# Paul Wilmott on Quantitative Finance, Chapter 3, First Stochastic Differential Equation finance chapter 3

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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#### Related Posts 12/03/2022 - 4:50 Chiều

Great job! 12/03/2022 - 4:50 Chiều

If this stuff works then why did so few people predict the crash of 2007-2008? 12/03/2022 - 4:50 Chiều

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? 12/03/2022 - 4:50 Chiều

What is the name of this equation, so that I can look up a more detailed derivation? 12/03/2022 - 4:50 Chiều

returns are NOT normally distributed in the real world. 12/03/2022 - 4:50 Chiều

its very hard to understand these processes so thank you for your take 12/03/2022 - 4:50 Chiều 12/03/2022 - 4:50 Chiều

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? 12/03/2022 - 4:50 Chiều

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)? 12/03/2022 - 4:50 Chiều

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. 12/03/2022 - 4:50 Chiều

You mean that was a joke almost everywhere. I lol'd. 12/03/2022 - 4:50 Chiều

Thanks for explain in that way!!!! I can understand now 🙂 12/03/2022 - 4:50 Chiều

Thanks! helped clear that up 12/03/2022 - 4:50 Chiều

meybe i can help u with that a litle bit.
u can talk with me on skype-
look for alonsela972 from israel
🙂 12/03/2022 - 4:50 Chiều

@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. 😉 12/03/2022 - 4:50 Chiều

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