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, $E[Y\_{t\_i, t\_{i+1}}Y\_{t\_j, t\_{j+1}}]=f(t\_{i},t\_{i+1},t\_{j},t\_{j+1})$, $i \neq j$. 



Correlation structure among the maximums of a Brownian motion
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
$$
E((Y\_{s,t})^2)=4t-3s.
$$


