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Taking a class in financial derivatives (book we use is Tomas Björk'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 non-differentiable 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?

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    $\begingroup$ Continuity is a weaker property than differentiability. $\endgroup$
    – quasi
    Apr 8, 2014 at 7:26

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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.

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  • $\begingroup$ 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? $\endgroup$ Apr 8, 2014 at 7:55
  • $\begingroup$ 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 ;) !! $\endgroup$ Apr 8, 2014 at 8:14
  • $\begingroup$ Thanks, will do. Believe the ones in my class will have use of this site as will I! $\endgroup$ Apr 8, 2014 at 8:35
  • $\begingroup$ +1: I think the example of the absolute function is a good one! $\endgroup$
    – vonjd
    Apr 9, 2014 at 15:48
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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)

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    $\begingroup$ this gif is really making me dizzy ^^ $\endgroup$ Apr 9, 2014 at 17:31
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    $\begingroup$ @Probilitator The gif actually looks like some old-school flight simulator in the mountain area :) $\endgroup$
    – Ilya
    Apr 14, 2014 at 9:38
  • $\begingroup$ @Ilya now that you say it - it can no longer be unseen :D $\endgroup$ Apr 14, 2014 at 10:47
  • $\begingroup$ 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? $\endgroup$
    – Ilya
    Apr 14, 2014 at 13:06
  • $\begingroup$ @Ilya: It is always good to be careful, right ;-) Yet I think you described the trivial case of self-similarity. It is a little bit like which function is its own derivative? Well, the exp-function, right? But there is also a trivial case: f(x)=0... Yet this isn't very helpful when you try to solve a differential equation... $\endgroup$
    – vonjd
    Apr 14, 2014 at 14:05
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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.

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    $\begingroup$ +1 well earned for making the answer to this question three-dimensional ;) $\endgroup$ Apr 10, 2014 at 13:02
  • $\begingroup$ @Probilitator: thanks. 3d - is some quant joke I fail to understand? $\endgroup$
    – Ilya
    Apr 10, 2014 at 15:47
  • $\begingroup$ nope - it refers to the three prespectives our answers provide ;) $\endgroup$ Apr 14, 2014 at 7:00
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I know this answer is years later but in case anyone looks this up, I'm hoping I can help.

I can go through all of the mathematical proofs, but instead I'll try to give some intuition.

By way of background, functions that are non-differentiable at any point but continuous at every point was first discovered by Karl Weierstrass. If you have trouble getting your arms around it, you have good company! Even other famous mathematicians decried them - Poincare called them 'monsters', Hermite called them a 'lamentable scourge'. Intuition said that if the function were continuous it would only be non-differentiable at only small parts of the domain.

But the basic ideas are:

A function is continuous at a point if being 'close' to the point means that the function value is close; f(x) is continuous at X if x being close to X implies that f(x) is close to f(X).

A function is differentiable at a point not only if it is 'close', but points close have to be approaching at a uniform speed; velocity is the derivative of position. Being differentiable implies being continuous.

So f(x) = absolute value of x is continuous at x = 0; f(0) is also 0; being close to x=0 means f(x) is also close to 0. But this function is not differentiable at x=0, since approaching 0 from the right comes in with a 'speed' of 1 but from the left is -1.

So what is going on with Brownian motion? The result is kind of intuitive - at each point it 'jumps' with a value from a standard normal. It will still be continuous, because the small interval is sufficient to keep it 'close' to the last point - the 'jump' multiplied by the infinitesimal interval is small enough for the function to stay 'close'. However, it isn't differentiable, because the size of the last 'jump' (on the left of the graph) won't be the same as the jump on the right side, so it isn't approaching at the same speed; on any interval, there will be lots of different 'speeds'.

Now to be a little more rigorous, Brownian motion is non differentiable at every point ALMOST SURELY and continuous ALMOST SURELY; without getting too technical, this means it happens with probability one (but may fail on a set of measure 0).

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  • $\begingroup$ I was reading the earlier answers , where it says that a continuous function (such as brownian motion) can be drawn without taking your pencil off the paper. Not to be pedantic, but is this true for brownian motion? Isn’t the arc length infinite ? So it would take you an infinite time. $\endgroup$
    – dm63
    Feb 2 at 4:13

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