Converting a Symmetric Covariance Sum into a Single Integral

Summary

The note derives how a double integral of a function of the time difference can be rewritten as a single integral weighted by the remaining horizon. The key condition is that the integrand depends on the absolute difference between the two time points, making it symmetric. Substituting the time difference as a new variable splits the integration range into positive and negative parts; symmetry lets both parts be expressed over nonnegative lags. Reordering the integrations then counts how much of the time domain contributes at each lag, producing the factor proportional to the horizon minus the lag.

The derivation uses a finite interval from zero to the horizon and indicator functions to express the integration limits before applying Fubini's theorem. This is a useful calculus step in stochastic volatility and covariance calculations. Its result depends on the symmetry assumption and the stated integration bounds; it does not by itself establish that a particular covariance model satisfies those conditions.

Key ideas

• The integrand must be symmetric in the time difference for the two sides of the integration domain to contribute equally.
• Changing variables to the time lag gives bounds that depend on the outer integration variable.
• Indicator functions make those variable limits explicit before interchanging the order of integration.
• The measure of time pairs at a given lag yields a weight proportional to the horizon minus that lag.

Tags

# Turning a covariance sum into an integral


# Turning a covariance sum into an integral












I am reading Lorenzo's Bergomi's book Stochastic Volatility Modeling, and I have come to this passage.

I just would like to understand the derivation between the first and the second equality. I guess I just have to "correctly" re-express the integral and then use Fubini's theorem so as to obtain an integral with just a $dt$/$du$/whatever term that turns into the $T - \tau$ term, but I can't figure how to do the right change of variables as $t - u$ is a function of $t$ and $u$. Any idea over there?

## Answer by Daneel Olivaw (score 2, accepted)

https://quant.stackexchange.com/a/54797

Note that the function $f$ only depends on $|t-u|$, meaning it is actually symmetric: $f(x)=f(-x)$. Doing the change of variable $\tau:=t-u$: $$\begin{align} \int_0^Tdu\int_0^Tf(t-u)dt &=\int_0^Tdu\int_{-u}^{T-u}f(\tau)d\tau \\ &=\int_0^Tdu\left(\int_0^{T-u}f(\tau)d\tau+\int_0^uf(\tau)d\tau\right) \\ &=\int_0^T{du \left(\int_0^T{f(\tau) \textbf{1}_{\tau \leq T - u} d\tau} + \int_0^T{f(\tau) \textbf{1}_{\tau \leq u}d\tau}\right)} \\ &=\int_0^T{f(\tau)d\tau \left(\int_0^T{ \textbf{1}_{u \leq T - \tau} du} + \int_0^T{\textbf{1}_{u \geq \tau}du}\right)} \\ &=\int_0^Tf(\tau)d\tau\left(\int_0^{T-\tau}du+\int_\tau^Tdu\right) \\ &=2\int_0^T(T-\tau)f(\tau)d\tau \end{align}$$ For the second equality, note that $0\leq u\leq T$ hence $-u\leq0$ and $0\leq T-u$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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