0
$\begingroup$

This video, from about 6 to 12 minutes: https://youtu.be/qdbkvD4N-us

I feel like I’m following him ok, but then at the end his f(t,B(t)) has become an f(t,x) and there is no B(t) in his result, so it appears to be suddenly deterministic? If he has made a mistake, I’d appreciate a correction, or if someone can explain to me how I’m supposed to understand the randomness one would expect at some X of t

$\endgroup$
  • $\begingroup$ At last you just see the definition of a function $f$ that transforms two Variables. If you set transform $t$ and $B_t$ according to $f$ you will get $X_t$ $\endgroup$ – Sanjay Jul 31 '19 at 0:59
  • $\begingroup$ I do accept that this is a better fit for math stackexchange, and I appreciate everybody’s time. $\endgroup$ – Wriiight Aug 1 '19 at 22:23
0
$\begingroup$

So for everyone's benefit, this is an MIT OCW video 21.Stochastic Differential Equation, and the professor is explaining the solution of the GBM SDE:

$dX(t)=\mu X(t)dt+\sigma X(t)dB(t)$

He guesses a solution of the form:

$X(t)=f\left(t,B(t)\right)$

And then applies the Ito's lemma etc, and then write the solution as:

$f(t,X)=X_0 e^{\sigma X +\left(\mu-\frac{1}{2}\sigma^2 \right)t}$

So seems like X is used in two different senses, and this is the context to the question. The solution should indeed be:

$X(t)=f\left(t, B(t)\right)=X_0 e^{\sigma B(t) +\left(\mu-\frac{1}{2}\sigma^2 \right)t}$

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, I’m just starting to get the hang of stochastic calc and having the Xt and Bt swapped out on me threw me rather hard, especially where earlier he is taking partials wrt x and t, and I’m sure it wasn’t really a partial wrt dBt at that point. $\endgroup$ – Wriiight Aug 1 '19 at 22:22
  • $\begingroup$ Steady is the way forward! keep it up! $\endgroup$ – Magic is in the chain Aug 3 '19 at 12:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.