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I was working with this explanation of the Stochastic Integral where both functions are Brownian Motion. $$\int_0^t B(x) \, \mathrm{d}(B)x$$ So they wanted to develop this expression, to show how the integration in this case works. $$\sum_{i=1}^{n-1} B_{x_i} [B_{x_{i+1}}-B_{x_i}] $$ So the next step is this equation: $$\sum_{i=1}^{n-1} B_{x_i} [B_{x_{i+1}}-B_{x_i}] =\frac{1}{2} B_{t_n}^2 -\frac{1}{2} \sum_{i=1}^{n-1} [B_{x_{i+1}}-B_{x_i}]^2 $$ Could anyone explain me how do we get from that left side to the expression on the right side?

Thank you a lot!

PrincessEev
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Vladan Marjanovic
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1 Answers1

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For the sake of simplicity of notation let $$ \Delta B_{x_i} = B_{x_{i+1}} - B_{x_i} $$ In general, if we consider a sequence $\{a_i\}$ on which we define $\Delta a_i := a_{i+1} - a_i$ then \begin{align*} \newcommand{\D}{\Delta} \newcommand{\para}[1]{\left( #1 \right)} \Delta \para{ a_i^2 } &= a_{i+1}^2 - a_{i}^2 \\ &= a_{i+1}^2 - a_{i}^2 \underbrace{+ 2a_{i+1} a_{i} - 2a_{i+1} a_{i } + 2a_{i}^2 - 2a_{i}^2 }_{=0} \\ &= a_{i+1}^2 - 2a_{i+1} a_{i } + a_{i}^2 + 2a_{i+1}a_{i } - 2a_{i}^2 \\ &= (a_{i+1} - a_{i})^2 + 2 a_{i } (a_{i+1} - a_{i }) \\ &= (\D a_{i})^2 + 2 a_{i } \D a_{i} \end{align*} so $$ a_{i} \D a_{i} = \frac{\D \para{ a_{i}^2 } - (\D a_{i})^2}{2} $$

Now note that $$ X := \sum_{i=1}^{n-1} B_{x_i} [B_{x_{i+1}}-B_{x_i}] = \sum_{i=1}^{n-1} B_{x_i} \Delta B_{x_i} $$ and use the previous identity to get $$ X = \frac 1 2 \sum_{i=1}^{n-1} \para{ B_{x_{i+1}}^2 - B_{x_{i}}^2 } - \frac 1 2 \sum_{i=1}^{n-1} \para{ B_{x_{i+1}} - B_{x_i} }^2 $$ The first sum telescopes and should you to the desired conclusion (provided you have an initial assumption of $B_{x_1}=0$ I believe).

Addendum: I chased myself through this calculation a few years ago for a final exam/project in a stochastic processes class. I was able to find my notes from the derivation back then and this is mostly a copy-and-paste/summary of them. However, in the notes I used the backwards difference in lieu of the forward difference you use. If seeing the difference in notation helps, e.g. if I made a reindexing error above, then the PDF is at this link, see Section $3.4$, example $5$.

PrincessEev
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