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We know that $\operatorname{Cov}(B_s,B_t)=\min(s,t)$ if $B_t$ is Brownian motion.

What is $\operatorname{Cov}(B_{f(s)},B_{f(t)})$ for some injective $f$?

How can I write $B_{f(t)}$ in an Ito process form $dX=\mu(X,t)dt+\sigma(X,t)dB_t$, or Ito integral?

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1  
cov $ = f(s) \wedge f(t)$, if possible want $\int \sqrt{f(s)^{\prime}} dB_s$ –  mike Dec 23 '12 at 23:55
    
sounds correct, can you please emphasize? –  Troy McClure Dec 25 '12 at 2:09

1 Answer 1

up vote 4 down vote accepted

Let $f: [0,\infty) \to [0,\infty)$ and $s,t \in [0,\infty)$. Then

$$\DeclareMathOperator{cov}{cov} \cov(B_{f(s)},B_{f(t)}) = \min \{f(s),f(t)\}$$

since $\cov(B_u,B_v)=\min\{u,v\}$ holds for all $u,v \in [0,\infty)$, so in particular for $u:=f(s)$, $v:=f(t)$.

Concerning your second question:

Theorem: Let $(X_t)_{t \geq 0}$ a continuous local martingale, $X_0=0$. Then there exists a Brownian motion $(W_t)_t$ (with respect to a filtration $(\mathcal{G}_t)_{t \geq 0}$) such that $X_t = W_{[X]_t}$ where $[X]$ denotes the compensator of $X$.

(You can find some more information here, it's theorem 1.)

In this case: If there exists a function $g$ such that $f(t)=\int_0^t g(s)^2 \, ds$, then we can define

$$X_t := \int_0^t g(s) \, dB_s$$

which gives $[X]_t = \int_0^t g(s)^2 \, ds=f(t)$. By applying the theorem we obtain that there exists a Brownian motion $(W_t)_{t \geq 0}$ such that $$X_t = \int_0^t g(s)^2 \, dB_s = W_{[X]_t} = W_{f(t)}$$

This works in particular if $f$ is differentiable and $f(0)=0$, put $g(s) := \sqrt{f'(s)}$. Of course that's not what you were looking for since you wanted to find a representation as an Itô process for a given Brownian motion.

The problem is the following: You could define (as above)

$$X_t := \int_0^t g(s) \, dB_s$$

where $g(s) := \sqrt{f'(s)}$, $f(0)=0$. Then $X_t$ is a centered Gaussian random variable with variance $f(t)$ and $X_t-X_s$ has variance $f(t)-f(s)$ which means that $X_t$ has the same distribution as the time-changed Brownian motion $B_{f(t)}$. But they are not necessarily the same processes! If you choose for example $f(t):=2t$ you obtain

$$X_t = \int_0^t \sqrt{2} \, dB_s = \sqrt{2} B_t$$

and this is clearly not the same process as $B_{f(t)} = B_{2t}$. On the other hand we know that $W_t := \sqrt{2} \cdot B_{\frac{t}{2}}$ is a Brownian motion and obviously $$X_t = \sqrt{2} \cdot B_{t} = \sqrt{2} \cdot B_{\frac{2t}{2}} = W_{2t} = W_{f(t)}$$


Remark:

  1. Let $X_t := \int_0^t f_s \, dB_s$ where $f: \Omega \times \mathbb{R} \to \mathbb{R}$ such that $\mathbb{E}\left(\int_0^t f(s)^2 \, ds\right)<\infty$. Then $(X_t)_t$ is a martingale and the compensator is given by $$[X]_t = \int_0^t f(s)^2 \, ds$$ If $f$ is deterministic, i.e. $f: \mathbb{R} \to \mathbb{R}$, we know therefore that $(X_t)$ has a deterministic compensator.
  2. Let's consider the process $(t,\omega) \mapsto B(f(t),\omega)$ where $f: [0,\infty) \to [0,\infty)$ is a deterministic injective function such that $f(0)=0$. In this case $f$ determines how fast you run through the trajectories $(t,\omega) \mapsto B(t,\omega)$. For example $f(t):=2t$ means that you run through the trajectories twice as fast.
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thanks!! wonderful answer! can you please give example for $f(x)=x^2$ so i'll understand more? –  Troy McClure Dec 25 '12 at 11:04
    
I also dont understand how you took the compensator. if it's only to vanish the mean, why do you define $X$ with $g$ rather than $g^2$? –  Troy McClure Dec 25 '12 at 11:13
    
Well, $f(x) = x^2$ is not that easy... you obtain $X_t = \int_0^t \sqrt{2s} \, dB_s$ and it's much more difficult to see how you can write this as $W_{t^2}$ for some Brownian motion $W$. Concerning the compensator: For $Y_t := \int_0^t h_s \, dB_s$ one can show that the compensator is given by $[Y]_t = \int_0^t h_s^2 \, ds$. The idea to define $X$ with $g$ is that this implies that $X_t$ has the same distribution as $B_{f(t)}$. –  saz Dec 25 '12 at 13:18
    
ok. let me ask you another question. let $f(x)$ run over all functions possible. so what are all possible $B_{f(t)}$? is this the set of all centered Ito processes? –  Troy McClure Dec 25 '12 at 23:22
    
For a fixed brownian motion $B$: No. If you consider the process $(t,\omega) \mapsto B(f(t),\omega)$ (for some deterministic function $f$), you modify the "speed" of your brownian motion. For example $f(t) := 2t$ means that you run through the trajectories twice as fast. –  saz Dec 26 '12 at 17:43

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