Law of the Unconscious Statistician

Suppose you know the distribution of XX and want E[X2]E[X^2], or E[eX]E[e^X], or the expected payoff of an option on XX. The obvious route is to derive the distribution of the transformed variable first. LOTUS says you never have to.

The law of the unconscious statistician
E[g(X)]=xg(x)P(X=x)E[g(X)]=g(x)fX(x)dxE[g(X)] = \sum_x g(x)\,P(X = x) \qquad E[g(X)] = \int_{-\infty}^{\infty} g(x) f_X(x)\,dx

The expectation of a function needs no new distribution: weight g(x) by the density you already have.

Weight g(x)g(x) by the density of XX, not by the density of g(X)g(X). The unwieldy name refers to statisticians using it without noticing it needs proof.

Why it saves so much work

Deriving the distribution of a transformed variable is genuinely painful: it requires inverting gg, tracking whether it is monotonic, and applying a Jacobian. For Y=X2Y = X^2 with XX standard normal, that route leads to a chi-squared density.

The rest of this lesson is for subscribers

Unlock every lesson in Fundamentals of Probability and Statistics, and every other premium course.

Subscribe to continue

Test your knowledge

Questions are only available to subscribers.

Keep reading Fundamentals of Probability and Statistics

41 lessons in this course, and every other premium course, on one subscription.

  • Every lesson in every course, with the worked examples and interactive simulators
  • Graded questions on every lesson, with explanations for the wrong answers as well as the right one
  • The trainers, timed assessments and brainteaser library that go with them