A stochastic integral computed using Itô's lemma

Question

I need some help with this question:

I have to check the following "identity" using Itô's lemma, but I can't see how to do it...

$$\bullet\int_{0}^{t}\int_{0}^{s}W(r)\;drds=-tW(t)+(t+1)\int_{0}^{t}W(s)\;ds$$

($W(t)$ is the Wiener process)

Thanks a lot for any help!

Edit: The "identity" finally was not true (as saz and Did proved)

Answer 1

This "identity" does not hold: Obviously,

$$t \mapsto \int_0^t \int_0^s W(r) \, dr \, ds$$

and

$$t \mapsto (t+1) \cdot \int_0^t W(s) \, ds$$

are both differentiable (for fixed $\omega \in \Omega$). In contrast,

$$t \mapsto t \cdot W_t$$

is almost surely nowhere differentiable. This means that the equality

$$\int_0^t \int_0^s W(r) \, dr \, ds - (t+1) \cdot \int_0^t W(s) \, ds = - t \cdot W_t$$

cannot hold.

Answer 2

Itô's lemma is quite irrelevant here (note there is no stochastic integral in the identity you suggest) and, as already mentioned by @saz, the identity cannot hold pointwisely because the LHS is a $C^1$ function of $t$ while the RHS is not. If only distributions are concerned, note that both sides are centered gaussian hence their distributions coincide if and only if their variances do.

The LHS $X$ is $$ X=\int_0^t(t-s)W(s)\;\mathrm ds, $$ and $E[W(s)W(u)]=\min(s,u)$ hence $$ \mathrm{var}(X)=2\int_0^t(t-s)\int_0^s(t-u)u\;\mathrm du\;\mathrm ds=\frac{t^5}{20}. $$ Likewise, the RHS $Y$ has variance $$ \mathrm{var}(Y)=t^3-2t(t+1)\int_0^ts\;\mathrm ds+2(t+1)^2\int_0^t\int_0^su\;\mathrm du\;\mathrm ds=\frac{(t^2-t+1)t^3}3. $$ Thus, the distributions of $X$ and $Y$ do not coincide.