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Almost Sure

@Almost_Sure • 8,927 subscribers

George Lowther, Author of Almost Sure blog, on maths, probability and stochastic calculus. Also on YouTube https://t.co/VyOijwbe9l

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Quick "proof" of Taylor expansion using operator exponential

Quick "proof" of Taylor expansion using operator exponential

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"everything is linear if you zoom in close enough..."
0:51
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"everything is linear if you zoom in close enough..."

Almost Sure

1,019,346 просмотров • 7 месяцев назад

Animating the solution to this surprisingly interesting fruity math problem:
1:41
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Animating the solution to this surprisingly interesting fruity math problem:

Almost Sure

88,444 просмотров • 3 месяцев назад

in quantum physics position X and momentum P can’t be simultaneously measured. But, linear combinations aX+bP 𝘢𝘳𝘦 observable, and their distributions can be used to back out the joint X,P probability density. The Wigner distribution but, it is 𝘸𝘦𝘪𝘳𝘥
1:34
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in quantum physics position X and momentum P can’t be simultaneously measured. But, linear combinations aX+bP 𝘢𝘳𝘦 observable, and their distributions can be used to back out the joint X,P probability density. The Wigner distribution but, it is 𝘸𝘦𝘪𝘳𝘥

Almost Sure

70,143 просмотров • 5 месяцев назад

What software do math YouTubers use to do the fancy math animations?
0:39
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What software do math YouTubers use to do the fancy math animations?

Almost Sure

274,318 просмотров • 2 лет назад

stylistic question. In YouTube, color coding terms in math formulas. Yes, or no?
0:52
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stylistic question. In YouTube, color coding terms in math formulas. Yes, or no?

Almost Sure

43,711 просмотров • 4 месяцев назад

Let X = exp(A²/2) and Y=exp(B²/2) where A,B are joint normal with mean 0, variance 1, and correlation ρ (between 0 and 1). Then, E[X|Y] = Y/ρ > Y E[Y|X] = X/ρ > X So X is greater than Y 'on average', and Y is greater than X on average. Here's a proof of that
1:16
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Let X = exp(A²/2) and Y=exp(B²/2) where A,B are joint normal with mean 0, variance 1, and correlation ρ (between 0 and 1). Then, E[X|Y] = Y/ρ > Y E[Y|X] = X/ρ > X So X is greater than Y 'on average', and Y is greater than X on average. Here's a proof of that

Almost Sure

31,334 просмотров • 10 месяцев назад

A paradox of conditional expectations applied to the Ornstein-Uhlenbeck process
1:36
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A paradox of conditional expectations applied to the Ornstein-Uhlenbeck process

Almost Sure

14,994 просмотров • 10 месяцев назад

From the Riemann zeta function to Xi and the functional equation
0:29
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From the Riemann zeta function to Xi and the functional equation

Almost Sure

13,201 просмотров • 9 месяцев назад

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
0:22
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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

Almost Sure

11,759 просмотров • 1 год назад

2025 as a sum of cubes: Nicomachus's theorem, an animated proof
1:06
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2025 as a sum of cubes: Nicomachus's theorem, an animated proof

Almost Sure

12,908 просмотров • 1 год назад

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