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If $X(t) = \int_0^t \! B(s)ds$ where $B(s)$ for $s > 0$ is a Brownian motion process.

Part a) what are the finite dimensional distributions of $X(t)$? (not an explicit formula, you don't need to evaluate integrals)

Part b) Compute the quadratic variation of $X(t)$.

Part c) is $X(t)$ a markov process?

My attempt:

Part a) I don't know how to do it without writing down the integral expression. Is it just Gaussian? will saying that be enough? I don't know what I am supposed to say for this part.

Part b) $X(t) = \int_0^t \! B(s)ds$. So $<X(t)> = \int_0^t \!1 <B(s)>ds = \int_0^t \! sds = 1/2t^2$ Is this correct ? (since quadratic variation of $B(s)$ is just $s$.

Part c) for this part is it enough to say $X(t)$ is a markov process because it is a function of another markov process ($B(s)$) and that function (integration) is invertible? I think I had seen this as a lemma somewhere.

I appreciate help/hints/suggestions.

Alex
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1 Answers1

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  1. Yes, they are Gaussian. Have a look at this question.
  2. No, this is not correct; the identity $$\langle \int B_s \, ds \rangle = \int \langle B_s \rangle \, ds$$ does not hold. How did you define the quadratic variation? Hint: The process $X(t)$ is of bounded variation.
  3. No, the process is not a Markov process, but one can show that the process $(X_t,B_t)_{t \geq 0}$ is a Markov process. To see that $(X_t)_{t \geq 0}$ is not a Markov process, calculate expressions of the form $$\mathbb{E}(f(X_t) \mid \mathcal{F}_s)$$ for $s \leq t$ and bounded, continuous functions $f$.

Lemma: $\langle X \rangle_t = 0$.

Proof: Note that

$$|X_t-X_s| = \left| \int_s^t B_r \, dr \right| \leq |t-s| \sup_{r \in [s,t]} |B_r|$$

for any $0 \leq s \leq t$. Consequently,

$$\begin{align*} \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^2 &\leq \sup_{i=0,\ldots,n} \sup_{r \in [t_i,t_{i+1}]} |B_r|^2 \cdot \sum_{i=0}^n |t_{i+1}-t_i|^2 \\ &\leq |T_n| \cdot \sup_{r \in [0,t]} |B_r|^2 \cdot \underbrace{\sum_{i=0}^n |t_{i+1}-t_i|}_{t}\end{align*}$$

for any partition $T_n$ of the interval $[0,t]$. Letting $|T_n| \to 0$, we find

$$\langle X \rangle_t = \lim_{|T_n| \to 0} \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^2 = 0.$$

Remark: The proof actually shows $$\langle X \rangle_t =0$$ for any process of the form $$X_t(\omega) = \int_0^t f(s,\omega) \, ds$$ where $f(\cdot,\omega)$ is a continous function for each fixed $\omega \in \Omega$.

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saz
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  • I know that for y any deterministic function $s$, $X(t) = \int_0^t ! s dB(s)$ has quadratic variation $\int_0^t ! s^2ds$.Also for $N(t) = \int_0^t ! sdX(s)ds$ the qudratic we have $<N(t)> = \int_0^t ! s^2d<X(s)>$ so can I write $B(t)$ as X(t) = \iint! dB(s)$ – Alex Feb 14 '15 at 18:57
  • @Alex What do you mean by "deterministic function $s$"? $s$ is simply a variable of integration. And who claims that this second identity (the one with $\langle N \rangle$) holds? – saz Feb 14 '15 at 19:02
  • I am sorry I am trying to edit my comment and it doesn't let me do that. – Alex Feb 14 '15 at 19:03
  • what I mean is if $f$ is any function of the variable of integration $s$. – Alex Feb 14 '15 at 19:04
  • In general, is there a computational shorthand for computing quadratic variation? it just seems difficult to use the definition directly. – Alex Feb 14 '15 at 19:06
  • I have the two identities in my notes, they are not from any book though so they might be wrong. – Alex Feb 14 '15 at 19:06
  • @Alex Well, perhaps you have something of the following form in your notes: If $f$ is a continuous function and $$X(t) := \int_0^t f(s) , ds,$$ then $\langle X \rangle_t =0$. (Concerning computation: Not many, no. There are certain rules for martingales, but the given process is not a martingale.) – saz Feb 14 '15 at 19:17
  • Well I just double checked and it just says $f$ and it is with respect to $dB(s)$ not $ds$. so $<\int_o^t f(x) dB(s)> = \int_o^t f(s)^2 ds$. – Alex Feb 14 '15 at 19:38
  • Then what is the asnwer for partb in this case? – Alex Feb 14 '15 at 19:39
  • @Alex Yeah, but that's something different; it's a stochastic integral. And the answer is $0$; see my previous comment. – saz Feb 14 '15 at 19:43
  • Oh I see, do you mind me asking what general theorem this identity follows from? – Alex Feb 14 '15 at 19:45
  • @Alex Actually this follows more or less directly from the definition of quadratic variation. But so far, you haven't told me how the quadratic variation was defined in your class. – saz Feb 14 '15 at 19:48
  • Here is how it is defined :$\lim_{n \to +\infty} \sum_{i=1}^{+\infty} |X_{t_{i+1}} - X_{t_i}|^2 $. for $X_t$ being a stochastic process and $[T_n]$ for $n = 1, 2, ..$ is subdivision of $[0, t]$ with $|T_n|$ goes to zero. – Alex Feb 14 '15 at 20:02
  • @Alex see my edited answer. – saz Feb 14 '15 at 20:36
  • I see, that makes sense, thanks alot! – Alex Feb 14 '15 at 20:44