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Let $B_t$ a standard real valued Brownian motion, its running minimum $s_t \triangleq \min \limits_{0 \leq s \leq t} B_s$ and running maximum $S_t \triangleq \max \limits_{0 \leq s \leq t} B_s$. From Revuz Yor p.111, we know that for any Borel set $E \in \mathcal{B}([a,b])$, $$ \mathbb{P}(B_t \in E, a \leq s_t < S_t \leq b) = \int_E k(x) {\rm d} x $$ where $$ k(x) \triangleq \frac{1}{\sqrt{2 \pi t}}\sum \limits_{j \in \mathbb{Z}} f_j(x,a,b) $$ and for each $j$ $$ f_j(x,a,b) \triangleq \exp \left (-\frac{1}{2t}g_j(x,a,b)^2 \right ) - \exp \left (-\frac{1}{2t}\left (g_j(x,a,b) - 2b \right )^2 \right ) $$ and $g_j(x,a,b) \triangleq x+2j(b-a)$.

Question: Can we find a function $\hat k(x,\mathfrak{z}_1,\mathfrak{z}_2)$ such that $$ k(x) = \int_a^\infty \int_{\mathfrak{z}_1}^b \hat k(x,\mathfrak{z}_1,\mathfrak{z}_2) {\rm d} \mathfrak{z}_2 {\rm d} \mathfrak{z}_1? $$ In that case, the function $\hat k$ is the joint density of the triple $(B_t,s_t,S_t)$.

heads up on (my) difficulties

Maybe I am missing something but it does not seem straightforward. If we proceed term by term in the series, we would like to find $\hat f_j(x,\mathfrak{z}_1,\mathfrak{z}_2)$ such that $$ f_j(x,a,b) = \int_a^\infty \int_{\mathfrak{z}_1}^b \hat f_j(x,\mathfrak{z}_1,\mathfrak{z}_2) {\rm d} \mathfrak{z}_2 {\rm d} \mathfrak{z}_1, $$ thus $$ -\frac{\partial f_j}{\partial a}(x,a,b) = \int_a^b \hat f_j(x,a,\mathfrak{z}_2) {\rm d} \mathfrak{z}_2. $$ If we can find a function $\lambda_j(x,a,b)$ such that $$ -\frac{\partial f_j}{\partial a}(x,a,b) = \lambda_j(x,a,b) - \lambda_j(x,a,a), $$ then $\hat f_j$ must be $\frac{ \partial \lambda_j}{\partial b}$. If I am not mistaken, the problem is to find such a function $\lambda_j$.

megaproba
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1 Answers1

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From the formula $$ \mathbb{P}(B_t \in E, a \leq s_t < S_t \leq b) = \int_E k(x,a,b) {\rm d} x $$ where $$ k(x,a,b) \triangleq \frac{1}{\sqrt{2 \pi t}}\sum \limits_{j \in \mathbb{Z}} f_j(x,a,b) $$ you deduce directly that the joint density is $$\hat{k}(x,a,b)=-\frac{\partial^2 k}{\partial a \, \partial b} (x,a,b) $$ for $a<x<b$, as long as this partial derivative is continuous (which it is).

Yuval Peres
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  • Thank you very much. But what happens when $a \geq x$ or $x \geq b$? If I am not mistaken $\hat k$ does not vanish. I am not sure but can we still write for any continuous and bounded function $F$ on $\mathbb{R}^3$ that $\mathbb{E}F(B_t,s_t,S_t) = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty F(x,a,b) \hat k(x,a,b) {\rm d} x {\rm d} a {\rm d} b$? – megaproba Jun 21 '22 at 07:56
  • I should have added that $\hat{k}(x,a,b)=0$ if $x \notin[a,b]$. Indeed the formula in your comment is correct, and follows directly from the first formula in your post if $F$ is the indicator of an interval, using the fundamental theorem of calculus twice. The general case then follows by approximation – Yuval Peres Jun 21 '22 at 14:47
  • Thank you very much. – megaproba Jun 22 '22 at 02:46
  • Do you know how to accept an answer? – Yuval Peres Jun 22 '22 at 06:55
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    Is it the green stuff? that should be ok now – megaproba Jun 22 '22 at 07:39