Transformations of a Random Variable

When you push a random variable through a function, its density stretches or compresses — you'll learn the Jacobian rule that tells you by how much.

By the end you'll be able to transform a density using the change-of-variables (Jacobian) formula for a monotone \(g\), and switch to the CDF method when \(g\) isn't monotone.

Predict: before you draw a sample, guess the shape of Y's density for Y = −ln(X)/λ — does it peak at 0 or somewhere in the middle? Then draw and check.

Draws of X ~ Uniform(0,1) flow down through a transform Y = g(X). Watch the flat input density get reshaped into the output density below — empirical bars from your draws vs. the theoretical curve, which is exactly f_X(g⁻¹(y))·|d/dy g⁻¹(y)|. Hover a bar for exact numbers.

No draws yet.
Change of variables — top: X draws; bottom: empirical (bars) vs theoretical (dashed) density of Y = g(X)
empirical Y theoretical f_Y(y)

For a monotone transformation \(Y = g(X)\), the density transforms by \(f_Y(y) = f_X(g^{-1}(y)) \cdot \left| \dfrac{d}{dy} g^{-1}(y) \right|\); the CDF method handles non-monotone cases.

Formal

This is called change of variables. For monotone \(g\): \(f_Y(y) = f_X(g^{-1}(y)) \cdot \left| \dfrac{d}{dy} g^{-1}(y) \right|\) — the derivative term (the Jacobian) corrects for how \(g\) stretches or compresses intervals. You take the absolute value so the density stays non-negative whether \(g\) is increasing or decreasing. Non-monotone \(g\) — or whenever you're not sure — fall back on the CDF method: write \(F_Y(y) = P(g(X) \le y)\), re-express the event in terms of \(X\), evaluate with \(F_X\), then differentiate.

Applied

Say you're writing simulation software: you can generate Uniform(0,1) draws easily, but you need Exponential losses for your model. The transform \(Y = -\ln(X)/\lambda\) (with \(X\) uniform) gives you exactly that — this "inverse transform method" is what underlies most Monte Carlo loss simulations you'll see in pricing and reserving. It's the same transform driving the interactive above.

Worked example — monotone
\(X \sim \text{Uniform}(0,1)\), \(Y = -\ln(X)\). \(g^{-1}(y) = e^{-y}\), \(\left| dg^{-1}/dy \right| = e^{-y}\). \(f_Y(y) = 1 \cdot e^{-y} = e^{-y}\), \(y>0\) → Y ~ Exponential(1).
Your turn — faded step

Same setup, but now \(Y = -2\ln(X)\) — that's λ = ½ in the interactive above. You know \(g^{-1}(y) = e^{-y/2}\). Fill in the missing step:

\(\left| dg^{-1}/dy \right| = \) ____, so \(f_Y(y) = \) ____

Reveal the missing steps

\(\left| dg^{-1}/dy \right| = \tfrac12 e^{-y/2}\), so \(f_Y(y) = 1 \cdot \tfrac12 e^{-y/2} = \tfrac12 e^{-y/2}\) for \(y>0\) — Y ~ Exponential(1/2). Set λ = 0.5 in the interactive and compare the dashed theoretical curve to this density.

Worked example — non-monotone (CDF method)
\(X \sim \text{Normal}(0,1)\), \(Y = X^2\). \(F_Y(y) = F_X(\sqrt{y}) - F_X(-\sqrt{y})\), so \(f_Y(y) = f_X(\sqrt{y})/\sqrt{y}\) for \(y>0\) — the chi-square(1) density.
More info — why the CDF method always works

The Jacobian formula is really just a shortcut for the CDF method when \(g\) is monotone — differentiating \(F_Y(y) = F_X(g^{-1}(y))\) with the chain rule gives you the same \(f_X(g^{-1}(y)) \cdot |dg^{-1}/dy|\). When \(g\) isn't monotone (like \(Y=X^2\) above), you can't write a single \(g^{-1}\), so you go back to the CDF definition itself — which is why the CDF method is the safe fallback whenever you're unsure a transform is monotone. See the jbstatistics link in Dive deeper below for another worked derivation.

Check your understanding

Question 1 of 4

X ~ Uniform(0,1) and Y = −ln(X). Using the change-of-variables formula, what is the distribution of Y?

Question 2 of 4

Why can't the single-branch Jacobian formula \(f_Y(y)=f_X(g^{-1}(y))|d/dy\, g^{-1}(y)|\) be used directly for Y = X² with X ~ Normal(0,1)?

Question 3 of 4

The interactive's Y = −ln(X)/λ transform is exactly the technique used to simulate losses from a named continuous family. Which family, and what's its rate parameter?

Question 4 of 4

In \(f_Y(y) = f_X(g^{-1}(y)) \cdot |d/dy\, g^{-1}(y)|\), why take the absolute value of the derivative instead of just multiplying by it directly?

Recap

  • Monotone \(g\): \(f_Y(y) = f_X(g^{-1}(y)) \cdot |dg^{-1}/dy|\) — the Jacobian rescales the density; always take the absolute value.
  • Non-monotone \(g\) (or whenever you're unsure): use the CDF method — write \(F_Y(y) = P(g(X) \le y)\) in terms of \(F_X\), then differentiate.
  • \(Y = -\ln(X)/\lambda\) turns a Uniform(0,1) draw into Exponential(λ) — the inverse-transform method used to simulate from named distributions.
  • The Jacobian formula is a shortcut for the CDF method under monotonicity — when that assumption breaks, fall back to the CDF definition itself.

Dive deeper

Sources

  • Transformations of a Random Variable