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...
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$\begingroup$ 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? $\endgroup$– Nate EldredgeSep 20, 2015 at 21:03
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2$\begingroup$ Hmmm. $\endgroup$– cardinalSep 20, 2015 at 22:34
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$\begingroup$ Why was this heavily upvoted? Seems to me to be rather offtopic for the site... $\endgroup$– DidSep 29, 2015 at 16:49
2 Answers
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 $\mathbb EX'_t=t$.
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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
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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
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$\begingroup$ @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.) $\endgroup$ Sep 18, 2015 at 21:19
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.
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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
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$\begingroup$ 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