Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer
    
Mathematica simplifies your first equation to Erf[x/Sqrt[2*t]], which seems simpler. Is this a valid simplification? –  barrycarter Nov 10 '10 at 3:50
    
Erf[x/Sqrt[2*t]] differs from my first equation only by a simple change of variable, hence it is not a simplification. However, the representation in terms of (the error function) Erf might be more useful, since Erf is a "common function". –  Shai Covo Nov 10 '10 at 13:58
    
Wasn't being critical, just hoping to find a closed form for the second, more general, integral. –  barrycarter Nov 10 '10 at 16:39
    
As I noted at the end of my edited answer, maybe one can simplify the second (double) integral. I will think about it a little. –  Shai Covo Nov 10 '10 at 17:19
1  
@ronaf: It is quite obvious that the process is supposed to be a Brownian motion. In fact, it is quite difficult to construct a counterexample (assuming that $X$ is continuous and $X_0 = 0$). –  Shai Covo Nov 10 '10 at 17:34

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.

share|improve this answer
    
What you did above seems to correspond to the density function (rather than distribution function) of the maximum between $t_1$ and $t_2$. This is why your final answer is in terms of one integral only (erf); however, for the distribution function the answer is still in terms of a double integral. –  Shai Covo Nov 10 '10 at 21:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.