Ito's Isometry for three factors

Question

Ito's Isometry states the following: If $\{W_t\}_{t\ge0}$ is a Brownian motion and $\{\phi_t\}_{t\ge0},\{\psi_t\}_{t\ge0}$ are two non-anticipative piecewise-continous processes with $\mathbb E[\int\phi_t^2dt]<\infty$, $\mathbb E[\int\psi_t^2dt]<\infty$ then $$\mathbb E\left[\int_0^t\phi_sdW_s \int_0^t \psi_sdW_s\right] = \mathbb E\left[\int_0^t \phi_s\psi_s ds\right]$$

Now I am wondering whether one can generalise this fact for three factors, that is, does there exist an identity somewhat similar to

$$\mathbb E\left[\int_0^t\phi_sdW_s \int_0^t \psi_sdW_s\int_0^t \gamma_sdW_s\right] = \mathbb E\left[\int_0^t \phi_s\psi_s\gamma_s ds\right]$$ ?

Answer 1

Using itô's lemma with the process $f(X,Y,Z)_t =X_t.Y_t.Z_t $ where $X_t=\int_0^t\phi_sdW_s$, $Y_t=\int_0^t \psi_sdW_s$, znd $Z_t=\int_0^t \gamma_sdW_s$ you will be able to prove that your intuition is actually wrong.

Less generally here is a trivial counterexample take $X_t=Y_t=Z_t= W_t$ (with integrands identically equal to one).

Then $E[W^3_t]=0 \not=E[\int_0^t 1.1.1 ds]=t$

Best Regards

Answer 2

Let me change the notations and assume that $\mathrm d(X_t,Y_t,Z_t)=(A_t,B_t,C_t)\mathrm dW_t$ (thus, your $\phi_t$ is now $A_t$, your $\psi_t$ is now $B_t$ and your $\gamma_t$ is now $C_t$). Consider the regular function $F$ defined by $F(x,y,z)=xyz$. Then Itô's formula applied to the process $U=F(X,Y,Z)$ yields $$ \mathrm dU_t=(A_tY_tZ_t+X_tB_tZ_t+X_tY_tC_t)\mathrm dW_t+(X_tB_tC_t+A_tY_tC_t+A_tB_tZ_t)\mathrm dt, $$ hence there exists indeed a process $D$ such that $$ E\left[\int_0^tA_s\mathrm dW_s\int_0^tB_s\mathrm dW_s\int_0^tC_s\mathrm dW_s\right]=E\left[\int_0^tD_s\mathrm ds\right], $$ but now, the process $D$ is given by the identity $$ D_t=X_tB_tC_t+A_tY_tC_t+A_tB_tZ_t. $$ Remark: To complete @TheBridge's quite relevant answer, note that if $X=Y=Z=W$ then $E[XYZ]=E[W^3]=0$, and $A=B=C=1$ hence $D=3W$ and $E[D]=0$.