Stochastic Calculus Simplified: Intro to Stochastic Differential Equations - Integration Method

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0:00 Intro
1:37 Couple of Book Recommendations
2:20 Roadmap
3:34 General Form of an SDE
5:41 Solution by Integration/Example 1
8:14 Two Properties of Variance
14:01 Example 2
17:26 Example 3
21:27 How to Verify a Solution
25:38 Exercise!

[@MonkeyDLuffy-gd6se]

this is a cool idea ill watch it later!

[@MathematicalToolbox]

Thank you, sir!

[@balvy06]

These videos are excellent. Can’t wait for the next videos!!!

[@MathematicalToolbox]

Thank you very much for your comment. I had lost some motivation for the series. I'll try to get the next one up soon!

[@yannledu4783]

Problem4:
Xt = 1 + 1/2 t + 1/3 t^3-1/2 Wt^2 + \int_0^t s dWs
E = 1 + 1/3 t^3
Var = 1/2 t^2 + 1/3 t^3
Very sorry to read you got demotivated while doing that series of videos, because it feels really good to just compute, and I think your series is a very fine computational introduction to stochastic calculus. It looks like your noisy pendulum video is a good application of that introductory series, I’ll check it out after.

[@MathematicalToolbox]

Thank you so much for the nice words of encouragement and for your interest in the series. We've got about three or four more videos left, variation of parameters, SODEs (basically the noisy pendulum again, but with less moving parts), and an SPDE.
Looks like your solution is correct, as well as the expectation. I think I made a mistake on my variance, but our solutions are very close, and I've been at it for an hour, so I am sure you're fine.
I definitely made a mistake in requiring you all to find the variance of the solution. I must have swapped out the problems at the last second. There's no way I would ask you guys to compute the variance of this. It's way too much algebra than what I would have wanted for a course like this. I apologize for this, and I respect your resolve.
Here's another (MUCH simpler) problem as an apology:
dX_t = (1/2)dt + (W_t)dW_t
X_0=1
Where the solution is:
X_t=W²_t/2+1
E[X_t]=1+t/2
Var(X_t)=t²/2

[@yannledu4783]

Good to know there are more videos brewing!
As for the variance, using the integral form I gave for Xt, I use the linearity of the variance, supposing all the annoying cross terms vanished, then the deterministic term bunch becomes zero, and I am left with the var(1/2 Wt^2) + var(integral), and the first term is easy, 1/2 t^2, and then by moving var inside the integral in var(\int_0^t s dWs) = \int s^2 var(dWs) because s acts like a constant, and var(dWs) = ds, so I get \int_0^t s^2 ds = t^3/3, thus my result.
The little problem you gave in your reply is indeed simpler.