Financial Mathematics · Lecture 2 of 12

Normal and Lognormal: The Shape of Returns

Prices can't go negative, but returns can. The normal distribution and its exponential cousin, the lognormal, and the moment-generating trick that produces the famous −σ²/2.

November 15, 2026By Diganta Bhattacharya5 min read

The normal distribution

$X\sim\mathcal{N}(\mu,\sigma^2)$ has density

$$f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\!\Big(-\frac{(x-\mu)^2}{2\sigma^2}\Big).$$

We write $N(x)$ for the standard normal CDF, $N(x)=\mathbb{P}(Z\le x)$ with $Z\sim\mathcal{N}(0,1)$. The Black–Scholes formula is literally built out of two evaluations of $N$. Useful facts:

Standardising: $X=\mu+\sigma Z$, so $\mathbb{P}(X\le x)=N\!\big(\tfrac{x-\mu}{\sigma}\big)$.
Symmetry: $N(-x)=1-N(x)$.
Sums of independent normals are normal: means add, variances add.

The moment-generating function

The single most useful identity in this series:

$$\mathbb{E}\big[e^{\theta X}\big]=\exp\!\Big(\theta\mu+\tfrac12\theta^2\sigma^2\Big)\qquad X\sim\mathcal N(\mu,\sigma^2).$$

Proof sketch. Complete the square in the exponent: $\theta x-\frac{(x-\mu)^2}{2\sigma^2}=-\frac{(x-\mu-\theta\sigma^2)^2}{2\sigma^2}+\theta\mu+\frac{\theta^2\sigma^2}{2}$. The remaining integral is a normal density shifted by $\theta\sigma^2$, which integrates to 1. ∎

Remember the shiftMultiplying a normal density by $e^{\theta x}$ just shifts its mean by $\theta\sigma^2$. That is Girsanov's theorem in miniature (Lecture 9), and it's where $d_1 = d_2+\sigma\sqrt T$ comes from (Lecture 11).

Why prices are lognormal

Suppose the continuously compounded return over $[0,T]$ is normal:

$$\ln\frac{S_T}{S_0}\sim\mathcal N\!\Big(\big(\mu-\tfrac12\sigma^2\big)T,\;\sigma^2T\Big).$$

Then $S_T=S_0e^{X}$ is lognormal. It's always positive, right-skewed, and can't fall below zero, which fits limited liability for a stock. Over many small periods, returns multiply, so log-returns add, and by the CLT a sum of small log-returns is roughly normal.

The −σ²/2 puzzle

Using the MGF with $\theta=1$:

$$\mathbb{E}[S_T]=S_0\,e^{(\mu-\frac12\sigma^2)T+\frac12\sigma^2T}=S_0e^{\mu T}.$$

So the mean price grows at $\mu$, but the median, $S_0e^{(\mu-\sigma^2/2)T}$, grows more slowly. A few huge winners pull the mean up while the typical path lags behind. This is volatility drag: a +50% year followed by a −50% year leaves you at 75%, not 100%.

Raise σ in the figure: the mean stays put while the median slides left and the right tail stretches out. The −σ²/2 term will reappear when we apply Itô's lemma to $\ln S$ in Lecture 8. Then it won't be an assumption but a consequence of the calculus.

Two quantities we'll need later

For $S_T$ lognormal as above, and with $d_2=\frac{\ln(S_0/K)+(\mu-\sigma^2/2)T}{\sigma\sqrt T}$:

$$\mathbb{P}(S_T \gt K)=N(d_2),\qquad \mathbb{E}\big[S_T\mathbf 1_{\{S_T \gt K\}}\big]=S_0e^{\mu T}N(d_2+\sigma\sqrt T).$$

The second follows from the "shifted mean" remark. Replace μ by the risk-free rate r, discount, and these two lines are the Black–Scholes formula. We'll derive that properly in Lecture 11.

Key takeaways

N(x), the standard normal CDF, is the building block of the Black–Scholes formula.
MGF: E[e^{θX}] = e^{θμ + θ²σ²/2}. Exponential tilting shifts a normal's mean.
Lognormal prices: mean grows at μ, median at μ − σ²/2. The gap is volatility drag.

Exercises

Prove N(−x) = 1 − N(x) from the symmetry of the density.
If S₀ = 100, μ = 8%, σ = 30%, T = 1, compute the mean, median and P(S_T > 120).
Derive E[S_T 1{S_T > K}] using the shifted-mean idea.