-1
$\begingroup$

Suppose $B_t$ is a standard Brownian motion in $\mathbb{R}^d$ and $X_t = |B_t|$. What is the easiest way to see that$$\langle X\rangle_t = t?$$I need this result for a simulation I am running...

$\endgroup$
3
  • $\begingroup$ Under the usual definition of "standard Brownian motion in $\mathbb{R}^d$", this isn't right: the quadratic variation $\langle X \rangle_t$ would be $td$, not $t$. Is this a typo or something more significant? $\endgroup$ Sep 20, 2015 at 21:03
  • 2
    $\begingroup$ Hmmm. $\endgroup$
    – cardinal
    Sep 20, 2015 at 22:34
  • $\begingroup$ Why was this heavily upvoted? Seems to me to be rather offtopic for the site... $\endgroup$
    – Did
    Sep 29, 2015 at 16:49

2 Answers 2

4
$\begingroup$

This isn't even right in one dimension. In one dimension, $B_t\sim N(0,t)$ and so $\mathbb EX_t=(2\pi t)^{-1/2}\int_{-\infty}^{\infty}|x|e^{-x^2/(2t)}\,dx=2(2\pi t)^{-1/2}\int_0^\infty xe^{-x^2/(2t)}\,dx=2t(2\pi t)^{-1/2}=\sqrt{2t/\pi}$. In general, you would expect square root growth of $X$ and not linear growth as "Brownian motion moves $\approx\sqrt t$ in time $t$".

Maybe you wanted $X_t'=|B_t|^2$. In that case, it's much easier: $|B_t|^2=\sum_{i=1}^d |B^i_t|^2$ where the $B^i_t$ are independent Brownian one-dimensional Brownian motions with volatility $1/d$. In this case $\mathbb E|B^i_t|^2=t/d$ and so $\mathbb EX'_t=t$.

$\endgroup$
3
  • 3
    $\begingroup$ I think he is asking about the quadratic variation, which is indeed $t$ because $|B_t| = L_t + W_t$ for some other Brownian motion $W$ and a bounded variation process (the local time of $B$ at the origin) $L$. $\endgroup$ Aug 26, 2015 at 9:17
  • 5
    $\begingroup$ I guess it's even easier than that -- we want to find a process $X_t$ such that $|B_t|^2 - X_t$ is a martingale but $|B_t|^2$ can be expressed, as in the answer, via one-dimensional Brownian motions, so this is easy. $\endgroup$ Aug 26, 2015 at 9:30
  • $\begingroup$ @MateuszWasilewski, you may want to post your comment as an answer---especially considering there is a bounty on this question. (Also, it would be more clear than having the answer in the comment to another answer which says the question is ill-formed.) $\endgroup$
    – Jason Rute
    Sep 18, 2015 at 21:19
2
+50
$\begingroup$

Prodded by Jason Rute, I decided to write up a short answer. We want to prove that $X_t^2 - t$ is a martingale. We know that $X_t^2 = \sum_{i=1}^{d} (B_{t}^{i})^2$, where $B_t^{i}$ are independent Brownian motions normalised so that $\mathbb{E} (B_t^{i})^2 = \frac{t}{d}$. Then $$X_t^2 - t = \sum_{i=1}^{d} \left((B_t^i)^2 - \frac{t}{d} \right),$$ is a martingale, being a sum of independent martingales.

$\endgroup$
2
  • 4
    $\begingroup$ This is an unusual normalization. The usual definition of "standard Brownian motion in $\mathbb{R}^d$" is that the components are independent standard Brownian motions. Under that definition, the quadratic variation of $|B_t|$ would not be $t$ but $td$. $\endgroup$ Sep 20, 2015 at 20:40
  • $\begingroup$ Yes, that puzzled me a bit also but I guess only OP knows, what he wanted to ask exactly. I should have stated it clearly in my answer that the normalisation I am using is indeed unusual for the standard Brownian motion in $R^{d}$ and I introduce it only to match the exact statement in the question. $\endgroup$ Sep 20, 2015 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.