Using Itô Isometry to Find the Variance of a Brownian Integral

Summary

The document asks how to compute the variance of a stochastic integral of Brownian motion over a unit-length interval. It applies Itô isometry to the second moment, reducing the calculation to the integral of the expected squared Brownian value. Since Brownian motion at time s has second moment s, this gives a second moment of t plus one half.

The question then argues that the integral has mean zero and proposes the same expression for its variance. The text does not include an accepted answer or further discussion, so it presents a candidate derivation rather than a verified explanation. The calculation relies on standard Brownian motion starting at zero and the adapted integrand being square-integrable over the interval.

Key ideas

• Itô isometry converts the second moment of a Brownian stochastic integral into an ordinary integral of the integrand's squared expectation.
• For standard Brownian motion, the second moment at time s equals s.
• The proposed integral has second moment t plus one half over the stated interval.
• The variance conclusion depends on the integral having mean zero, and the document does not provide a verified answer.

Tags

# Ito isometry of variance of integral on interval [t, t+1]


# Ito isometry of variance of integral on interval [t, t+1]












Can I use Ito isometry to calculate $Var(Y_t)$ where $Y_t= \int_{t}^{t+1}B_sdB_s$ and $B_s$ is Brownian motion, in this way? Is my reasoning correct? $$Var(Y_t)=E[Y_t^2]-(E[Y_t])^2$$ Applying Ito isometry

$E[\int_{t}^{t+1}B_sdB_s]^2=E[\int_{t}^{t+1}B_s^2ds]=\int_{t}^{t+1}E[B_s^2]ds=\int_{t}^{t+1}sds=\ldots=t+\frac{1}{2}$

Knowing that $E[B_t]=0$ $$(E[\int_{t}^{t+1}B_sdB_s])^2=(\int_{t}^{t+1}E[B_s]dB_s)^2=0$$

$$Var(Y_t)=t+\frac{1}{2}?$$

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