26829th poster gets a cookie (cookie thread (Part 7)) (Part 11)

General
#23917

All differentiable functions are Liepschtiz continuous, and all Liepschtiz continuous functions are continuous.
Brownian paths are continuous but not Liepschtiz continuous.

#23918

We are proving Brownian paths are not Liepschitz continuable via proof by contradiction:

There is some s between 0 and 1 (inclusive) such that |B_t - B_s| <= C \times |t - s| for all t such that |t - s| <= 3/n

#23919

P(A_1 \cup A_2\cup A_3…\cup A_n) = 0
A_1 \subset A_2 \subset A3…\subset A_n

Prove P(A_n) \to 0 \quad\text{as } n \to \infty.

We don’t know what P(A_n) is; we can estimate it with Monte Carlo, but there’s a better way.

A_n <= G_n
0 <= P(A_n) <= P(G_n); we know P(G_n) converges to 0, so P(A_n) must also converge to 0 via the squeeze theorem.

#23920

y_{k,n} = \max\left\{ \left|B\left(\frac{k}{n}\right) - B\left(\frac{k-1}{n}\right)\right|, \left|B\left(\frac{k+1}{n}\right) - B\left(\frac{k}{n}\right)\right|, \left|B\left(\frac{k+2}{n}\right) - B\left(\frac{k+1}{n}\right)\right| \right\}

#23921

I will retroactively LaTeX all of this once I get back to my dorm.

1 Like
#23922

G_n = \left\{\omega : y_{k,n}(\omega) \leq \frac{5C}{n} \text{ for at least one } k\right\}

#23923

this is sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeew

3 Likes
#23924

“We’re saying probability of 1 because I want my proof to sound more rigorous.” (slightly paraphrased?)

1 Like
#23925

@iamagummybear

#23926

P(G_n) <= \sum_{k=1}^{n-2} P\left(y_{k,n} \leq \frac{5C}{n}\right)

#23927

l.m.a.o

#23928

P(\max(X_1, X_2, X_3) <= a) = P(X_1 <= a, X_2 <= a, X_3 <= a)
If X_1, X_2, X_3 are independent of each other, then we can multiply them by each other.
But if they’re the same distribution, they are identical:
P(X_1 <= a) \times P(X_2 <= a) \times P(X_3 <= a) = (P(X_1 <= a))^3

#23929

God help me on the integral he just drew.

#23930

this is crooked

#23931

\int_{-1/n}^{1/n} \frac{1}{\sqrt{2\pi/n}} e^{\frac{z^2}{2/n}}

#23932

This somehow breaks down to (\frac{2}{\sqrt{n}})^3.

#23933

P(G_n) <= \sum_{k=1}^{n-2} P\left(y_{k,n} \leq \frac{5C}{n}\right) = nP(y_k,n <= \frac{5C}{n}) <= n \times (\frac{2}{\sqrt{n}})^3 = \frac{8}{\sqrt{n}}, which converges to 0.

#23934

this is honestly heartbreaking

4 Likes
#23935

The curse of playing mafia irl.

#23936

Ok, we’re taking a break.
If people leave, I will probably slip out.