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.
Best Answer
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.