Watch a stock trade at \$100. Over the next year it might double to \$200, or halve to \$50 — but it can never go below zero. A limited-liability share is worth, at worst, nothing. That one-way floor immediately disqualifies the normal distribution as a model for the price itself: a normal random variable spills probability across the entire real line, so a "normal price" would assign real, positive probability to the absurd event S < 0.

Notice, too, that "double" and "halve" felt like equally natural moves — a jump from 100 to 200 and a drop from 100 to 50 are the same size move multiplicatively, even though they are +100 and -50 additively. On the log scale they really are equal and opposite: \ln 2 \approx 0.69 up versus \ln\tfrac12 \approx -0.69 down. So the fix is not to abandon the normal — it is to move it to the log scale. Model the logarithm of the price as normal, and the price itself becomes finance's favourite distribution:

A positive random variable X satisfying this is lognormal with parameters \mu and \sigma^2. Because e^{Z} is always positive, X lives on (0, \infty) — it can never be zero or negative, exactly like a price. This is the shape the Black–Scholes model assumes for a stock at every future date, so a quant meets it on day one. Note carefully: the parameters \mu, \sigma describe the normal on the log scale, not the mean and standard deviation of X itself — a distinction that will keep earning its keep below.

The density, derived line by line

We know everything about Y = \ln X \sim N(\mu, \sigma^2); we want the density of X = e^{Y}. The cleanest route is through the cumulative distribution function (CDF) — accumulate probability for X, then differentiate to recover the density. Write F_X(x) = \mathbb{P}(X \le x) for the CDF of X, and let f_Y and F_Y be the density and CDF of the normal variable Y.

Step 1 — turn an event about X into one about Y. Fix x > 0. Because \ln is strictly increasing, the inequality X \le x is equivalent to \ln X \le \ln x — taking logs of both sides preserves the direction. So the two events are the same event:

Step 2 — recognise the normal CDF. The right-hand side is just F_Y evaluated at \ln x, and we can standardise it using the standard normal CDF \Phi:

Step 3 — differentiate to get the density. The density is the derivative of the CDF, f_X(x) = \frac{d}{dx}F_X(x). The right-hand side is a composition, so the chain rule applies: differentiate F_Y at its argument, then multiply by the derivative of the inner function \ln x. Since F_Y' = f_Y and \frac{d}{dx}\ln x = \frac{1}{x}:

That 1/x from the chain rule is the whole story of the lognormal — it is the Jacobian of the log transform, and it is what bends the symmetric bell into a skewed curve: it inflates the density near 0 (where 1/x is huge) and squashes it far to the right (where 1/x is tiny), while the long right tail survives because the normal exponent decays only in \ln x, glacially slowly in x.

Step 4 — substitute the normal density. Now plug in f_Y, the N(\mu, \sigma^2) density, evaluated at \ln x (replace the normal's argument y by \ln x):

Step 5 — multiply by 1/x. Carrying the chain-rule factor through gives the lognormal density, valid for x > 0 (and defined to be 0 for x \le 0, since X is positive):

Worked example: the simplest lognormal

Take the cleanest possible case: Z \sim N(0,1) and X = e^{Z}, the standard lognormal (\mu = 0, \sigma = 1). Where does X sit?

The median is easy. Half the time Z < 0, half the time Z > 0 (the standard normal is symmetric about zero). Since e^{z} is increasing, Z < 0 happens exactly when X = e^{Z} < e^{0} = 1. So

Now the mean — and here is the surprise. Naively you might guess \mathbb{E}[X] = e^{\mathbb{E}[Z]} = e^{0} = 1, matching the median. But the true value is

nearly 65% bigger than the median. Why? Because e^{z} is convex — it curves upward — so equal and opposite moves in Z do not produce equal and opposite moves in X. Feel it with a two-point caricature: suppose Z were just \pm 1 with equal probability. Then X is either e^{1} \approx 2.718 or e^{-1} \approx 0.368. The up-move gains 1.718 above 1; the down-move loses only 0.632. Average them: \tfrac12(2.718 + 0.368) \approx 1.543 > 1. The symmetric coin on the log scale becomes a lopsided coin on the price scale, and the average tilts up. That is Jensen's inequality in action — \mathbb{E}[e^{Z}] > e^{\mathbb{E}[Z]} for any non-degenerate Z — and the exact size of the tilt for a normal is the famous e^{\sigma^2/2} factor.

And the mode — the peak of the density — sits below both, at e^{\mu - \sigma^2} = e^{-1} \approx 0.368. The single most likely neighbourhood of values is down near 0.37, the typical (median) value is 1, and the long-run average is 1.65. Three different answers to "where is X?" — that spread of answers is the skew.

Mean, median, mode — and the right skew

The general lognormal repeats that story with \mu and \sigma in place of 0 and 1. The curve is right-skewed: it rises from zero, peaks, then trails off with a long tail to the right. The mean and median are no longer equal — the long right tail drags the mean above the median:

The extra \tfrac12\sigma^2 in the mean is the fingerprint of the skew: more spread on the log scale pushes the average further above the typical (median) value. The three landmarks always line up in the same order from left to right:

Because e^{x} is increasing, that ordering is just \mu - \sigma^2 < \mu < \mu + \tfrac12\sigma^2 exponentiated — true for any \sigma > 0. The mean sits furthest right, pulled out by the tail; the mode (the peak) sits furthest left. That gap is the visual signature of skew.

Concrete numbers. Take \mu = 1, \sigma = 0.5. Then:

Peak at 2.12, halfway point at 2.72, average at 3.08 — a 13% gap between mean and median from a modest \sigma = 0.5. Crank \sigma to 1.5 and the mean e^{\mu + 1.125} is over three times the median e^{\mu}: the \sigma^2 inside an exponential compounds ferociously.

Watch the skew

Slide the log-scale parameters and watch the landmarks you just computed move. The whole curve stays on the positive side x > 0, hugging zero on the left and stretching into a long tail on the right. Raising \mu slides the bulk multiplicatively to larger values (the median is e^{\mu}, so each +0.5 of \mu scales the whole picture by e^{0.5} \approx 1.65). Raising \sigma does something more violent: the peak (the mode e^{\mu-\sigma^2}) slumps left towards zero while the tail fattens to the right — the two motions that pry mean and median apart. Push \sigma to 1 and the curve barely looks bell-descended at all.

A good exercise: set \mu = 0, \sigma = 1 (the standard lognormal from the worked example) and eyeball the peak near 0.37, then imagine the median tick at 1 and the mean out at 1.65 — in the flattest part of the curve, held up only by the tail.

Why prices are lognormal: the CLT on logs

Two facts make this the natural model for a stock price S_t. First, a price cannot go negative — and the lognormal lives exactly on (0, \infty). Second, returns compound multiplicatively. Chop a year into n short periods and let R_k be the growth factor over period k (R_k = 1.003 for a +0.3\% day, and so on). The year-end price is the starting price times the product:

The logarithm turns the product into a sum of log-returns — and sums of many small, roughly independent terms are exactly what the central limit theorem eats for breakfast. Whatever the (reasonable) distribution of a single day's log-return, the sum \sum \ln R_k is approximately normal for large n. Normal log \Rightarrow lognormal price. The CLT explains sums, so it explains additive quantities as normal — and multiplicative quantities as lognormal. Heights and measurement errors (sums of small effects) go normal; prices, populations and incomes (products of small growth factors) go lognormal.

Run the clock continuously and this heuristic becomes the exact model: in geometric Brownian motion — the engine of the Black–Scholes model — the time-T price is

so \ln S_T \sim N\!\bigl(\ln S_0 + (\mu - \tfrac12\sigma^2)T,\; \sigma^2 T\bigr): a lognormal at every horizon, with log-variance growing linearly in T. And look at the drift — there is our \tfrac12\sigma^2 again, now with a minus sign. It is placed there precisely so that the convexity lift e^{+\sigma^2 T/2} from taking the expectation cancels it:

Worked example. A stock at S_0 = \$100 with drift \mu = 8\% and volatility \sigma = 20\% per year, at T = 1:

• median = 100\,e^{0.06} \approx \$106.18 — the "typical" year;
• mean = 100\,e^{0.08} \approx \$108.33 — the average, lifted by the lucky tail;
• mode = 100\,e^{0.06 - 0.04} = 100\,e^{0.02} \approx \$102.02 — the single most likely neighbourhood.

Half of all sample paths finish below \$106.18, yet the average finish is \$108.33 — the mean is propped up by the minority of paths that soar. Over long horizons the gap widens without bound: with \sigma = 20\% over T = 25 years the convexity factor is e^{\sigma^2 T/2} = e^{0.5} \approx 1.65, so the average portfolio outcome is 65% above the median one. A retirement projection quoting the mean is quietly describing a better-than-typical future.