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Given a Brownian motion $(W_t)_{t\geq 0},$ it is well-known that $W_t^3$ is not a Brownian motion as its SDE $$d(W_t^3) = 3W_t^2 dW_t + 3W_t dt$$ contains a nonzero drift term. To make it to be a martingale, one can consider $$W_t^3 - 3\int_0^t W_s ds.$$ On the other hand, this post shows that $W_t^3 - 3tW_t$ is a martingale.

Question: Is it true that $$\int_0^t W_s ds = tW_t?$$

I have a feeling that they are no equal as LHS is deterministic whereas RHS is random.

Community
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Idonknow
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2 Answers2

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Both are random. However, LHS is differentiable whereas RHS is not, hence they are not equal.

Will
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Applying integration by parts (which is a standard results in Stochastic Calculus) to $t W_t$ we obtain:

$$ d(t W_t) = t dW_t + W_tdt + d[t,W_t],$$ where $[t,W_t]$ is the quadratic covariation of $t$ and $W_t$. Therefore, expressing the above equation in integral form we obtain

$$ t W_t = \int_0^t s dW_s + \int_0^tW_sds + [t,W_t]$$ Quadratic covariation of a process of finite variation $t$ with a Brownian motion (or more generally with an Ito process) is $0$. Thus, we obtain that

$$ t W_t = \int_0^t s dW_s + \int_0^tW_sds$$

Note that this last equation implies that $t W_t - \int_0^tW_sds$ is a martingale. But being a martingale, does not mean that $t W_t - \int_0^tW_sds=0$.

Mdoc
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