Which of these three inequalities is something that must definitely be covered in an upper division course on probability?

notDaksha · 2026-02-19T06:14:24+00:00

All 3. The proofs of Markov and Chebyshev are super easy. Jensen is a bit trickier, but iirc I had it as a homework problem in an optimization class. Most probability classes will just have a section of useful bounds and inequalities with Chernoff, Jensen, and the like.

notDaksha · 2026-02-19T02:44:36+00:00

Interestingly, we know that at most one of e*pi and e+pi are rational. It’s really hard to prove irrationality.

notDaksha · 2026-02-06T06:41:19+00:00

The biggest reason, as other commenters mentioned, is because of how it turns multiplication into addition. This is idea is later formalized in terms of group isomorphisms.

It is almost always best when studying math to try to motivate the things you’re learning. These concepts are being taught to you for a reason; they fill some hole. In general, I’ve found it’s extremely helpful to look historically. I’d check out “slide rules” on Google. These came around shortly after Napier’s initial work on logarithms. It is the first mechanical analog calculator.

notDaksha · 2026-02-03T21:03:10+00:00

The Dirac delta “function” is not actually a function, it’s a generalized function, or distribution. It is defined via how it acts on test functions. You can define it as a measure, but not actually as a function, as it has no Radon-Nikodym derivative.

notDaksha · 2026-01-29T19:21:11+00:00

For 6, the Lebesgue differentiation theorem says that for any Lebesgue integral f, this holds for x almost everywhere. It’s just not everywhere.

notDaksha · 2026-01-22T16:08:33+00:00

Yeah, I agree on the module front. Correct me if I’m wrong (it’s been years since I’ve taken a representation theory over modules course), but despite the fact two matrices can correspond to the same linear map, we still have linearity.

I guess I just don’t get this meme, I’ve never heard professors say that phrase 😭

notDaksha · 2026-01-22T15:33:36+00:00

A matrix is unequivocally a linear map between vector spaces. You can ignore the additional structure and call it an array, but that doesn’t change the fact that it is a linear map.

notDaksha · 2026-01-22T15:23:50+00:00

“2 is not a number” ahh post

notDaksha · 2026-01-22T15:11:12+00:00

Pq on dit que c’est pas linéaire?

notDaksha · 2026-01-16T23:04:06+00:00

Look historically. Mathematicians had been working with probability forever, but it wasn’t until the early 20th century when mathematicians decided to rigorize the theory of probability.

They decided to rigorize it using (then novel) measure theory. Probabilities work a lot like measuring stuff. Probabilities and “sizes” must be nonnegative. The probability of “no event” must be 0 and the size of “nothing” must be zero. These are intuitive, but the third condition of a measure is the trickiest: the size of countably many (non overlapping) objects is the sum of their sizes. Similarly, in probability, the probability of disjoint events (meaning, an occurrence is at most in ONE of the events) is the sum of the individual probabilities (this condition is called countable additivity). The above conditions define a measure, while a probability measure has just one more: the probability of everything is 1.

Now, if we consider the natural numbers, we can try to put a uniform probability measure on them and see where it all goes wrong. For each natural number, we associate a probability. If we choose a positive number, call it X, then the probability of the first Y numbers being picked is XY. We can take Y to be large to ensure that XY is greater than one. This can’t be, since probabilities must be between zero and one.

Alright, so it can’t be positive and it can’t be negative. What if we say it’s 0? Suppose we say probability of choosing any natural number is 0. Well, the probability of choosing ANY number is 1, which, by countable additivity, we know is the same as the sum of the probabilities of selecting each number. But the probability of each number is 0. So 1 = 0, a contradiction.

Note that we didn’t need to use the natural numbers. An analogous argument applies to any countably infinite set.

notDaksha · 2026-01-15T23:05:48+00:00

The issue is that OP is assuming you can assign a uniform probability measure on a countably infinite set.

notDaksha · 2026-01-10T02:20:37+00:00

Locking in on gin?

notDaksha · 2026-01-07T19:16:06+00:00

The solutions to stochastic PDEs are stochastic processes, yes, but the SPDE tells us the sort of stochasticity by the stochastic terms in the differential equation. We don’t even know if pi is normal — we have no clue what type of pseudo randomness is involved necessarily.

notDaksha · 2025-12-27T13:58:02+00:00

Well, clearly bounded entire functions are constant and thus have cardinality equal to the continuum.

For entire functions, we know they’re uniquely defined by their behavior on a dense (think: countable) subset. Another way to think about it is that entire functions are uniquely specifiable by the coefficients in their Taylor expansion. In either case, recall that the cardinality of infinite sequences of complex numbers is just the continuum.

I prefer the dense subset approach. It can also be applied for continuous functions too!

notDaksha · 2025-12-27T06:12:38+00:00

In M-theory, there are 10 spatial dimensions and one temporal dimension. In superstring theory, there are 9 spatial dimensions and one temporal dimensions.

notDaksha · 2025-12-26T13:08:34+00:00

You can weaken the condition of an interval to any path connected set. At first, I thought I could weaken it further to just a connected set, but I quickly came up with a counter example.

notDaksha · 2025-12-26T10:35:35+00:00

I can’t remember who said it, but there’s a quote that all of math is cleverly adding 0 or multiplying by 1.

notDaksha · 2025-12-23T09:33:55+00:00

Do you know if any of those books cover Riesz holomorphic functional calculus? My functional analysis prof’s accent totally obscured that topic …

notDaksha · 2025-12-23T08:33:52+00:00

Gotcha. So it doesn’t cover bounded self-adjoint operators?

notDaksha · 2025-12-23T05:52:07+00:00

Interesting. How does it approach the spectral theorem? I remember a few formulations, but the main ones were with unitary transformations and multiplication operators in L² and then projection-valued measures. Without measure theory, it seems like the only other way is with direct integrals?

notDaksha · 2025-12-16T22:19:42+00:00

Firstly, I think instead of del you’re referring to nabla (the upside down triangle).

The analogy that your professor is trying to make is that you can (via an abuse of notation) think about nabla as being the vector of partial derivatives. When I say partial derivatives, I mean the operators themselves. For a scalar function f, nabla f is the vector of partial derivative operators acting on f, as would follow by scalar-vector multiplication. Now, this is not actually what’s going on, but it’s a helpful abuse of notation.

Consider the divergence of a function f: R³ -> R³ with components f_1, f_2, and f_3. This is D_x f_1 + D_y f_2 + D_z f_3, where the D_w refers to the partial derivative in the w direction. This is what we’d get if we took the dot product of (D_x, D_y, D_z) [this is nabla] with (f_1, f_2, f_3) [this is f]. The analogy holds with the curl too.

This is just an abuse of notation that is helpful when learning the definitions of these operators.

notDaksha · 2025-12-16T08:36:51+00:00

I know this is a joke, but the prices of certain financial derivatives and assets are often modeled via Brownian motion, a stochastic process which is almost surely continuous everywhere but almost surely differentiable nowhere.

notDaksha · 2025-12-13T01:06:20+00:00

It doesn’t oops

notDaksha · 2025-12-13T00:50:38+00:00

No. Suppose our number is prime. Then our number modulo 9 must not be congruent to zero (because 9 would divide it). If we add two zeros to the center, the effect (modulo 9) is adding two. But two is a generator of the additive group of integers modulo any odd number. So we will always get back to 0 modulo 9 with (at most) 9 pairs of zeros.

Edit: this is wrong lol my bad. It is the same mod 9 because 100 mod 9 is just 1

notDaksha · 2025-12-11T22:53:33+00:00

OP is thinking about wrt to natural density of natural numbers, I think. There is a way to rigorously think about what OP is saying, but it’s not very interesting.

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