Skip to content
All library documents

Finding the Variance of a Deterministic Brownian Integral

Article Quant Q&A · Author: user22715
Read the original

Summary

The document finds the variance of a stochastic integral with respect to Brownian motion when its integrand is a deterministic function of time. Since the integral has mean zero, its variance equals its second moment. Applying Itô's isometry reduces that second moment to the ordinary integral of the squared integrand over the time interval.

For the specified integrand, the remaining integral is the integral of arctangent divided by time. Expanding arctangent in its power series and integrating term by term yields the alternating sum over odd squared denominators, identified as Catalan's constant. The response reports an approximate value as well. This example illustrates a general variance technique, while its series evaluation relies on a valid term-by-term integration on the stated interval; it is a stochastic-calculus exercise rather than a trading application.

Key ideas

  • A deterministic-integrand Brownian integral has mean zero.
  • Itô's isometry makes its variance equal to the integral of the squared integrand.
  • The resulting arctangent integral can be evaluated by integrating its power series term by term.
  • The series is identified with Catalan's constant.

Tags

Statistics
Full text
# What's the variance of this Ito integral?


# What's the variance of this Ito integral?












I am reading stochastic calculus and I have understood that the process $$X=\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t$$ has normal distribution with mean zero. How can I find the variance of $X$?

## Answer by user16651 (score 10, accepted)

https://quant.stackexchange.com/a/29557

$$\mathbb{E^P}\left[\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t\right]=0 $$ thus $$\sigma^2=\mathbb{Var^P}\left(\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t\right)=\mathbb{E^P}\left[\left(\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t\right)^2\right] $$ By application of Ito's isometry, we have $$\sigma^2=\mathbb{E^P}\left[\int_{0}^{1}\left(\sqrt{\frac{\tan^{-1}t}{t}}\right)^2dt\right]=\mathbb{E^P}\left[\int_{0}^{1}\frac{\tan^{-1}t}{t}dt\right]=\int_{0}^{1}\frac{\tan^{-1}t}{t}dt\tag 1$$ we know (See Maclaurin Series of $\tan^{-1}x$ in wolfram) $$\tan^{-1}t=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}t^{2n-1}$$ hence $$\frac{\tan^{-1}t}{t}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}t^{2n-2}$$ and $$I=\int_{0}^{1}\frac{\tan^{-1}t}{t}dt=\int_{0}^{1}\left(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}t^{2n-2}\right)dt=\sum_{n=1}^{\infty}\int_{0}^{1}\frac{(-1)^{n+1}}{2n-1}t^{2n-2}dt$$ therefore $$I=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n+1}}{(2n-1)^2}t^{2n-1}\Big{|}_{0}^{1}\right]=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)^2}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^2}=\color{red}{G}\tag 2$$ where $G$ is Catalan's constant.

$(1)$ and $(2)$ $$\sigma^2=G\simeq 0.916$$

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.