Suppose that $W_{t}$ is a Wiener process.

Assume $W_{0}=0.$ Is it true that $\int_{t=0}^{T}dW_{t}=W_{T}$? If so, why?

Is one preferred to the other?


1 Answer 1


yes, by definition. The integral is by definition the limit $$ \int_0^T f_s dW_s = lim_{ n \to \infty } \sum_{ s_i \in \mathcal{P}_n } f_{s_i} ( W_{ s_{i+1} } - W_{s_i} ), $$ where $\mathcal{P}_n$ is a partition of $n$ points of the interval $[0,T]$. In your case, for any partition, the sum above is telescopic and always evaluates to $W_T$. so the limit does too and the equality follows.

Once this idea is establish to use one or the other is a matter of taste and notation.


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