In Chapter 3 I learned how to model asset prices using normal returns, both for discrete time and for continuous time using a Wiener process. The first stochastic differential equation! .
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Paul Wilmott on Quantitative Finance, Chapter 3, First Stochastic Differential Equation
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Paul Wilmott on Quantitative Finance, Chapter 3, First Stochastic Differential Equation
finance chapter 3
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16 comments
Great job!
If this stuff works then why did so few people predict the crash of 2007-2008?
Why do you need to do every other day? Why wouldn't you just do two time steps?
S = stock price
u = mean return
dt = time step
g = standard deviation of return
p = randomness
R=(S2-S1)/S1, where R is the return
Then,
S2=S1+S1(u*dt+g*p*dt)
S3=S2+S2(u*dt+g*p*dt)
What is the point of changing R, and adding in the square root of two?
Also, (2dt)^1/2 = sqrt(2)sqrt(dt), not sqrt(2)dt, whats up with that?
What is the name of this equation, so that I can look up a more detailed derivation?
returns are NOT normally distributed in the real world.
its very hard to understand these processes so thank you for your take
nice video, and the comments from you guys are very helpful.
This is neat to know. What's the proven application of stochastic calculus to trading and making investment decisions? How can it help make profitable decisions?
why go the long way to prove that Ri^ = Ri + R(i+1)?
Can we not simply break [S(i+2) – S(i)]/S(i) to S(i+2) – S(i+1) + S(i+1) – S(i)]/S(i) i.e. simply R(i+1) + R(i)?
A different example: dx^2 versus dx. Which is bigger? As dx goes to zero, the dx^2 gets really small. The dx just gets small at the normal rate. So the dx dominates the dx^2.
Similarly dt^1/2 dominates dt.
You mean that was a joke almost everywhere. I lol'd.
Thanks for explain in that way!!!! I can understand now 🙂
Thanks! helped clear that up
meybe i can help u with that a litle bit.
u can talk with me on skype-
look for alonsela972 from israel
🙂
@alonsela972 Thanks, that actually makes some sense. Although I still *almost surely* don't understand all the technical requirements of Wiener processes. That was almost a joke. 😉
hi
u have an answer for u:
if z is a variable distrebuting normaly with 0 mean and std 1then:
x=z*dt^0.5 E[x] =E[z*dt^0.5]=E[z]*E[dt^0.5]=0
and VAR[z]=VAR[z*dt^0.5]=E[(z*dt^0.5)^2]-E^2[z*dt^0.5]=E[z^2*dt]-0=E[z^2]*[dt]=1*dt=dt
therefor x is a wiener process whith E[x]=0 and V[x]=dt
i hope it helps u
alon