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Is there a known correlation structure among the maximums of a Brownian motion on disjoint intervals ?

Let $(W_t)_{t\geq 0}$ be a one-dimensional standard Brownian motion, and take the partition $0=t_0< t_2 <...< t_i, t_j <...t_N=T$ and define the maximum of the Brownian motion on the interval $[t_i,t_{i+1})$ as $Y_{t_i,t_{i+1}}:=\sup_{t_i\leq t < t_{i+1}}W_t.$

My question is to find the function $f(t_i,t_{i+1},t_j,t_{j+1})$ characterizing the correlation between $Y_{t_i,t_{i+1}}$ and $Y_{t_j,t_{j+1}}$, for all $i,j$. That is, $E[Y_{t_i, t_{i+1}}Y_{t_j, t_{j+1}}]=f(t_{i},t_{i+1},t_{j},t_{j+1})$, $i \neq j$.

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Hi, as $Y_{t_i,t_{i+1}}=|B_{t_{i+1}−t_i}|+Y_{t_i}$ (equality in law) the expectation of the product shoud be tractable, shouldn't it ? –  The Bridge Mar 8 '11 at 18:05

1 Answer 1

up vote 7 down vote accepted

You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ One first computes $E(Y_{s,t})$. For every $t\ge0$, one introduces $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. For every $x\ge0$, Désiré André's reflection principle yields $$ P(M_t\ge x)=2P(B_t\ge x)=P(|B_t|\ge x). $$ One gets $E(M_t)=E(|B_t|)=\sqrt{2t/\pi}$. Since $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, one deduces that $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, one can use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but one will not need this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+E(B_{t-s}M_{t-s}). $$ The computation of $E(B_tM_t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=g_t(\max(2x-y,y))\mathrm{d}y, $$ for every $x\ge0$, where $g_t$ is the centered Gaussian density of variance $t$. Now, $$ E(B_tM_t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ A (carefully executed) interversion of the order of integration yields $$ E(B_tM_t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$ Special cases are $$ \mathrm{Cov}(Y_{s,t},B_{u})=\frac12(s+t),\quad \mathrm{Cov}(B_{t},Y_{u,v})=t, $$ and the same method yields $$ E((Y_{s,t})^2)=4t-3s. $$

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Nice. $ $ $ $ $ $ –  Louigi Addario-Berry Mar 19 '11 at 4:04
    
@Louigi Thanks. –  Did Mar 21 '11 at 11:28
    
This looks good !!! –  kakuritsu Mar 21 '11 at 13:34

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