Compute the stochastic differential for $t\cdot B_t$ using Ito's Lemma

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 $$d(tB_t)$$?

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*}

• There is no need to use Ito here. It's a simple product rule. Feb 22, 2022 at 10:55
• @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. Feb 22, 2022 at 14:02
• 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. Feb 22, 2022 at 19:47