Determining distribution of maximum of dependent normal variables I have a stochastic variable x with this property: if it's measured at 
t1 and again at t2, then x(t2)-x(t1) has a normal distribution with 
mean 0 and standard deviation Sqrt[t2-t1]. 
I want to find the distribution of the maximum value this variable 
reaches between t1 and t2, or confirm my intuition that this is not 
well-defined. 
My approach: break [t1,t2] into multiple intervals and integrate the 
following (PDF[NormalDistribution[x,y]] = normal probability 
distribution with mu of x and standard deviation of y): 
PDF[NormalDistribution[0,Sqrt[1/4]]][x0]* 
PDF[NormalDistribution[x0,Sqrt[1/4]]][x1]* 
PDF[NormalDistribution[x1,Sqrt[1/4]]][x2]* 
PDF[NormalDistribution[x2,Sqrt[1/4]]][x3] 
where each xi is integrated from -Infinity to m. 
This specific example computes the probability that the maximum on 
[0,1] is less than m by breaking [0,1] into 4 parts. 
Breaking [0,1] into more parts should yield more accurate results, 
although I slightly suspect that the limit diverges. 
Mathematica slows to a crawl even breaking [0,1] into 5 or more parts. 
I've tried replacing the normal distribution with others (uniform, 
DeltaDirac, C/(1+x^2), etc), with no better luck. 
Googling yields many results (this appears to be a "Wiener Process"), 
but I can't find the actual distribution of the maximum anywhere (nor 
does it say anywhere that such a maximum doesn't exist). 
Ultimate goal is to price box options:
https://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option 
 A: You have a stochastic process $\lbrace{X(t):t \geq 0\rbrace}$ with the property that $X(t_2)-X(t_1) \sim N(0,t_2 - t_1)$, which is quite obviously supposed to be Brownian motion (BM). Suppose first that you want to find the distribution function of the running maximum $M(t)=\mathop {\max }\limits_{0 \le s \le t} X(s)$ (the maximum exists, since BM has continuous sample paths). There is a very simple formula for that, namely:
$$
{\rm P}(M(t) \le x) = \sqrt {\frac{2}{{\pi t}}} \int_0^x {e^{ - u^2 /(2t)} {\rm d}u}, \;\; x \geq 0. 
$$
The situation is a little more complicated if you want to find the distribution function of $
M(t_1 ,t_2 ) = \mathop {\max }\limits_{t_1  \le s \le t_2 } X(s)$ (i.e., the maximum of $X$ over the time interval $[t_1,t_2]$). For this purpose, we need to condition on the initial value $X(t_1)$. Since $X(t_1) \sim N(0,t_1)$, it has density function $f(u;t_1) = \frac{1}{{\sqrt {2\pi t_1 } }}e^{ - u^2 /(2t_1 )}$, and by the law of total probability we have
$$
{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty  {{\rm P}(M(t_1 ,t_2 ) \le x|X_{t_1 }  = u)f(u;t_1) {\rm d}u}. 
$$
Now, as follows from basic properties of BM, conditioned on $X_{t_1}=u$, $M(t_1 ,t_2 )$ can be replaced by $u + M(0,t_2 - t_1)$, i.e. by $u + M(t_2 - t_1)$ (more precisely, by $u$ plus an independent copy of $M(t_2 - t_1)$, which is independent of $X_{t_1}$). This leads to
$$
{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty  {{\rm P}(M(t_2  - t_1 ) \le x - u)f(u;t_1 ){\rm d}u}. 
$$
Finally, since $M(t_2  - t_1 )$ cannot be negative, we have to integrate only from $-\infty$ to $x$. That is,
$$
{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^x {{\rm P}(M(t_2  - t_1 ) \le x - u)f(u;t_1 ) {\rm d}u}, \;\; x \in {\bf R}.
$$
So, we have a double integral with an elementary integrand. Maybe one can simplify it. Also, maybe one can find the result in the literature. 
A: OK, I think I've found a closed form (someone double check my math here).
If the CDF of x (for x>0) is:
$\text{erf}\left(\frac{x}{\sqrt{2} \sqrt{t}}\right)$
the PDF (for x>0) is clearly:
$\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t}}}{\sqrt{t}}$
To compute the probability that the maximum between t1 and t2 is exactly x, we compute the probability that x[t1] == y (normal distribution w SD of Sqrt[t1]):
$\frac{e^{-\frac{y^2}{2 \text{t1}}}}{\sqrt{2 \pi } \sqrt{\text{t1}}}$
and then multiply the probability that the max will be x-y between t1 and t2 (only valid for y<=x):
$\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(x-y)^2}{2
(\text{t2}-\text{t1})}}}{\sqrt{\text{t2}-\text{t1}}}$
So we integrate this wrt y from -Infinity to x:
$\frac{e^{-\frac{(x-y)^2}{2 (\text{t2}-\text{t1})}-\frac{y^2}{2\text{t1}}}}{\pi  \sqrt{\text{t1}} \sqrt{\text{t2}-\text{t1}}}$
and finally get:
$\frac{e^{-\frac{x^2}{2 \text{t2}}} \left(\text{erf}\left(\frac{x\sqrt{\frac{1}{\text{t1}}-\frac{1}{\text{t2}}}}{\sqrt{2}}\right)+1\right)}{\sqrt{2 \pi } \sqrt{\text{t2}}}$
I realize all of this was in the previous post, but I wanted to play around with LaTex.
