Square Integrability Assumptions for Stochastic Stock Volatility
Summary
The note explains why stock price models often require the volatility process to be square integrable so that its stochastic integral with Brownian motion is well defined in the standard Itô framework. It distinguishes the condition that the expected integrated variance is finite from pathwise finiteness of integrated variance, which appears in descriptions of Itô processes.
The answer says Itô integration can be defined under broader conditions, but the usual square-integrability assumption makes applications such as pricing more tractable. Whether it is realistic depends on whether stock return variance is finite: Mandelbrot argued for infinite variance, while the answer says finite variance is the more common contemporary belief. It also cautions that pricing models are often calibrated under a risk-neutral measure to value options, rather than to reproduce stock-price behavior perfectly. The note offers intuition, not a formal derivation or empirical evidence resolving the finite-variance question.
Key ideas
- The standard Itô integral requires an integrability condition on the volatility process.
- Square integrability is commonly assumed because it simplifies applications such as pricing.
- The realism of the assumption depends on whether stock return variance is finite.
- Risk-neutral pricing models may prioritize option valuation over accurate stock-price description.
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# Why is the stochastic process of the volatility of a stock price square integrable? # Why is the stochastic process of the volatility of a stock price square integrable? I am taking a course in financial mathematics(Ito-Integrals, Black-Scholes,...) and there is something that is not immediately clear to me. When constructing our stock price model, the integral $\int_0^t \sigma_s\;dW_s$, $\sigma_t$ being the volatility stochastic process, was the reason we had to construct the Ito-Integral. So we set up all the theory and here in my course notes it is stated that $\int_0^t \sigma_s\;dW_s$ really is an Ito-Integral which means that $E\big[\int_0^t\sigma_s^2 ds\big] < \infty$. So far so good. Right now we are dealing with Ito-processes because our stock price model seems to be an Ito-process. According to my lecture notes for the stock price model to really be an Ito-process the volatility $\sigma_t$ needs to be square integrable meaning $\int_0^t\sigma_s^2 ds < \infty$. Is this normally the case? In my lecture notes it isn't stated anywhere so maybe this is a trivial implication from some of our assumptions which I overlooked. ## Answer by fes (score 4) https://quant.stackexchange.com/a/76047 You can define the Ito integral without square integrability but this makes working with applications like pricing more complicated, so the assumption is typically made in practice. The question of whether this actually holds is pretty complicated and boils down to an old question of whether stock return variance is finite or not. Famously, Mandelbrot (https://www.jstor.org/stable/168611) argued it is not, implying that square integrability is unrealistic. However, I think most people nowadays believe variance is actually finite so the condition would be reasonable. Another issue is that these types of models are typically used for pricing and applied under the risk neutral measure. Hence the goal is often to provide a realistic model for option pricing rather than model stock prices as accurately as possible.
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