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I need to calculate the following stochastic integrals, assuming $B_{t}$ is a standard Brownian motion: $$\int_{0}^{T}{t}\ dB_{t}$$ $$\int_{0}^{T}{B_t^{2} }\ dB_{t}$$

I don't know how to go about this with the facts that I know.

First, I'd like to clarify if Ito's formula and Ito's lemma refer to the same understanding I have. For a transformation of a stochastic process $X_{t}$, $Y_{t}=f(t,X_{t})$ (side questions: why is this expressed as a function of $t$ as well as $X_{t}$? I understand $X_{t}$ can be more general than a standard Brownian motion $B_{t}$, but in what ways?), $$dY_{t}=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial X_{t}}dX_{t}+\frac{1}{2}\frac{\partial^2f}{\partial X_{t}^2}dX_{t}\cdot dX_{t}$$

Using the fact that $dt \cdot dt=0$,$dt \cdot dB_{t}=0, dB_{t} \cdot dB_{t}=t$ (side question: why/how are these true?), the above can be changed to $$dY_{t}=\frac{\partial f}{\partial t}dt+a(t,X_{t})\frac{\partial f}{\partial X_{t}}+b(t,X_{t})\frac{\partial f}{\partial X_{t}}dB_{t}+\frac{1}{2}b^2(t,X_{t})\frac{\partial^2f}{\partial X_{t}^2}dt$$ (side question: how does this work? I have some notion of $a(t, X_{t})$ and $b(t,X_{t})$ representing "drift" and "variability" coefficients, respectively, but what does this mean?)

Then, for a standard Brownian motion, the above simplifies to $$dY_{t}=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial X_{t}}dB_{t}+\frac{1}{2}\frac{\partial^2f}{\partial X_{t}^2}dt$$ (side question: I understand the "drift" o Brownian motion is 0, so $a(t,X_{t})=0$, but how do $b(t,X_{t})$ and $b^2(t,X_{t})$ =1?)

This formula is now where I'm stuck and don't know how to proceed with the evaluation of the above integrals.

two_hearted_river
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1 Answers1

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Please avoid to ask too many questions at the same time in future posts. Let's tackle the side questions first.

First side question. I mean, why not ? If you have a stochastic variable $X_t$ itself depending on time $t$, nothing prevents us to consider a transformation involving both variables, such as $Y_t = f(t,X_t) = e^{-t}X_t$ for instance.

Second side question. See this answer of mine.

Third side question. The drift represents the deterministic (i.e. non-random) variation of a stochastic variable, when the variability is responsible for its random change. Their names are historical and come from concrete applications of stochastic models. You can think of the drift as the wind effect on a boat. Note that the variability $-$ a notion from finance, I guess $-$ is usually called diffusion, because all this stochastic calculus stuff originates from the study of diffusive phenomena at the very beginning.

Fourth side question. $a = 0$ and $b = 1$ for Brownian motion, because $\mathrm{d}B_t = 0 \cdot \mathrm{d}t + 1 \cdot \mathrm{d}B_t$.

Main question. Taking $X_t = B_t$ (hence $a = 0$ and $b = 1$, as said above) and $f(t,x) = tx$, one has $\mathrm{d}(tB_t) = \mathrm{d}f(t,B_t) = B_t\mathrm{d}t + t\mathrm{d}B_t$ thanks to Itô's lemma, hence $$ \int_0^T t \mathrm{d}B_t = \int_0^T\mathrm{d}(tB_t) - \int_0^t B_t \mathrm{d}t = TB_T - \int_0^t B_t \mathrm{d}t, $$ which cannot be simplified further. Using the same procedure with $f(t,x) = x^3$, one finds $\mathrm{d}(B_t^3) = \mathrm{d}f(t,B_t) = 3B_t^2\mathrm{d}B_t + 3B_t\mathrm{d}t$, hence : $$ \int_0^T B_t^2 \mathrm{d}B_t = \frac{1}{3} \int_0^T \mathrm{d}(B_t^3) - \int_0^T B_t \mathrm{d}t = \frac{1}{3}B_T^3 - \int_0^T B_t \mathrm{d}t. $$

Abezhiko
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