(\int_{\mathbb{R}} f(x)\,dx=1\), and, for every Borel set \(A\subseteq\mathbb{R}\), \(\mathbb{P}(X\in A)=\int_A f(x)\,dx\). A function \(g:\mathbb{R}\to\mathbb{R}\) is measurable if its inverse images of Borel sets are measurable; this ensures \(g(X)\) is a random variable and the integrals below are defined.

The integral is over the real line, but only the support of \(X\), where \(f\) is nonzero, contributes. If \(g\) is nonnegative, an infinite integral means the expectation is \(+\infty\). If \(g\) is signed, absolute integrability is the condition that makes the positive and negative contributions finite and their difference well-defined. In applications, the integral may be evaluated as an ordinary improper integral when the usual integrability conditions justify doing so.

Proof. For a Borel set \(A\), the definition of a density gives \(\mathbb{E}[\mathbf{1}_{\{X\in A\}}]=\mathbb{P}(X\in A)=\int_A f(x)\,dx\). Thus, if \(s=\sum_{k=1}^m a_k\mathbf{1}_{A_k}\) is a nonnegative simple function with disjoint measurable sets \(A_k\) and coefficients \(a_k\ge0\), linearity for a finite sum gives \(\mathbb{E}[s(X)]=\sum_{k=1}^m a_k\mathbb{P}(X\in A_k)=\int_{\mathbb{R}}s(x)f(x)\,dx\). For any nonnegative measurable \(g\), choose nonnegative simple functions \(s_n\) increasing pointwise to \(g\). The monotone convergence theorem states that the integral of an increasing sequence of nonnegative measurable functions converges to the integral of its pointwise limit. Applying it on the probability space to \(s_n(X)\), and on the real line with measure \(f(x)\,dx\) to \(s_n(x)\), gives \(\mathbb{E}[g(X)]=\lim_n\mathbb{E}[s_n(X)]=\lim_n\int s_n(x)f(x)\,dx=\int g(x)f(x)\,dx\), including the possibility \(+\infty\). The monotone convergence theorem is a standard result of Lebesgue integration; its measure-theoretic proof is beyond this tutorial. For signed \(g\), apply the nonnegative result separately to \(g^+\) and \(g^-\). The expectation is the difference of these two expectations when at least one is finite; it is finite exactly when both are finite, equivalently when \(\int |g|f<\infty\). If both are infinite, their difference is not defined. \(\square\)

Using the Formula

The calculation has three basic steps: identify a valid density, substitute the function of interest into the integrand, and evaluate over the support. The integrand is \(g(x)f(x)\), not just \(g(x)\). A useful special case is an indicator: taking \(g(x)=\mathbf{1}_{A}(x)\) gives \(\mathbb{P}(X\in A)=\int_A f(x)\,dx\). Thus event probabilities can be obtained by integrating the density over the event’s values.

Nothing in LOTUS requires \(g\) to be one-to-one or continuous. It can be piecewise-defined or an indicator, provided it is measurable and the expectation is interpreted under the stated integrability conditions. In particular, discontinuities at individual points do not cause difficulty for a density: a single point has probability zero, so changing \(g\) at that point does not change the expectation.

Unbounded Functions and Integrability

The function \(g\) need not be bounded for LOTUS to apply. The integral itself determines whether its expectation is finite. For a nonnegative \(g\), one can first integrate away from a problematic endpoint and then take a limit. If that limit is finite, the expectation is finite; if it diverges, the expectation is infinite. For a signed function, it is not enough for positive and negative areas to appear to cancel in a symmetric calculation. The positive and negative parts must be considered separately.

A Common Mistake: Dropping the Density

For the uniform density in the first example, a wrong calculation is \(\mathbb{E}[(X-1)^2]=\int_0^3(x-1)^2\,dx=3\). This omits the factor \(f(x)=\tfrac13\), so it is not an integral with respect to the probability law of \(X\). The correction is to use the full integrand \((x-1)^2f(x)\): the integral is \(\tfrac13(3)=1\). More generally, a density is a weight in the expectation formula; it cannot be discarded just because the integration limits match the support.

A second warning is that a symmetric improper integral is not a definition of expectation. For example, let \(X\) have the Cauchy density \(f(x)=\tfrac{1}{\pi(1+x^2)}\) on \(\mathbb{R}\). For every \(R>0\), symmetry gives \(\int_{-R}^R x f(x)\,dx=0\). But the positive part has integral \(\int_0^\infty \tfrac{x}{\pi(1+x^2)}\,dx=\lim_{R\to\infty}\tfrac{1}{2\pi}\log(1+R^2)=\infty\), and the negative part has the same infinite integral by symmetry. Thus \(\mathbb{E}[X]\) is undefined: the symmetric value \(0\) results from cancelling two divergent contributions and is not an expectation.

LOTUS matters because many quantities of interest are functions of observed measurements: costs, indicators, losses, or nonlinear scores. Integrating the function against the original density can avoid a separate derivation of the law of \(g(X)\), especially when that transformed law has several branches or atoms. The formula does not remove the need to check measurability and integrability, but it gives a direct route once those conditions are understood.