Almost Sure
@almostsure.bsky.social
George Lowther. Author of Almost Sure blog on maths, probability and stochastic calculus
https://almostsuremath.com
Also on YouTube: https://www.youtube.com/@almostsure
https://almostsuremath.com
Also on YouTube: https://www.youtube.com/@almostsure
Pinned
What does Riemann Zeta have to do with Brownian Motion?
YouTube video by Almost Sure
youtu.be
New YouTube video uploaded on connections between Riemann zeta and Brownian motion!
What does Riemann Zeta have to do with Brownian Motion?
youtu.be/YTQKbgxbtiw
What does Riemann Zeta have to do with Brownian Motion?
youtu.be/YTQKbgxbtiw
New YouTube video uploaded on connections between Riemann zeta and Brownian motion!
What does Riemann Zeta have to do with Brownian Motion?
youtu.be/YTQKbgxbtiw
What does Riemann Zeta have to do with Brownian Motion?
youtu.be/YTQKbgxbtiw
What does Riemann Zeta have to do with Brownian Motion?
YouTube video by Almost Sure
youtu.be
November 9, 2025 at 8:19 PM
New YouTube video uploaded on connections between Riemann zeta and Brownian motion!
What does Riemann Zeta have to do with Brownian Motion?
youtu.be/YTQKbgxbtiw
What does Riemann Zeta have to do with Brownian Motion?
youtu.be/YTQKbgxbtiw
It's a while since my last YT video upload. Nothing is happening, I am busy working on the next one. Just taking longer than expected (been very busy recent weekends).
Will be on connections between Riemann zeta and Brownian motion.
ETA a few days to a week.
Will be on connections between Riemann zeta and Brownian motion.
ETA a few days to a week.
November 2, 2025 at 1:15 PM
It's a while since my last YT video upload. Nothing is happening, I am busy working on the next one. Just taking longer than expected (been very busy recent weekends).
Will be on connections between Riemann zeta and Brownian motion.
ETA a few days to a week.
Will be on connections between Riemann zeta and Brownian motion.
ETA a few days to a week.
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?...
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
September 23, 2025 at 8:59 PM
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?...
This looks at weird properties of conditional expectations, such as two random variables being bigger than each other 'on average'
youtu.be/4ZwRXVVepj8?...
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.
If you were so stupid to do that, then 𝙮𝙤𝙪 𝙬𝙤𝙪𝙡𝙙 𝙣𝙤𝙩 𝙚𝙫𝙚𝙣 𝙗𝙚 𝙖𝙗𝙡𝙚 𝙩𝙤 𝙖𝙙𝙙 𝙧𝙖𝙣𝙙𝙤𝙢 𝙫𝙖𝙧𝙞𝙖𝙗𝙡𝙚𝙨 𝙩𝙤𝙜𝙚𝙩𝙝𝙚𝙧.
September 18, 2025 at 8:36 PM
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.
If you were so stupid to do that, then 𝙮𝙤𝙪 𝙬𝙤𝙪𝙡𝙙 𝙣𝙤𝙩 𝙚𝙫𝙚𝙣 𝙗𝙚 𝙖𝙗𝙡𝙚 𝙩𝙤 𝙖𝙙𝙙 𝙧𝙖𝙣𝙙𝙤𝙢 𝙫𝙖𝙧𝙞𝙖𝙗𝙡𝙚𝙨 𝙩𝙤𝙜𝙚𝙩𝙝𝙚𝙧.
New video released on mixing quantum and classical probabilities.
The Algebra of Mixed Quantum States
youtu.be/K6h62Gr0nwg
The Algebra of Mixed Quantum States
youtu.be/K6h62Gr0nwg
The Algebra of Mixed Quantum States
YouTube video by Almost Sure
youtu.be
August 23, 2025 at 3:56 PM
New video released on mixing quantum and classical probabilities.
The Algebra of Mixed Quantum States
youtu.be/K6h62Gr0nwg
The Algebra of Mixed Quantum States
youtu.be/K6h62Gr0nwg
New video released on the Gaussian correlation inequality.
This unexpected proof shocked mathematicians!
youtu.be/WJGR1oc6Gxo?...
This unexpected proof shocked mathematicians!
youtu.be/WJGR1oc6Gxo?...
This unexpected proof shocked mathematicians
YouTube video by Almost Sure
youtu.be
July 3, 2025 at 12:47 PM
New video released on the Gaussian correlation inequality.
This unexpected proof shocked mathematicians!
youtu.be/WJGR1oc6Gxo?...
This unexpected proof shocked mathematicians!
youtu.be/WJGR1oc6Gxo?...
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?
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?
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.
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.
April 17, 2025 at 9:14 AM
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?
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?
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.
April 16, 2025 at 2:02 AM
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.
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.
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.
April 15, 2025 at 7:20 PM
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.
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.
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})
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})
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.
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.
April 14, 2025 at 12:03 AM
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})
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})
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.
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.
April 12, 2025 at 5:33 PM
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.
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.
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...
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...
April 12, 2025 at 4:50 PM
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...
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...
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)
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)
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
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
April 12, 2025 at 2:51 PM
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)
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)
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
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
April 12, 2025 at 2:49 PM
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
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
Reposted by Almost Sure
Brownian Motion - A Beautiful Monster
YouTube video by Almost Sure
youtu.be
April 6, 2025 at 2:52 PM
Brownian Motion - A Beautiful Monster
YouTube video by Almost Sure
youtu.be
April 6, 2025 at 2:52 PM
“Reciprocal tariffs”
April 4, 2025 at 9:18 AM
“Reciprocal tariffs”
New Youtube short: Infinite complexity of Brownian motion
An extreme zoom in to Brownian motion
youtube.com/shorts/EvP9G...
An extreme zoom in to Brownian motion
youtube.com/shorts/EvP9G...
Infinite complexity of Brownian motion #maths #probability #stochastic
YouTube video by Almost Sure
youtube.com
March 29, 2025 at 4:25 PM
New Youtube short: Infinite complexity of Brownian motion
An extreme zoom in to Brownian motion
youtube.com/shorts/EvP9G...
An extreme zoom in to Brownian motion
youtube.com/shorts/EvP9G...
Maybe you’ve heard that Henri Poincaré said that Weierstrass' work (on continuous functions without derivatives) is "an outrage against common sense".
That’s what Wikipedia, and many other sources say. But, no!
Looking up the original source, he's actually being positive about Weierstrass' work!
That’s what Wikipedia, and many other sources say. But, no!
Looking up the original source, he's actually being positive about Weierstrass' work!
March 21, 2025 at 4:57 PM
Maybe you’ve heard that Henri Poincaré said that Weierstrass' work (on continuous functions without derivatives) is "an outrage against common sense".
That’s what Wikipedia, and many other sources say. But, no!
Looking up the original source, he's actually being positive about Weierstrass' work!
That’s what Wikipedia, and many other sources say. But, no!
Looking up the original source, he's actually being positive about Weierstrass' work!
Fractional Brownian motion, varying the Hurst parameter H between 0 and 1. H=0.5 corresponds with standard Brownian motion, and the path has fractal dimension 2-H
March 21, 2025 at 1:22 AM
Fractional Brownian motion, varying the Hurst parameter H between 0 and 1. H=0.5 corresponds with standard Brownian motion, and the path has fractal dimension 2-H
Constructing Weierstrass' function.
Continuous, nowhere differentiable, unbounded variation, no intervals of increase or decrease. Fractal dimension 3/2.
All things it has in common with mathematical Brownian motion
Continuous, nowhere differentiable, unbounded variation, no intervals of increase or decrease. Fractal dimension 3/2.
All things it has in common with mathematical Brownian motion
March 21, 2025 at 1:15 AM
Constructing Weierstrass' function.
Continuous, nowhere differentiable, unbounded variation, no intervals of increase or decrease. Fractal dimension 3/2.
All things it has in common with mathematical Brownian motion
Continuous, nowhere differentiable, unbounded variation, no intervals of increase or decrease. Fractal dimension 3/2.
All things it has in common with mathematical Brownian motion
Happy Feigenbaum approximation day!
14/3 = 4.66…
14/3 = 4.66…
March 14, 2025 at 11:13 AM
Happy Feigenbaum approximation day!
14/3 = 4.66…
14/3 = 4.66…
“They don’t know I’m universal for chaotic maps”
February 25, 2025 at 9:12 PM
“They don’t know I’m universal for chaotic maps”
Reposted by Almost Sure
The American Mathematical Society has also started a page to coordinate support for professional mathematics, so far focusing on executive orders impacting the National Science Foundation: www.ams.org/government/g...
AMS :: Take Action
www.ams.org
February 22, 2025 at 2:59 PM
The American Mathematical Society has also started a page to coordinate support for professional mathematics, so far focusing on executive orders impacting the National Science Foundation: www.ams.org/government/g...