Correlation structure among the maximums of a Brownian motion 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, $E[Y_{t_i, t_{i+1}}Y_{t_j, t_{j+1}}]=f(t_{i},t_{i+1},t_{j},t_{j+1})$, $i \neq j$.  
 A: 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
$$
E((Y_{s,t})^2)=4t-3s.
$$
