Why does the density of Y = e^X carry a factor of 1/y when X is normal?

## Densities are per-unit-length, and transformations change length A probability density is not a probability. `f_X(x)` is a rate — probability per unit of `x` — so `f_X(x) * dx` is the probability of a tiny interval. When you apply a transformation `Y = g(X)`, the probability sitting in a tiny interval must be preserved, but the *width* of that interval changes. If `g` stretches a small interval, the same probability is now spread over more room, so the density there must be smaller. The correction factor that accounts for this stretching is the Jacobian. ## The rule for a monotone map Let `g` be strictly monotone and differentiable, with inverse `h = g_inverse`. Then `f_Y(y) = f_X(h(y)) * |h'(y)|`. Two pieces: `f_X(h(y))` asks "which `x` produced this `y`, and how dense was `X` there?", and `|h'(y)|` converts from `x`-length to `y`-length. The absolute value matters because a decreasing `g` has a negative derivative, and a density can never be negative — for a decreasing map the CDF derivation produces a minus sign that the absolute value absorbs. ## The exponential map, step by step Take `Y = e^X` with `X` normal, mean `m`, variance `s^2`. Then `h(y) = ln y`, defined for `y > 0`, and `h'(y) = 1/y`. So `f_Y(y) = f_X(ln y) * (1/y) = (1 / (y * s * sqrt(2*pi))) * exp(-((ln y) - m)^2 / (2 * s^2))` for `y > 0`, and zero otherwise. That is the log-normal density, and the `1/y` is not decoration. The exponential map crushes the far-left tail of `X` into a tiny sliver just above zero and blows up the far-right tail into an enormous stretch of the positive axis. Near `y = 0` the same probability is packed into a much narrower interval, so the density there is *multiplied* by a large `1/y`; far out to the right it is divided down. Note also that `Y` is automatically supported only on the positive numbers, because the exponential of a real number is always positive. ## The CDF route, which needs nothing memorised If you cannot recall the Jacobian rule, derive it. Because the exponential is increasing, `F_Y(y) = P(Y <= y) = P(e^X <= y) = P(X <= ln y) = F_X(ln y)` for `y > 0`. Differentiate both sides with respect to `y`; the chain rule contributes the derivative of `ln y`, which is `1/y`: `f_Y(y) = f_X(ln y) * (1/y)`. The Jacobian is exactly the chain-rule factor. This route also handles decreasing maps gracefully: there the event flips to `P(X >= h(y))`, the derivative comes out negative, and the sign is absorbed by the absolute value. ## The check that catches a dropped Jacobian Any candidate density must integrate to 1. Substituting `ln y` into a normal density without the `1/y` gives a function whose integral over `y > 0` is not 1, so the omission is detectable without any theory. Making that check a reflex is worth more than memorising the formula. ## Non-monotone maps The rule as stated needs invertibility. For a map like `Y = X^2`, two values of `X` produce each positive `y`, so you split the domain into monotone branches and add their contributions: `f_Y(y) = [f_X(sqrt(y)) + f_X(-sqrt(y))] * (1 / (2 * sqrt(y)))` for `y > 0`, where `1 / (2*sqrt(y))` is the Jacobian of each branch. The general recipe is always the same: go back to the CDF, describe the event `{g(X) <= y}` as a union of intervals in `x`, and differentiate. ## Discrete variables need no Jacobian Probability mass functions assign probability to points, not to length, so nothing is stretched. For a discrete `X`, `P(Y = y) = sum of P(X = x)` over every `x` with `g(x) = y`. Applying a derivative factor to a PMF is a category error; the Jacobian exists only because continuous probability is spread over intervals. ## A transformation worth knowing One consequence of the change-of-variables idea deserves separate billing. If `X` is continuous with strictly increasing CDF `F`, then `U = F(X)` is Uniform on `(0,1)` — the probability integral transform. Running it backwards, if `U` is Uniform on `(0,1)` then `F_inverse(U)` has distribution `F`, which is the basis of inverse-transform sampling: turn uniform numbers into draws from any distribution whose quantile function you can evaluate. ## Why interviewers ask It tests whether you know what a density *is*. Candidates who think of `f(x)` as "the probability of x" substitute and stop; candidates who think of it as probability per unit length know a stretching factor must appear, and can produce it from the CDF in two lines even under pressure.