LOTUS for Continuous Random Variables

Why Integrate the Function Against the Density?

A random variable $X$ may have a simple density even when the quantity of interest is a nonlinear function of it. For example, if $X$ is a waiting time, we might want the expected discount factor $e^{-X}$; if $X$ is a measurement error, we might want the expected squared error $X^2$. In each case, the target is $E[g(X)]$, not necessarily $E[X]$.

For a continuous random variable, the Law of the Unconscious Statistician (LOTUS) gives a direct calculation: integrate $g(x)$ against the density of $X$. This avoids first finding the distribution of $g(X)$. The integral must still use the support of $X$, and it must exist in the relevant sense.

Definition: Suppose $X$ has density $f$ with respect to Lebesgue measure on $\mathbb{R}$, and $g:\mathbb{R}\to\mathbb{R}$ is Borel measurable. For nonnegative $g$, its expectation is allowed to be $+\infty$, and $$ E[g(X)]=\int_{\mathbb{R}} g(x)f(x)\,dx. $$ For signed $g$, if $$ \int_{\mathbb{R}} |g(x)|f(x)\,dx<\infty, $$ then $E[g(X)]$ is finite and is given by the same formula. If both the positive and negative parts have infinite expectations, the expectation is undefined.

The function $g$ must be measurable so that $g(X)$ is a random variable and the integral is defined. The absolute-integrability condition is the standard guarantee of a finite expectation for signed functions. For nonnegative $g$, no finiteness assumption is needed to write LOTUS, but the answer may be infinite.

The Continuous LOTUS Formula

A density $f$ satisfies $P(X\in A)=\int_A f(x)\,dx$ for every Borel set $A$. LOTUS applies this density representation to a function of $X$. It does not say to integrate over the values that $g(X)$ can take using $f$; $f$ is the density of $X$, so the integration variable is $x$.

Theorem (LOTUS for a continuous random variable): Let $X$ have density $f$, and let $g:\mathbb{R}\to[0,\infty]$ be Borel measurable. Then $$ E[g(X)]=\int_{\mathbb{R}}g(x)f(x)\,dx, $$ where either side may be $+\infty$. If $g:\mathbb{R}\to\mathbb{R}$ is Borel measurable and $\int_{\mathbb{R}}|g(x)|f(x)\,dx<\infty$, then the same identity holds with a finite expectation.

Proof. First take $g$ to be the indicator of a Borel set $A$. The definition of the density gives $$ E[\mathbf{1}_{\{X\in A\}}]=P(X\in A)=\int_{\mathbb{R}}\mathbf{1}_{\{x\in A\}}f(x)\,dx. $$ By adding these identities with nonnegative coefficients, the formula holds for every nonnegative simple Borel function $s$. For any nonnegative Borel $g$, there is an increasing sequence of nonnegative simple Borel functions $s_k$ with $s_k(x)\uparrow g(x)$ for each $x$. The monotone convergence theorem gives $$ E[g(X)]=\lim_{k\to\infty}E[s_k(X)] =\lim_{k\to\infty}\int_{\mathbb{R}}s_k(x)f(x)\,dx =\int_{\mathbb{R}}g(x)f(x)\,dx. $$ For signed $g$ with finite $\int |g|f$, apply the nonnegative result to $g^+=\max(g,0)$ and $g^-=\max(-g,0)$. Both integrals are finite because $g^++g^-=|g|$, so subtracting them gives the stated formula for $g=g^+-g^-$. $\square$

Support and Integrability

In applications, a density is often zero outside a particular interval or region. If $f(x)=0$ almost everywhere outside a Borel set $S$, the integral can be restricted to $S$. This is a statement about where the density places probability, not about the range of $g$. Endpoints do not affect a density integral because individual points have Lebesgue measure zero.

Result (support restriction and finiteness): Suppose $X$ has density $f$, with $f=0$ almost everywhere outside a Borel set $S$, and $g$ is Borel measurable. For nonnegative $g$, $$ E[g(X)]=\int_S g(x)f(x)\,dx. $$ For signed $g$, the expectation is finite whenever $\int_S|g(x)|f(x)\,dx<\infty$, and in that case $E[g(X)]=\int_Sg(x)f(x)\,dx$.

Proof. By LOTUS, $E[g(X)]=\int_{\mathbb{R}}g(x)f(x)\,dx$ for nonnegative $g$. Since $f=0$ almost everywhere on $\mathbb{R}\setminus S$, the integral over that complement is zero, leaving the integral over $S$. For signed $g$, the same argument applied to $|g|$ shows that $\int_{\mathbb{R}}|g|f=\int_S|g|f$. Thus finiteness of the latter is exactly the absolute-integrability condition in LOTUS, and restricting the signed integral to $S$ gives the result. $\square$

This criterion is useful before doing algebra: it can show whether a proposed expectation is finite, even if the integral has no convenient closed form. It also prevents a common error: an integral that looks well behaved over a few values of $x$ may still diverge near a boundary or in a tail.

Worked Calculations

Worked Example: A Squared Uniform Random Variable

Let $X\sim \operatorname{Uniform}(0,2)$, with density $f(x)=\tfrac12$ for $0\le x\le2$ and zero elsewhere. Find $E[X^2]$.

The function $g(x)=x^2$ is nonnegative, and the density is supported on $[0,2]$. First check the density is normalized: $$ \int_0^2\frac12\,dx=\frac12(2)=1. $$ Then apply LOTUS: $$ E[X^2]=\int_0^2x^2\frac12\,dx =\frac12\left[\frac{x^3}{3}\right]_0^2 =\frac12\cdot\frac{8}{3} =\frac{4}{3}. $$

The answer is $4/3$, not $8/3$: the factor $1/2$ from the density must be included. The calculation also uses the support of $X$, $[0,2]$, rather than the range $[0,4]$ of $X^2$.

Worked Example: An Exponential Discount Factor

Let $X$ have an exponential distribution with rate $2$, so $f(x)=2e^{-2x}$ for $x\ge0$ and zero for $x<0$. Find $E[e^{-X}]$.

The function $g(x)=e^{-x}$ is nonnegative, and on the support $x\ge0$ it is at most $1$. Thus its expectation is finite. LOTUS gives $$ E[e^{-X}] =\int_0^\infty e^{-x}2e^{-2x}\,dx =2\int_0^\infty e^{-3x}\,dx. $$

For $b>0$, $\int_0^\infty e^{-bx}\,dx=1/b$, since the antiderivative is $-e^{-bx}/b$ and $e^{-bx}\to0$ as $x\to\infty$. Therefore $$ E[e^{-X}]=2\left(\frac13\right)=\frac23. $$

As a check, $0<e^{-X}\le1$, so an expectation of $2/3$ is within the possible range.

Worked Example: A Normal Exponential Function

Let $X\sim N(0,1)$, with density $f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. For a fixed real number $t$, find $E[e^{tX}]$ and check that it is finite.

The integrand is nonnegative. Completing the square gives $$ -\frac{x^2}{2}+tx =-\frac{(x-t)^2}{2}+\frac{t^2}{2}. $$ Hence $$ E[e^{tX}] =\int_{-\infty}^{\infty}e^{tx}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx =e^{t^2/2}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-(x-t)^2/2}\,dx. $$

The last integral is $1$: after the substitution $u=x-t$, it is the integral of the standard normal density over the real line. Thus $$ E[e^{tX}]=e^{t^2/2}, $$ which is finite for every real $t$. This calculation illustrates that an unbounded function $g$ can still have a finite expectation; finiteness depends on its integral against the density.

Worked Example: A Function on a Bounded Support

Let $X\sim\operatorname{Uniform}(-1,1)$, and take $g(x)=1/(1+x^2)$. Find $E[g(X)]$.

The density is $1/2$ on $[-1,1]$, and $g$ is bounded there by $1$, so the expectation is finite. Applying LOTUS, $$ E\left[\frac{1}{1+X^2}\right] =\frac12\int_{-1}^1\frac{1}{1+x^2}\,dx =\frac12\left[\arctan(x)\right]_{-1}^{1} =\frac12\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right) =\frac{\pi}{4}. $$

The result is approximately $0.785$, which is between the minimum $1/2$ and maximum $1$ of $g$ on the support.

Common Mistakes and Why LOTUS Matters

LOTUS matters because many statistical quantities are expectations of functions: squared error, transformed measurements, and nonlinear scores are common examples. The density of $X$ often makes their expectations accessible even when describing the full distribution of $g(X)$ would require additional work.

Integrating over the wrong variable. In $\int g(x)f(x)\,dx$, $x$ is a value of $X$. The integral is not taken over the range of $g(X)$ using $f$.
Forgetting the density factor. The integral is not usually just $\int g(x)\,dx$. For the uniform example, omitting the factor $1/2$ would incorrectly double the answer.
Ignoring the support. A density that is zero outside its support contributes no probability there. Use the support of $X$ to set the integration limits.
Assuming every integral is finite. For a signed function, check $\int |g(x)|f(x)\,dx<\infty$ before claiming a finite expectation. For a nonnegative function, the expectation may be $+\infty$.

Key takeaway: For a continuous random variable with density $f$, compute $E[g(X)]$ by integrating $g(x)f(x)$ over the support of $X$. Check measurability and, for a finite signed expectation, absolute integrability.

Check Your Understanding

Try each question by identifying the support, writing the LOTUS integral, and checking any needed finiteness condition.

If $X\sim\operatorname{Uniform}(0,3)$, write the integral for $E[\sqrt{X}]$ and evaluate it.
If $X$ has density $3x^2$ on $0\le x\le1$, is $E[1/X]$ finite? Justify your answer by writing the relevant integral.
For an exponential random variable with rate $1$, write and evaluate the LOTUS integral for $E[Xe^{-X}]$.
Explain why the density of $X$, rather than the range of $g(X)$, appears in the integral for $E[g(X)]$.
State the condition that guarantees a finite expectation when $g$ can take both positive and negative values.