Quadratic Variation

Here is a question that sounds trivial and is not: how far did the price move today? Not where it ended up — how much travelling did it do? The obvious answer is to add up the sizes of all the little moves. Try it on real data and something disturbing happens. Hourly bars give one number, minute bars a bigger one, every tick bigger still, with no sign of stopping. The distance travelled by a price is not a well-defined quantity at all; it depends entirely on how closely you look.

There is, however, exactly one way of measuring a jagged path that does settle down: instead of adding up the moves, add up their squares. That number converges, and essentially all of continuous-time finance is built on the fact that it does.

The analogy: measuring a coastline

How long is the coast of Britain? A 100-kilometre ruler gives one answer. A 1-kilometre ruler picks up every bay you skipped, so the answer grows. A metre stick grows it again. The coastline has no length in the ordinary sense — every refinement finds more detail.

A price path is the same kind of object. But watch what changes when you square each measurement before adding. Squaring punishes small pieces enormously: halve a step and its square drops by a factor of four, while you only gain twice as many steps. For a smooth curve the squares therefore shrink to nothing.

A random path is different, and this is the crux: halve the time interval and a random step does not halve. It shrinks by $\sqrt{2}$, because randomness accumulates as the square root of time. Square that and you get a factor of two, exactly cancelled by having twice as many steps. The total stays put. Squared moves are the natural unit of a random path, the way ordinary lengths are the natural unit of a smooth one.

Writing it down

Take a process $X_t$ over the interval from $0$ to $T$. Chop that interval into $n$ pieces at times $0 = t_0 < t_1 < \cdots < t_n = T$. Over each piece the process moves by $X_{t_i} - X_{t_{i-1}}$. The quadratic variation of $X$ up to time $T$, written $[X]_T$, is what the sum of the squares of those moves converges to as the pieces get arbitrarily small:

$$[X]_T = \lim_{n\to\infty} \sum_{i=1}^{n}\big(X_{t_i} - X_{t_{i-1}}\big)^2 .$$

In words: chop the path finely, square every step, add them up, and see where that total lands as your chopping gets finer. Three facts do all the work:

• For any smooth path — anything with a derivative — $[X]_T = 0$.
• For a standard Brownian motion $W$, $[W]_T = T$ exactly, and it is not random. Every single path gives the same answer.
• For a process with volatility $\sigma$, $[X]_T = \sigma^2 T$. The drift contributes nothing at all.

That middle fact is usually abbreviated to the shorthand $(dW)^2 = dt$: a squared Brownian increment behaves like plain elapsed time, not like a negligible second-order crumb. It is the reason Itô's lemma carries a term that ordinary calculus does not.

Worked example 1: watching it converge, by hand

Take one year, so $T = 1$, and chop it into $n$ equal steps of length $\Delta t = 1/n$.

The Brownian case. Each step $\Delta W$ is normal with standard deviation $\sqrt{\Delta t}$, so its square has expectation $\Delta t = 1/n$. There are $n$ steps, so the expected total is $n \times (1/n) = 1$. The total's variance also has a clean form: for a standard normal $Z$, $\operatorname{Var}(Z^2) = 2$, and since $(\Delta W)^2 = \Delta t\, Z^2$, each squared step has variance $2(\Delta t)^2$. Adding $n$ independent ones gives $2n(\Delta t)^2 = 2/n$, so the total's standard deviation is $\sqrt{2/n}$.

The smooth case. Take the straight line $f(t) = t$. Every step is exactly $\Delta t = 1/n$, so every square is $1/n^2$, and $n$ of them sum to $1/n$.

Ordinary distance travelled, for contrast. The average size of a Brownian step is $\sqrt{2/\pi}\,\sqrt{\Delta t} \approx 0.798/\sqrt{n}$, and $n$ of those sum to $0.798\sqrt{n}$.

steps $n$	squared steps, Brownian	squared steps, straight line	total distance, Brownian
10	1.00 ± 0.45	0.100	2.5
100	1.00 ± 0.14	0.010	8.0
10,000	1.00 ± 0.014	0.0001	79.8
1,000,000	1.00 ± 0.0014	0.000001	797.9

Three behaviours in one table. The middle column dies. The right column diverges — a Brownian path really does travel infinitely far in a finite year. Only the second column locks onto a number, and that number is $T = 1$, with the wobble around it shrinking like $1/\sqrt{n}$.

See the same thing by eye. Set the explorer to Brownian with zero drift and redraw several times. The paths end up all over the place — the level is wildly random. But their roughness is visibly identical every time. That constant roughness is the quadratic variation, and it is the part you can measure.

Worked example 2: realised variance from five days of data

This is quadratic variation as a trading desk actually computes it. Five daily log returns:

$$+1.20\%,\quad -0.80\%,\quad +0.30\%,\quad -1.50\%,\quad +0.60\%.$$

Step 1 — square each one (as decimals, so 1.20 percent is 0.0120):

$$0.000144,\quad 0.000064,\quad 0.000009,\quad 0.000225,\quad 0.000036 .$$

Step 2 — add them. The total is $0.000478$. That is the realised variance over the five days. Note what you did not do: you never subtracted a mean. Quadratic variation squares raw increments, and over short horizons the drift is far too small to matter.

Step 3 — put it on a yearly footing. Per day the variance is $0.000478/5 = 0.0000956$. Multiply by 252 trading days: $0.0000956 \times 252 = 0.024091$. Take the square root: $\sqrt{0.024091} = 0.1552$, so 15.5 percent annualised volatility.

Step 4 — look at where it came from. The single $-1.50\%$ day contributed $0.000225$ out of $0.000478$ — 47 percent of the entire week's variance, from one day out of five. This is not an artefact of the example; it is what squaring does, and it is why realised volatility is jumpy.

Step 5 — price a trade with it. A variance swap pays the difference between realised and strike variance. Say you are long with a vega notional of $50,000 per volatility point at a strike of 14. The variance notional is $50{,}000/(2 \times 14) = 1{,}785.71$ per point of variance, and the payoff is

$$1{,}785.71 \times \big(15.52^2 - 14^2\big) = 1{,}785.71 \times (240.9 - 196.0) \approx 80{,}100 .$$

Roughly $80,100. Notice that the contract settles on the sum of squared returns and on nothing else — it is a direct trade on quadratic variation.

What this means in practice

• Volatility is observable; return is not. Pinning down an expected return takes decades of data, but a single day of intraday prices gives a decent volatility estimate — because quadratic variation is a property of the path, not an average over scenarios.
• It is why Itô calculus exists. Since $(dW)^2 = dt$ survives instead of vanishing, a second-order term appears in the chain rule, and that term is where the $\tfrac12\sigma^2$ in Black-Scholes comes from.
• It is what a delta hedge accumulates. A hedged option bleeds theta and earns gamma against realised squared moves; the P&L is a bet on quadratic variation against implied variance.
• Jumps break the tidy version. A gap adds its square all at once, so realised variance in a jumpy market mixes diffusion and jumps, and separating them is its own research area.