Why asset prices are log-normal

Why not model prices as normal?

Because a normal variable can take any value, including negative ones, and a share price cannot fall below zero. A log return, $\ln(S_T \div S_0)$, has no such floor: it heads toward minus infinity as the price heads toward zero. So the normal fits log returns, and prices need a distribution that stops at zero.

	Normal	Log-normal
Range	−∞ to +∞	0 to +∞
Shape	symmetric	skewed right
Used for	log returns (continuously compounded)	prices
Parameters	its own mean and variance	the mean and variance of $\ln Y$

The normal distribution is the bell curve of the previous lesson. The log-normal distribution is defined through it: a variable $Y$ is log-normal when its natural log, $\ln Y$, is normal, and the reverse also holds. The name tells you the test: take the log and you get a normal.

Note the last row of the table. A log-normal distribution is described by the mean $\mu_X$ and variance $\sigma_X^2$ of $\ln Y$, the associated normal variable, not by the mean and variance of $Y$ itself. Keep track of both sets of numbers.

Drag the volatility up and watch the price curve. The left edge stays at zero while the right tail stretches, and the mean drifts above the median. Switch the drift adjustment on and off to see the projected price move, and use the worked-example button to match the numbers later in this lesson.

Price after T periods$S_T = S_0\,e^{r}$

• Log return Memorize normal, can be negative without limit
  • Sum of period returns adds across time
  • Normal the modeling assumption
• Price Memorize log-normal: bounded at 0, skewed right
  • Mean above median the long right tail pulls the mean up

How do log returns turn into prices?

A price equals today’s price times $e$ raised to the log return, and the log return is the natural log of the price ratio.

$$S_T = S_0\,e^{r_{0,T}}, \qquad r_{0,T} = \ln\left(\frac{S_T}{S_0}\right)$$

The log return , also called the continuously compounded return, over several periods is the sum of the single-period figures, because growth factors multiply and logs turn products into sums. This is a recap of Learning Module 1.

$$r_{0,T} = r_{0,1} + r_{1,2} + \cdots + r_{T-1,T}$$

The same relationship, rearranged, gives the price from a return or the starting price from an ending price.

Why do normal returns give log-normal prices?

If the log return $r_{0,T}$ is normal, then $S_T = S_0\,e^{r_{0,T}}$ is the exponential of a normal variable, which is log-normal by definition. So a price cannot go below zero and its distribution leans right.

Three features make log returns the natural choice for prices.

• Log returns add across periods and reflect continuous compounding, so they suit long horizons.
• A simple return $S_1/S_0 - 1$ modeled as normal could fall below −100%, which no share can do. A normal log return can never push the price below zero.
• A product of independent log-normal variables is log-normal, so a price built from many period growth factors stays log-normal.

The Black–Scholes option model assumes log-normal prices and normal log returns, which is why this lesson leads into option pricing.

How do you project a price?

If the expected return (the drift) is $\mu$ and the volatility is $\sigma$, the log of the price grows by $\mu - \sigma^2/2$ a year. The drift adjustment $\sigma^2/2$ is subtracted. The standard deviation of the log return over $T$ years is $\sigma\sqrt{T}$, the square-root-of-time rule of Learning Module 5.

$$\ln S_T = \ln S_0 + \left(\mu - \tfrac{\sigma^2}{2}\right)T, \quad S_T = S_0\,e^{(\mu - \sigma^2/2)T}$$

$$\sigma_T = \sigma\sqrt{T}$$

The $\sigma^2/2$ term is the volatility drag: the more volatile the share, the lower the typical (median) outcome for the same expected return.

Exam tip When a question gives you the drift and a volatility and asks for a future price from the log-normal model, apply the $\mu - \sigma^2/2$ projection, even when the question calls that price the expected price. The two wrong options are almost always the price with no adjustment and the price with $\sigma^2/2$ added.

Go deeper Mean, variance and median of a log-normal variable

The mean of a log-normal variable is larger than $e^{\mu_X}$ by a factor of $e^{\sigma_X^2/2}$. Here $\mu_X$ and $\sigma_X^2$ are the mean and variance of the associated normal variable.

$$E[Y] = \mu_Y = e^{\mu_X + \sigma_X^{2}/2}$$

With a mean log return of 6% and a standard deviation of 30%, a price of 100 has a median of $100 \times e^{0.06}$ = 106.18 and a mean of $100 \times e^{0.105}$ = 111.07. A larger $\sigma$ adds big upside outcomes while the downside stays capped at a price of zero, so the gap between mean and median widens.

Mean price on your calculator

TI BA II Plus HP 12C

1. 0.3x²÷2+0.06= 0.105, the exponent
2. 2NDe^x×100= 111.07

x² square · 2ND second function (the shifted label above a key) · e^x e to the power x

1. 0.3ENTER×2÷0.06+ 0.105, the exponent
2. ge^x100× 111.07

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

The variance has a formula of its own.

$$\sigma_Y^2 = \left(e^{\sigma_X^2} - 1\right)e^{2\mu_X + \sigma_X^2}$$

For $\mu_X$ = 0.06 and $\sigma_X$ = 0.30: (e^0.09 − 1) × e^0.21 = 0.094174 × 1.233678 = 0.1162, so the standard deviation is 0.3409.

Variance on your calculator

TI BA II Plus HP 12C

1. 0.3x²=STO1 0.09
2. 2NDe^x−1=STO2 0.094174
3. 2×0.06+RCL1=2NDe^x×RCL2= 0.116181

x² square · STO store · 2ND second function (the shifted label above a key) · e^x e to the power x · RCL recall a stored number

1. 0.3ENTER×STO1 0.09
2. ge^x1−STO2 0.094174
3. 0.06ENTER2×RCL1+ge^xRCL2× 0.116181

ENTER enter (on the HP, puts the number on the stack) · STO store · g blue shift key · e^x e to the power x · RCL recall a stored number

The skewness, $(e^{\sigma^2} + 2)\sqrt{e^{\sigma^2} - 1}$, is positive for any σ (0.95 at σ = 0.30), which is the right skew in numbers.

The same idea explains the projected price. With a drift μ, the projected price S₀e^((μ − σ²/2)T) is the median of the log-normal price, and the mean price is S₀e^(μT). In the worked example the median is 42.26 and the mean is 44.21. Both numbers are correct; they answer different questions.

How do simulations move a price one step?

Know the idea

A simulation moves a price with geometric Brownian motion : $S_t = S_0 \times \exp\left[(\mu - \sigma^2/2)t + \sigma W_t\right]$. Here $W_t$ is a Wiener process (also called Brownian motion): normal noise with mean 0 and variance $t$. A simulation repeats four steps: draw a uniform number $u$, convert it to a standard normal draw $z$ with the inverse normal cumulative function (the previous lesson), plug $z$ into the formula, and repeat.

Learning Module 9 builds full simulations on this step.