The shortcut you didn't know you needed

You know $X$ is a fair die roll (1–6). What's the expected value of $X^2$ ?

Your instinct: find the distribution of $Y = X^2$ (so $Y$ takes values 1, 4, 9, 16, 25, 36), figure out the probabilities, then compute $E[Y]$ . That works, but there's a much easier way.

What do you think?

To find E[X²] for a fair die, do you need to first find the PMF of Y = X²?

The law of the unconscious statistician

LOTUS (Discrete)

If $X$ is a discrete random variable and $g$ is any function, then: $E[g(X)] = \Large\sum_x g(x) \cdot P(X = x)$ You compute E[g(X)] using the distribution of $X$ , not the distribution of $g(X)$ .

The name is playful — you use this rule "unconsciously" because it feels obvious. But it's not trivial: it says that to find the expected value of a transformed variable, you don't need to work out the new distribution first.

LOTUS is the bridge between knowing E[X] and computing E[g(X)] for arbitrary functions g. It saves you from the often painful step of finding the PMF of g(X).

See it in action

Pick a distribution for $X$ and a transform $g$ . The table shows each contribution $g(x) \cdot P(X=x)$ , and their sum is $E[g(X)]$ via LOTUS.

LOTUS Explorer

Distribution of X

Transform g(X)

x	P(X=x)	g(x)	g(x)·P(X=x)
-3	0.1429	9.0000	1.2857
-2	0.1429	4.0000	0.5714
-1	0.1429	1.0000	0.1429
0	0.1429	0.0000	0.0000
1	0.1429	1.0000	0.1429
2	0.1429	4.0000	0.5714
3	0.1429	9.0000	1.2857

E[X]

0.0000

E[g(X)] via LOTUS

4.0000

Example: variance via LOTUS

Variance itself is a LOTUS application! Recall:

$\text{Var}(X) = E[(X - \mu)^2] = E[X^2] - (E[X])^2$

The second form uses LOTUS twice: $E[X^2] = \sum x^2 P(X = x)$ .

Variance of a fair die via LOTUS

E[X] = \frac{1+2+3+4+5+6}{6} = 3.5

First moment: the mean.

Step 1 of 3

X ~ Bernoulli(0.4). What is E[X²]? (decimal, e.g. 0.42)

X is uniform on {1,2,3,4}. What is E[1/X]? (Round to 3 decimals) (decimal to 3 places, e.g. 0.456)

The continuous version

LOTUS extends to continuous random variables:

LOTUS (Continuous)

If $X$ is continuous with PDF $f(x)$ : $E[g(X)] = \Large\int_{-\infty}^{\infty} g(x) \cdot f(x)\, dx$

Same idea: weight $g(x)$ by the density of $X$ , integrate.

What do you think?

X ~ Unif(0,2). What is E[X³]?

Enter a decimal or whole number

Why not just find g(X)'s distribution?

For simple transforms, finding the new distribution is fine. But consider:

$g(X) = X^2$ when $X$ is continuous — you'd need the distribution of $X^2$ , which requires a change-of-variables formula
$g(X) = \max(X, 0)$ — finding the distribution involves splitting into cases
$g(X) = e^X$ — the new distribution might not have a standard name

LOTUS sidesteps all of this. You stay in the original distribution's world.

X ~ Expo(1). What is E[X²]? (Hint: Var(X) = E[X²] − (E[X])², and for Expo(1), E[X]=1, Var=1) (whole number)

X is uniform on {−2,−1,0,1,2}. What is E[|X|]? (decimal, e.g. 0.42)

Summary

Concept	Formula
LOTUS (discrete)	$E[g(X)] = \sum_x g(x) P(X=x)$
LOTUS (continuous)	$E[g(X)] = \int g(x) f(x)\, dx$
Key benefit	No need to find the distribution of $g(X)$
Common use	Computing $E[X^2]$ , variance, moment generating functions

Whenever you need E[g(X)], go straight to LOTUS. Don't waste time deriving the distribution of g(X) unless you actually need that distribution for something else.

What's next

LOTUS computes one moment at a time. Is there a single function that encodes all moments at once? Yes — the moment generating function.