Bipower Variation And Jump Detection

A pedometer counting your daily steps can't tell a smooth walking stride from one giant leap over a puddle — both just register as "movement." But if you instead multiply each step's length by the previous step's length, that giant leap stands out clearly: it's huge compared to its ordinary neighbors on either side, while a normal continuous stride times another normal stride stays modest. Bipower variation applies exactly this trick to price returns, using consecutive returns multiplied together to build a measure that stays nearly blind to isolated jumps.

The formula, defined

$$BV_t = \frac{\pi}{2} \sum_{i=2}^{n} |r_i|\,|r_{i-1}|$$

In words: instead of squaring each return individually and summing (that's ordinary realized variance $RV_t$, see realized volatility estimators), bipower variation multiplies each return's absolute size by its immediate neighbor's absolute size, sums those products, and rescales by the constant $\pi/2 \approx 1.5708$ so that in a pure continuous-diffusion world (no jumps at all), $BV_t$ converges to the same quantity as $RV_t$. The key property: a single enormous jump return $r_i$ gets multiplied by its neighbor $r_{i-1}$, which is an ordinary-sized continuous return — so the jump contributes only proportionally to $BV$, roughly like an ordinary large return would, instead of contributing its own huge value squared, which is what happens to it in $RV_t$.

Worked example 1 — computing RV and BV side by side

A stock has five consecutive returns: +0.5%, -0.3%, +4.0% (a jump), +0.2%, -0.4%. Ordinary $RV = 0.005^2 + 0.003^2 + 0.040^2 + 0.002^2 + 0.004^2 = 0.000025 + 0.000009 + 0.0016 + 0.000004 + 0.000016 = 0.001654$ — the jump return alone (0.0016) accounts for nearly 97% of total RV. Bipower variation uses consecutive pairs: $|r_1||r_2| + |r_2||r_3| + |r_3||r_4| + |r_4||r_5| = (0.005)(0.003) + (0.003)(0.040) + (0.040)(0.002) + (0.002)(0.004) = 0.000015 + 0.000120 + 0.000080 + 0.000008 = 0.000223$, times $\pi/2$: $BV = 1.5708 \times 0.000223 \approx 0.000350$. BV is barely a fifth of RV — because the jump, paired with its ordinary-sized neighbors on each side, contributes far less to the sum than it did to RV where it was squared against itself.

Worked example 2 — isolating the jump component

The jump contribution to total variance is estimated as $J_t = \max(RV_t - BV_t,\ 0)$. From example 1: $J = 0.001654 - 0.000350 = 0.001304$. That's the estimated variance attributable to the discontinuous jump, versus $BV = 0.000350$ attributable to ordinary continuous price movement. A formal jump test doesn't stop here — it standardizes this difference by its own estimated statistical noise (using a ratio statistic and known asymptotic distribution) to get a z-score and decide if $J_t$ is significantly different from zero, rather than just eyeballing that $RV > BV$, since some gap between the two is expected from estimation noise even with no true jump.

What this means in practice

Separating jump variance from continuous variance matters because the two behave differently for forecasting and risk purposes — continuous variance tends to be persistent and predictable (feeding models like HAR-RV, see HAR-RV model), while jump variance is closer to a one-off shock that shouldn't be extrapolated forward the same way. Trading desks and researchers use jump detection to decide whether a sharp move is "part of the ongoing volatility regime" or "an isolated event," which changes how much weight it should get in a forecast.

Practice

1. Given four consecutive returns +0.4%, +0.3%, -3.0%, +0.5%, compute both RV and BV (using the $\pi/2$ scaling), and estimate the jump contribution $J = \max(RV-BV, 0)$.
2. Why does pairing each return with its immediate neighbor, rather than squaring it against itself, make bipower variation much less sensitive to a single large jump?