Quadratic Variation and Variance Are Different Quantities
Summary
The document addresses a common confusion in stochastic calculus: an Itô process can have zero quadratic variation while still having nonzero variance at a fixed time. In the example, integrating Brownian motion with respect to time yields a process whose differential has no Brownian increment term, so its quadratic variation is zero. That does not imply its variance is zero.
The answer distinguishes quadratic variation, which is generally a random variable, from variance, which is a deterministic quantity at a given time. For a martingale, the expected squared value can be related to expected quadratic variation, with the stated simplification when the process starts at zero. The relationship does not hold in general for non-martingales; the example’s integrated process is identified as falling outside the martingale case. The discussion gives a conceptual explanation rather than a full derivation for the integral example.
Key ideas
- Quadratic variation and variance describe different properties of a stochastic process.
- Quadratic variation is generally random, while variance at a fixed time is deterministic.
- For a martingale starting at zero, variance can equal expected quadratic variation.
- The martingale relationship does not generally apply to non-martingales.
- A zero quadratic variation does not imply zero variance.
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Full text
# Can I always use quadratic variation to calculate variance? # Can I always use quadratic variation to calculate variance? Suppose we have a Brownian Motion $BM(\mu,\sigma)$ defined as $X_t=X_0 + \mu ds + \sigma dW_t$ The quadratic variation of $X_t$ can be calculated as $dX_t dX_t = \sigma^2 dW_tdW_t = \sigma^2 dt$ where all lower order terms have been dropped, therefore the quadratic variation (also the variance of $X_t$) $[X_t,X_t]=\int_0^t \sigma^2 ds=\sigma^2 t$ I was trying to use the same tech solve the problem posted in Integral of Brownian Motion w.r.t Time If I start as differential form $dX_t = W_tdt$ and calculate $dX_t dX_t$. After drop all lower order terms, I have $dX_tdX_t=0$. This means the quadratic variation is zero. Hence we have the variance is zero? I understand this isn't correct. But I really want to know what, prevents me doing this problem as previous one? I am pretty new to SDE and any help will be appreciated! Thanks a lot! ## Answer by M. Jeunesse (score 14, accepted) https://quant.stackexchange.com/a/31862 Quadratic variation and variance are two different concepts. Let $X $ be an Ito process and $t\geq 0$. Variance of $X_t$ is a deterministic quantity where as quadratic variation at time $t $ that you denoted by $[X,X]_t $ is a random variable. What is confusing you is the fact that when $X $ is a martingale then $X^2_t-[X,X]_t$ is a martingale thus you have $$E (X_t^2)=E ([X,X]_t)+E (X^2_0) $$ In the case where $X_0=0$ ( and thus $E (X_t)=0$ because $X $ is a martingale) You have $$Variance (X_t)=E ([X,X]_t) $$ In the general case, it is not true. Your example is the case where $X $ is not a martingale and thus it is not true. Please comment for any further explanation.
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