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I'm asked to consider three Ito processes $(X(t), t \ge 0)$, $(Y(t), t \ge 0)$, and $(Z(t), t \ge 0)$.

I am asked to show:

$$d(X(t)Y(t)Z(t)) = X(t)Y(t)dZ(t) + X(t)Z(t)dY(t) + Y(t)Z(t)dX(t) + X(t)dY(t)dZ(t) + Y(t)dX(t)dZ(t) + Z(t)dX(t)dY(t)$$

My work: I let $A(t) = X(t)$ and $B(t) = Y(t)Z(t)$ the product of two Ito processes is also Ito, so I just used Ito's product formula on $d(A(t),B(t))$ and $d(Y(t),Z(t))$ and plugged in the latter. However, doing so I get an extra term, $$dX(t)dY(t)dZ(t)$$ and I'm not sure why this necessarily vanishes.

Eddie
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    It is because the lowest term (in terms of powers of dt) is $dt^{3/2}$. Which we neglect in Ito formulation. – Chinny84 Dec 03 '14 at 18:33
  • A $dX(t)$ term doesn't just have $dt$ terms, it can also have $dW(t)$ terms right? So how do we argue that those cancel out as well? – Eddie Dec 03 '14 at 18:37
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    Because the lowest order term wold occur from the product of the $dW_i(t)$ terms which is ~$dW^3\approx dt^{3/2}$. Does that help? I will write up properly when I finish the commute if no one else has submitted an answer. – Chinny84 Dec 03 '14 at 18:40
  • So can you write up an answer and accept :). To close off this question with an answer. – Chinny84 Dec 04 '14 at 08:40
  • @Chinny84 I wouldn't want to take credit for your work, but sure if you write an answer, I'll accept it. – Eddie Dec 05 '14 at 15:35

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