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Consider $Y_t=B_t^3$, $t \geq 0$ where $(B_t)_{t \geq 0}$ is standard Brownian Motion.

Here, $dB_t = 0+1\cdot dB_t$ and $G(x,t)=x^3$, What is $dY_t$?

Using Ito's Lemma, we can calculate the partial derivatives and get:

$$\begin{align*} dY_t &= [3B_t^3\cdot 0+0\cdot \frac{1}{2}\cdot 6B_t\cdot 1^2]dt + 3B_t^2\cdot 1\cdot dB_t \\ &= 3B_t^2dt + 3B_t^2dB_t \end{align*}$$

If $Y_t = t\cdot B_t$, how can I use this same logic to get $d(tB_t)$?

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You need to use Itô's lemma with $g(x,t) = xt$. Notice that $g_x(x,t) = t, g_t(x,t) = x, g_{xx}(x,t) = 0$, so that \begin{align*} dY_t &= dg(B_t, t) \\ &= g_t(B_t, t) dt + g_x(B_t, t) + \frac{1}{2}g_{xx}(B_t,t)(dB_t)^2 \\ &= B_t dt +t dB_t \end{align*}

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  • $\begingroup$ There is no need to use Ito here. It's a simple product rule. $\endgroup$
    – Tobsn
    Commented Feb 22, 2022 at 10:55
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    $\begingroup$ @Tobsn Sure, although you may wish to note two things: (i) the OP asked to exhibit Itô's lemma in this question, and (ii) the product rule is a direct consequence of Itô's lemma, so even if I use the product rule, Itô's lemma is carrying most of the weight behind the scenes. $\endgroup$ Commented Feb 22, 2022 at 14:02
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    $\begingroup$ Regarding (i) sure, I've seen that. But not every question is a good question. (ii) I find the product rule far more elemental then Ito's lemma and I don't like thinking of it as a corollary thereof. To some extend that is like proving the convential product rule in calculus using the chain rule. Sure, it works, but I don't find it instructive. $\endgroup$
    – Tobsn
    Commented Feb 22, 2022 at 19:47

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