Scaling Invariance Forces Integrated Variance to Be Linear

Summary

The document considers a process whose integrated squared volatility has the same distribution after time is scaled by a positive factor and the integral is normalized by that factor. It asks why shrinking the scale leads to an integrated-variance process that is linear in time. The key step is a change of variables: the normalized integral over the shortened interval becomes an integral over the original time range of volatility squared evaluated at scaled times.

If volatility is continuous at zero, the integrand converges to its value at zero as the scale tends to zero. The limiting integral is therefore the initial squared volatility multiplied by elapsed time, yielding the stated distributional conclusion when combined with the assumed equality in distribution. This reasoning relies on continuity at the origin and on interpreting the process-level distributional identity and limit appropriately; the note does not elaborate on technical conditions for exchanging limits and distributions.

Key ideas

• A change of variables rewrites the normalized short-interval integral over a fixed time range.
• Continuity of volatility at zero makes the rescaled integrand approach its initial value.
• The limiting integrated variance is initial squared volatility multiplied by time.
• The conclusion depends on the assumed scaling identity and continuity condition.

Tags

# equality in distribution


# equality in distribution












I encounter the following problem :

I have the equality in distribution:

for all $\lambda >0, ((1/\lambda)*\int_{0}^{\lambda t}\sigma_{u}^{2}du,t\geq0)=(\int_{0}^{t}\sigma_{u}^{2}du,t\geq0)$

where $(\sigma_{t})$ is a predictable process.

Now I don't understand that when $\lambda->0$ and when we use the continuity of $(|\sigma_{u}|,u\geq0)$ at 0 then we get: $(\int_{0}^{t}\sigma_{u}^{2}du,t\geq0)=(c^{2}t,t\geq0)$ (in distribution)

I try to recognize a derivative but I don't get it... Thank you

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/17700

It appears that we need only to observe the following: \begin{align*} \lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\int_0^{\lambda t}\sigma^2_u du &= \lim_{\lambda\rightarrow 0}\int_0^{ t}\sigma^2_{\lambda u} du\\ &= \int_0^{ t}\sigma^2_{0} du \\ &=\sigma^2_{0} t. \end{align*}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.