# Question about obtaining Ito differential of stochastic integral

How do you obtain the Ito differential of $$X_t = \int_0^t W_s ds$$, $$Y_t = t\exp{(\int_{0}^{t} W_s ds)}$$, and $$Z_t = t\exp{(\int_{0}^{t} W_s dW_s)}$$? I'm not sure how to even go about applying Ito's lemma to stochastic integrals - any help would be appreciated!

I might be getting confused on notation, but is it the case that if $$X_t = \int_{0}^{t} W_s ds$$, then $$dX_t = W_t dt$$ and if $$Y_t = \int_{0}^{t} W_s dW_s$$, then $$dY_t = W_t dW_t$$? I have also seen that $$\int_0^t W_s ds = tW_t -\int_0^t s dW_s = \int_0^t (t-s) dW_s$$ and $$\int_0^t W_s d W_s = \frac{1}{2} (W_t^2 - t)$$. However, I'm not sure if/how to incorporate this.

• Commented Apr 19, 2022 at 10:31

I think you have to recall that an equality of Itô differentials actually means after it's Itô integration. And you remind that $$\int_0^t dA_s = A_t - A_0$$ for any Itô process $$A_t$$. So, by the definition of the Itô differential and the definition of $$X_t := \int_0^tW_sds$$, $$dX_t = W_tdt$$.
First one can easily verify that $$(dX_t)^2 = 0, dtd(\exp(X_t)) = 0$$. Hence $$d\exp(X_t) = \exp(X_t)dX_t$$, and $$d(t\exp(X_t)) = \exp(X_t)dt + td\exp(X_t)$$. Since $$Y_t = t\exp(X_t)$$, we obtain \begin{align*} dY_t &= d(t\exp(X_t)) \\ &= \exp(X_t)dt + td\exp(X_t) \\ &= \exp(X_t)dt + t\exp(X_t)dX_t \\ &= \exp(\int_0^tW_sds)(t+W_t)dt. \end{align*} Similarly, we can compute $$dZ_t$$ as follows: As you computed, $$A_t : = \int_0^t W_sdW_s = \frac{1}{2}(W_t^2 - t)$$. Then $$dA_t = W_t dW_t, (dA_t)^2 = W_t^2 dt$$. Hence, by Itô's formula, $$d(\exp(A_t)) = \exp(A_t)dA_t + \frac{1}{2}\exp(A_t)(dA_t)^2 = \exp(A_t)W_tdW_t + \frac{1}{2}\exp(A_t)W_t^2dt.$$ Thus \begin{align*} dZ_t &= d(t\exp(A_t)) \\ &= \exp(A_t)dt + td(\exp(A_t)) \\ &= \exp(A_t)dt + t\exp(A_t)W_tdW_t + \frac{1}{2}\exp(A_t)W_t^2dt\\ &= t\exp(A_t)W_tdW_t + \exp(A_t)(\frac{1}{2}W_t^2+1)dt \\ &= t\exp(\frac{1}{2}(W_t^2 - t))W_tdW_t + \exp(\frac{1}{2}(W_t^2 - t))(\frac{1}{2}W_t^2+1)dt \end{align*} I think no further calculations are necessary.