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I'm trying to understand the calculation of $$Var \left (\int_{a}^{b} r_{s} ds\right) $$ where $r_{s} = \sigma W_{s}$ and $W_{s}$ is a standard 1-d Brownian motion. $\sigma$ is assumed to be constant.

and the solution is, $$\sigma^2 (b-a)^2(a+\frac{b-a}{3})$$

if I were to calculate, I would proceed like this: $$ Var\left (\int_{a}^{b} r_{s} d_{s}\right) = Var \left (\sigma \int_{0}^{b} \int_{0}^{s} dW_{s} d_{s} - \sigma \int_{0}^{a} \int_{0}^{s} dW_{s} d_{s}\right) $$

How do we proceed further?

Prateek Bhatnagar
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1 Answers1

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One way of doing it is rewriting the integral $$I = \int_a^b \sigma W_s \mathrm d s$$ using the Ito lemma. $$\mathrm d f(t, W_t) = \frac{\partial f}{\partial t} (t, W_t) \mathrm d t + \frac{\partial f}{\partial x}(t, W_t) \mathrm d W_t + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t, W_t) \mathrm d t$$ Let $f(t,x) = \sigma tx$, we have that $$ \sigma \mathrm d( t W_t) = \sigma W_t \mathrm d t + \sigma t \mathrm d W_t$$ Integrate between $a$ and $b$, and use Ito isometry :)

Riccardo Sven Risuleo
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