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we can see that $\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_t \right\rangle_t = 0$

However if I am to use the expression $$\int_0^t \! W_s \, \mathrm{d} s= t W_t - \int_0^t \! s\, \mathrm{d} W_s$$ then $$\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_{t}\right\rangle_t =\left\langle t W_t - \int_0^t \! s\, \mathrm{d} W_s ,W_t \right\rangle_t$$ thus $$\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_t\right\rangle_t=t\left\langle W_t ,W_t\right\rangle - \left\langle \int_0^t \! s\, \mathrm{d} W_s ,W_t\right\rangle_t$$ which means that $$\left\langle \int_0^t \! W_s \, \mathrm{d} s ,W_t\right\rangle_t=t^2-\int_0^t \! s \, \mathrm{d} s=\frac {t^2} {2}$$

What is the flaw with my argument?

Many thanks in advance

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1 Answer 1

The flaw is that you cannot pull the process "$t$" outside the quadratic covariation. $$\left\langle t W_t , W_{t}\right\rangle \neq t\left\langle W_t , W_t \right\rangle.$$

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