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Almost Sure @almostsure.bsky.social · 4d

I posted this as a YouTube short, so I’ll refer to my description there

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Almost Sure @almostsure.bsky.social · 5d

Helter skelter

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Almost Sure @almostsure.bsky.social · 16d

New YouTube video: Wild Expectations!

This looks at weird properties of conditional expectations, such as two random variables being bigger than each other 'on average'

youtu.be/4ZwRXVVepj8?...

Wild Expectations

YouTube video by Almost Sure

youtu.be

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Almost Sure @almostsure.bsky.social · 21d

Whatever you do, 𝙛𝙤𝙧 𝙩𝙝𝙚 𝙡𝙤𝙫𝙚 𝙤𝙛 𝙂𝙤𝙙, do not ask random variables to be Lebesgue measurable!

If you were so stupid to do that, then 𝙮𝙤𝙪 𝙬𝙤𝙪𝙡𝙙 𝙣𝙤𝙩 𝙚𝙫𝙚𝙣 𝙗𝙚 𝙖𝙗𝙡𝙚 𝙩𝙤 𝙖𝙙𝙙 𝙧𝙖𝙣𝙙𝙤𝙢 𝙫𝙖𝙧𝙞𝙖𝙗𝙡𝙚𝙨 𝙩𝙤𝙜𝙚𝙩𝙝𝙚𝙧.

Absolutely continuous rvs summing to singular rv

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Almost Sure @almostsure.bsky.social · Aug 26

initial->unitial (autocorrect!)

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Almost Sure @almostsure.bsky.social · Aug 26

where unit element of each sub-algebra is not necessarily unit of the full algebra (just a projection in general)

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Almost Sure @almostsure.bsky.social · Aug 26

There's the question of if the freely generated product exists, according to the different types of independence.

It does for commutative & free products, as products of initial algebras.

Doesn't look like it does for boolean independence. Not as until algebras though.

Maybe as non-initial ones

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Almost Sure @almostsure.bsky.social · Aug 26

interesting. I'm only familiar with commutative & free independence

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Almost Sure @almostsure.bsky.social · Aug 23

New video released on mixing quantum and classical probabilities.

The Algebra of Mixed Quantum States

youtu.be/K6h62Gr0nwg

The Algebra of Mixed Quantum States

YouTube video by Almost Sure

youtu.be

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Almost Sure @almostsure.bsky.social · Jul 3

also, how can they tell it was due to a tyre blowing out? It was at night and the car was wrecked and burned out. And, I heard that these cars have run-flat tyres. I think it’s a simple case of losing control at speed

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Almost Sure @almostsure.bsky.social · Jul 3

I’ve been wondering what happened. I had a tyre blow out on the motorway before, no big deal - but they were ‘run-flat’ so I could keep driving almost normally.

Would the same thing with normal tyres result in loss of control, or was this a result of reckless driving?

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Almost Sure @almostsure.bsky.social · Jul 3

new-ish…actually a few weeks old now. I forgot to post here when I initially launched it on YouTube

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Almost Sure @almostsure.bsky.social · Jul 3

New video released on the Gaussian correlation inequality.

This unexpected proof shocked mathematicians!

youtu.be/WJGR1oc6Gxo?...

This unexpected proof shocked mathematicians

YouTube video by Almost Sure

youtu.be

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Almost Sure @almostsure.bsky.social · Apr 17

Unfortunately its not as nice as I first thought. And I made a mistake in the writeup - when you multiply diffusions then what you get need not be Markov.

Maybe there is a more natural way of fitting martingales.

bsky.app/profile/almo...

Almost Sure @almostsure.bsky.social · Apr 17

I am not sure if the distribution is symmetric under

X(mu,t)->1-X(1-mu,t).

It is for individual times, as it matches Dirichlet distribution, but probably not for the entire paths wrt t. Which is disappointing. Maybe it can be modified?

Almost Sure @almostsure.bsky.social · Apr 15

Simulating the martingales X(μ,t) which are beta distributed and martingale wrt t.

X(μ,t) ~ Beta(μ(1-t)/t, (1-μ)(1-t)/t)

The (1-t)/t scaling is so that on range 0<t<1 we cover entire set of Beta distributions.

For each t, X(μ_{i+1},t)-X(μ_i,t) have Dirichlet distribution.

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Almost Sure @almostsure.bsky.social · Apr 17

Also I think X(mu,t) is not markov for individual mu (it is for mu as a whole). Rather, it is the ratios of the jumps wrt mu: X(mu-,t)/X(mu+,t) which are markov (i.e., diffusions)

Almost Sure @almostsure.bsky.social · Apr 17

I am not sure if the distribution is symmetric under

X(mu,t)->1-X(1-mu,t).

It is for individual times, as it matches Dirichlet distribution, but probably not for the entire paths wrt t. Which is disappointing. Maybe it can be modified?

Almost Sure @almostsure.bsky.social · Apr 15

Simulating the martingales X(μ,t) which are beta distributed and martingale wrt t.

X(μ,t) ~ Beta(μ(1-t)/t, (1-μ)(1-t)/t)

The (1-t)/t scaling is so that on range 0<t<1 we cover entire set of Beta distributions.

For each t, X(μ_{i+1},t)-X(μ_i,t) have Dirichlet distribution.

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Almost Sure @almostsure.bsky.social · Apr 16

bsky.app/profile/almo...

Almost Sure @almostsure.bsky.social · Apr 16

Here's the method of simulation, and also shows that the joint distribution of X(μ,t) is uniquely determined if we impose independent ratios property wrt μ.

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Almost Sure @almostsure.bsky.social · Apr 16

Here's the method of simulation, and also shows that the joint distribution of X(μ,t) is uniquely determined if we impose independent ratios property wrt μ.

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Almost Sure @almostsure.bsky.social · Apr 15

here's another plot (more 'mu' points, fewer time points.

And, gamma(1) process scaled to hit 1 at time 1 (time parameter mu to compare). Corresponds to t=0.5 in the surface plot. You can see its dominated by a few large jumps.

gamma(40) process is shown in the 3rd plot, corresponds with t=0.024

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Almost Sure @almostsure.bsky.social · Apr 15

Simulating the martingales X(μ,t) which are beta distributed and martingale wrt t.

X(μ,t) ~ Beta(μ(1-t)/t, (1-μ)(1-t)/t)

The (1-t)/t scaling is so that on range 0<t<1 we cover entire set of Beta distributions.

For each t, X(μ_{i+1},t)-X(μ_i,t) have Dirichlet distribution.

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Almost Sure @almostsure.bsky.social · Apr 14

Probability fact:

A sequence X_0,X_1,X_2,…,X_i,… of Gamma(a_i) rv’s has independent increments

X_1-X_0,X_2-X_1,…

if and only if it has independent ratios

X_0/X_1,X_1/X_2,…

in which case a_{i+1}>=a_i and,

X_{i+1}-X_i~Gamma(a_{i+1}-a_i)
X_i/X_{i+1}~Beta(a_i,a_{i+1})

Almost Sure @almostsure.bsky.social · Apr 12

Probability fact:

If X,Y are independent Gamma(a), Gamma(b) random variables then

X/(X+Y), X+Y

are independent Beta(a,b), Gamma(a+b) rvs.

Equivalently: if X,Y are independent Beta(a,b), Gamma(a+b) random variables then

XY, (1-X)Y

are independent Gamma(a), Gamma(b) rvs.

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Almost Sure @almostsure.bsky.social · Apr 13

Either way, it’s lightning fast using my new Schrödinger’s Cat8 cables

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Almost Sure @almostsure.bsky.social · Apr 12

Probability fact:

If X,Y are independent Gamma(a), Gamma(b) random variables then

X/(X+Y), X+Y

are independent Beta(a,b), Gamma(a+b) rvs.

Equivalently: if X,Y are independent Beta(a,b), Gamma(a+b) random variables then

XY, (1-X)Y

are independent Gamma(a), Gamma(b) rvs.

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Almost Sure @almostsure.bsky.social · Apr 12

My extension of Hermite-Hadamard:

If c = (pa+qb)/(p+q) for p,q > 0 then

f(c)≤M(t)≤(pf(a)+qf(b))/(p+q)

where M(t) is the average value of f under Beta(qt,pt) distribution scaled to interval [a,b].

M(t) is ctsly decreasing from

M(0)=(pf(a)+qf(b))/(p+q)
to
M(∞) = f(c)
pbs.twimg.com/media/GoSUpd...

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Almost Sure @almostsure.bsky.social · Apr 12

Ok, I managed to prove this!

So Beta(at,bt) is decreasing in the convex order and can find reverse martingale (continuous Itô diffusion)

X(t)=E[X(s) | {X(u),u >=t}]
(all s < t)

with marginals

X(t)~Beta(at,bt)

Almost Sure @almostsure.bsky.social · Apr 12

Question: fixed a, b > 0, are distributions Beta(at,bt) decreasing in convex order over t > 0?

Equivalent to existence of a reverse-time martingale X_t ~ Beta(at,bt).

Equivalently:

-(d/dt)E[(x-X_t)_+] >=0

for all 0 < x < 1.
Plots suggest so: parameterised as s=a+b,mean=a/s

Plot of derivatives showing positive, as required for convex order

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