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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

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2 Answers 2

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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.

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  • $\begingroup$ Mathematica simplifies your first equation to Erf[x/Sqrt[2*t]], which seems simpler. Is this a valid simplification? $\endgroup$
    – user2469
    Commented Nov 10, 2010 at 3:50
  • $\begingroup$ 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". $\endgroup$
    – Shai Covo
    Commented Nov 10, 2010 at 13:58
  • $\begingroup$ Wasn't being critical, just hoping to find a closed form for the second, more general, integral. $\endgroup$
    – user2469
    Commented Nov 10, 2010 at 16:39
  • $\begingroup$ 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. $\endgroup$
    – Shai Covo
    Commented Nov 10, 2010 at 17:19
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    $\begingroup$ @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 $X_0 = 0$). $\endgroup$
    – Shai Covo
    Commented Nov 10, 2010 at 17:34
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Shai Covo
    Commented Nov 10, 2010 at 21:13

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