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Chapter 9

C

Chirag Wadhwa

Member
Can anyone explain what filtration and natural filtration imply and represents?

What do drift coefficients and diffusion mean in terms of Brownian Motion?

In the proof of non-differentiability of Sample paths, can anyone explain the case for t>s?
How (W_t-W_s)/(t-s)-dW_t/dt~N(-dW_t/dt,1/t-s) and the explanation afterwards?

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Unless the acted notes have changed a lot since I took the exams, I seem to recall there being quite a good explanation of filtrations in there. A natural filtration is just something that records the history of the stochastic process. Note when I did the exams, I got the distinct impression you don't really need to understand these things properly as you never get asked questions about them in the exams. Perhaps this has changed - but I suspect not because I don't think most student actuaries are comfortable enough with measure theory to really understand all this stuff. I wouldn't worry about it.

A Brownian motion is a continuous random walk that wiggles along going up and down over time. If the drift is positive, then on average over time the Brownian motion wiggles upwards. It might have long periods where it wiggles down, but in the long run it will wiggle its way up. The diffusion just controls how big the wiggles tend to be, but not which the average long run direction in which the process wiggles.

The case for t>s shows that the thing you want to be smaller than an arbitrary epsilon when t-s approaches 0 has a normal distribution with variance that gets massive, infinite in fact, as t-s approaches 0. The probability of a normally distributed random variable with essentially infinite variance having smaller magnitude than any epsilon is 0. That's what the notes mean when they say the expression holds 'almost surely' (which is a technical term meaning "with probability 1").

I think you asked somewhere else on the forum a similar question about how some expression involving W terms had normal distribution with certain parameters. Have a look at my answer there to see if it clarifies things - and also check again you understand the definition of a Wiener process / Brownian motion and what that has to do with the normal distribution.

If you'd like more explanation it would be helpful to understand which specific bits you don't understand, as I am not really sure whether in attempting to provide explanations to broader questions I am just repeating explanations which are already there in the acted notes or not.
 
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CapitalActuary said:
Unless the acted notes have changed a lot since I took the exams, I seem to recall there being quite a good explanation of filtrations in there. A natural filtration is just something that records the history of the stochastic process.
Click to expand...

Hi -thanks to @CapitalActuary for simple explanations above.

Two related questions on filtrations from the end of that Chapter:

Question 1: Page 18 introduces the concept of "Ft-measurable" by specifying that "a stochastic process, Xt is "Ft-measurable" if the value of the process is known at time t."

Does this mean the current value of the process or every historic value of the process (including the current value), which would fit better with the quoted generic definition of filtration.

Question 2: Final few sentences of the chapter's core reading says that "the stochastic process [exp(Lambda*Wt - 0.5*t*Lambda^2)] is a Ft^W - martingale." What does this expression 'Ft^W - martingale' mean? (There's no Acted explanation)
 
Thanks @CapitalActuary for the great post about Brownian motion.
Question 1: Ft will contain the entire history of the process up to and including time t, so every historical value will also be included.
Question 2: This phrase means that the future expected value of the process, conditional on the filtration generated by W, is equal to the current value. Notice that if the conditional expectation was executed with respect to the filtration of another process then the result would be different. Apart from this instance in the Core Reading, I think that it is always assumed that Ft is generated by the process that's under investigation.
 
The solutions in Acted notes to Chapter 9 Practice Question 9.1(ii) uses the MGF of a normal distribution, M(t) = exp(mu*t+0.5*sigma^2*t^2) "calculated at point -2μ to determine the expectation" (ie. E[exp-(2mu*(B_t-B_s)] = exp(0.5*(t-s)*(4mu^2)

Please can someone explain this step -
  1. Seems like we're inserting mu=0, t= -2*mu and sigma^2 = t-s into the generic formula M(t) = exp(mu*t+.5*sigma^2*t^2). Unsure why t is calculated as -2*mu?2.
  2. Why is the MGF of a normal distribution even relevant for the expectation of geometric brownian motion? I noticed a different approach, based on E(X) of a logNormal, is used in Subject 103, April 2001, Question 7(ii) brought in the Acted tutorials/online classroom.
  3. What would the equivalent approach be to solve the expectation of Arithmetic brownian motion?
Thanks very much in advance.
 
The MGF of a random variable X is E[exp(tX)]. X is normal if and only if exp(tX) is lognormal. Hence the link between the mean of a lognormal and the MGF of a normal.

Geometric Brownian motion results in lognormally distributed moves, which I think the course notes probably prove. (If not, apply Ito’s lemma to log(S) where S is a GBM to get the result.)
 
I’m afraid I don’t have any tips to share. It’s a bit unfortunate, but these are the standard parameters in the literature about this stuff.

Sometimes when I’ve written stuff out in the past I use other letters to keep track of things better, but I’m not even consistent with what I use.

The only thing I can say is that getting more familiar with the concepts helps to spot these issues. E.g. it’s something I’ve internalised that a GBM with drift a and dispersion b has lognormal moves with parameters a-b^2/2 and b. Hence I see a someone thinking the lognormal parameters are a and b it just sticks out to me.

I don’t know if this will help or confuse things, but there is a nice discussion of geometric bs arithmetic returns here: https://qoppac.blogspot.com/2017/02...ins the weird title,them to pay higher costs. which is related to GBM vs BM.
 
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