Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

Taking a class in financial derivatives (book we use is Tomas Bjork´s Arbitrage theory in continuous time) but can´t understand the exact meaning of how the Wiener process is defined. In the book one can read: "the Wiener process will be a continuous function of time which is nondifferentiable at every point. This a typical trajectory is a continuous curve consisting entirely of corners and it is of course quite impossible to draw a figure of such an object." Just by looking at a trajectory of a Wiener process I´ll say it is nowhere differentiable and thus non-smooth and non-continuous but here he says it is still continuous?

share|improve this question
4  
Continuity is a weaker property than differentiability. –  quasi Apr 8 at 7:26
    
Ah, that easy. Thanks! –  user2069136 Apr 8 at 7:52
    
@user2069136: If one of the answers were helpful to you please accept one of them - Thank you :-) –  vonjd Apr 13 at 11:00
1  
They all were! Rather new here and didn´t know that they needed to be accepeted. Thanks! –  user2069136 Apr 13 at 12:58

3 Answers 3

up vote 2 down vote accepted

The most basisc understanding of continuity of curve is:

You can draw it with a pen/pencil without lifting your hand. Thus the curve has no jumps that will force you to raise/shift your palm in order to continue drawing.

The function $f(x)=|x|$ is continuous but not differentiable at the origin. If you look at the relevant Wikipedia entries on continuity and differentiability the difference will become clear.

Also note that the continuity of the Brownian-Motion is the main reason why hedging in continuous time works so well theoretically. If one introduces jumps thus making the paths no longer continuous the theory becomes much more elaborate and complicated.

share|improve this answer
    
Thanks, should have googled it better but thought there was something special about it. With jumps, you mean like if introducing dividends or just sudden larger jumps in share price? –  user2069136 Apr 8 at 7:55
    
Dividends can work like jumps if you don't pay them out continously. E.g. if dividends are only paid once a year they will have the effect of a jump - the stock price will go from one value to another with no values in between. You can also always introduce stochastic jumps - just google "Jump diffusion models for option pricing" - !! also spread the word about this site ;) !! –  Probilitator Apr 8 at 8:14
    
Thanks, will do. Believe the ones in my class will have use of this site as will I! –  user2069136 Apr 8 at 8:35
    
+1: I think the example of the absolute function is a good one! –  vonjd Apr 9 at 15:48

Think of the Wiener process as a curve into which you could zoom in ever deeper and deeper and it will still be completely wiggly (= a fractal). That means that even if you tried to put a tangent line onto it, it would find no stable support (= no differentiability) - yet the whole curve is completely closed, i.e. could be drawn without raising your pen (= continuity):

enter image description here

(Source: http://en.wikipedia.org/wiki/Wiener_process)

share|improve this answer
3  
this gif is really making me dizzy ^^ –  Probilitator Apr 9 at 17:31
1  
+1 for mentioning fractals ;) –  Probilitator Apr 9 at 17:31
1  
@Probilitator The gif actually looks like some old-school flight simulator in the mountain area :) –  Ilya Apr 14 at 9:38
    
@Ilya now that you say it - it can no longer be unseen :D –  Probilitator Apr 14 at 10:47
    
I would be careful with such explanation, though. For example, a straight line is extremely self-similar on various scales, however it is perfectly smooth. Also, one can say that smoothness is exactly local similarity to straight lines, isn't it? –  Ilya Apr 14 at 13:06

There are many examples of "non-random" curves that are continuous everywhere, and yet differentiable nowhere. For example, the one defined by the formula $$ f(x) = \sum_{k=0}^\infty 2^{-k}\cos(2^kx). $$ You may think of it as a limit of partial sums $f_n(x) =\sum_{k=0}^n 2^{-k}\cos(2^kx)$. Each $f_n$ is differentiable, and consists of a combination of sinusoids with different frequency and magnitude. As $n\to\infty$ there are more and more frequencies present, which makes tangent lines (very much related to differentiability) behave poorer and poorer. In the limit, there is no single point which admit a tangent line.

share|improve this answer
1  
+1 well earned for making the answer to this question three-dimensional ;) –  Probilitator Apr 10 at 13:02
    
@Probilitator: thanks. 3d - is some quant joke I fail to understand? –  Ilya Apr 10 at 15:47
    
nope - it refers to the three prespectives our answers provide ;) –  Probilitator Apr 14 at 7:00

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.