Variance of a Deterministic Integral Against Brownian Motion
Summary
The note derives the second moment of a time integral formed by multiplying standard Brownian motion by a deterministic integrable function. Expanding the square and using the Brownian covariance, which depends on the earlier of two time points, gives a double-integral expression for the variance. The answer then rearranges that expression into a single integral involving the squared tail integral of the deterministic function.
An alternative derivation uses integration by parts to express the original time integral as an Itô integral, after which the Itô isometry yields the same result. The response also states that when the deterministic function is continuous, including piecewise continuous cases, the resulting integral is normally distributed. These statements concern deterministic integrands and require the integrability conditions needed for the manipulations; the note does not address random or adapted functions.
Key ideas
- The variance can be expressed as a double integral using the covariance of Brownian motion at two times.
- For deterministic integrable functions, the expression can be reduced to the integral of a squared tail integral.
- Integration by parts gives an Itô integral representation of the time integral.
- The Itô isometry provides an alternative route to the same second-moment formula.
- With a continuous deterministic function, the time integral is stated to be normally distributed.
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Full text
# Variance of a time integral with respect to a Brownian Motion function
# Variance of a time integral with respect to a Brownian Motion function
Let process $$I_t = \int_0^t f(s) W_s \,\mathrm d s $$ where $W_s$ is standard Brownian motion. My question are the following:
We know that $\mathbb{E} (I_{t})=0$ for all $t$ and $f$ a integrable function. Is there a general formula for the second-order moment i.e. $\mathbb{E}(I_{t}^2)$ ?
Thank you in advance for any comments, help, remarks or references related to this issue.
## Answer by Canardini (score 6)
https://quant.stackexchange.com/a/50464
Using Fubini's argument, assuming that $f$ is deterministic
$$E(I_t^2) = E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)=\int_0^t\int_0^t{f(s)f(u)min(s,u)duds}$$
If $f$ is continuous(even piece wise) you can prove that $I_t$ is normally distributed.
## Answer by Gordon (score 6)
https://quant.stackexchange.com/a/50466
As @Canardini pointed out, \begin{align*} E\big(I_t^2\big) &= E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)\\ &= \int_0^t\!\int_0^t f(s)f(u)\min(s,u)dsdu\\ &= \int_0^t\left(\int_0^u f(s)f(u) s ds + \int_u^t f(s)f(u) u ds \right)du\\ &= \int_0^t \int_0^u sf(s) f(u)ds du + \int_0^t \int_u^t uf(u)f(s) ds du\\ &=2\int_0^t uf(u) \int_u^t f(s) ds du\\ &=-u\left(\int_u^t f(s) ds\right)^2\Big|_0^t+\int_0^t\left(\int_u^t f(s) ds\right)^2 du\\ &=\int_0^t \left(\int_s^t f(u)du\right)^2 ds. \end{align*} Alternatively, note that \begin{align*} \int_0^t f(s) W_s ds &= W_t \int_0^t f(s) ds - \int_0^t \int_0^s f(u)du\, dW_s\\ &=\int_0^t f(s) ds\int_0^t dW_s - \int_0^t \int_0^s f(u)du\, dW_s\\ &=\int_0^t \int_s^t f(u)du\, dW_s. \end{align*} Then \begin{align*} E\big(I_t^2\big) &= \int_0^t \left(\int_s^t f(u)du\right)^2 ds. \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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