# closed form for $p(B_1>x>B_2)$ where $[B_1, B_2]'$ follows a bivariate lognormal dist?

Is there a closed form for $p(B_1>x>B_2)$ where $[B_1, B_2]'$ follows a bivariate lognormal dist:

$$[B_1, B_2]' \sim \text{lognorm} (\boldsymbol \mu, \boldsymbol \Sigma)$$

where $\boldsymbol \mu$ and $\boldsymbol \Sigma$ are known.

I'm trying to find:

$$x = \text{argmax}_x \; p(B_1>x>B_2)$$

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Got something from the answer below? – Did Feb 9 '13 at 10:31

Recall that each $X_i=\log B_i$ is normal, call $f$ the density of $(X_1,X_2)$, and note that $\mathbb P(B_1\gt x\gt B_2)$ is maximal when $\mathbb P(X_1\gt\log x\gt X_2)$ is.
The function $u:t\mapsto u(t)=\mathbb P(X_1\gt t\gt X_2)$ has derivative $$u'(t)=\int_t^{+\infty}f(s,t)\mathrm ds-\int_{-\infty}^tf(t,s)\mathrm ds.$$ The unique point or the several points where $u$ is maximum solve $u'(t)=0$. To actually compute the zeroes of $u'$ analytically does not seem doable. Even uniqueness, in the general case, does not seem easy to establish.
did you mean "... is maximal when $\mathbb P(X_1\gt\log x\gt X_2)$"? – shuaiyuancn Feb 11 '13 at 18:36
$B_1$ and $B_2$ are correlated. Does the equation $u'(t)$ consider that? – shuaiyuancn Feb 11 '13 at 18:39
One should read $P(X_1\gt\log x\gt X_2)$, obviously. // Yes the formula for $u'(t)$ takes into account the fact that $(B_1,B_2)$ are not independent (note the density $f(s,t)$, which is not a priori a product $f_1(s)f_2(t)$). – Did Feb 11 '13 at 18:46