Gaussian Process

algorithmic trading

[Gaussian Process @GaussianProcess]

@danrobinson @danielvf Yep both are almost surely 0.5 in the limit (but also I think many claimed solutions for interpretation B are bad, because they would imply that the expectation is 0.5 at finite steps, which is wrong)

[Gaussian Process @GaussianProcess]

@danrobinson @danielvf Vaguely similar energy to the question 'what is the average bitcoin block interval'

If you sample a random interval, it's 10 minutes

If you sample a random point in time, the average length of the interval you land in is 20 minutes

[Gaussian Process @GaussianProcess]

@danrobinson @danielvf I think it's a lot more subtle than this, and depends on the assumptions you make!

https://t.co/U6aPBzLQDm

[Gaussian Process @GaussianProcess]

The answer depends on your interpretation.

Interpretation A: We sample the fraction after each child is born. Then we have a series where each element is independently boy or girl, so by symmetry the expected fraction of girls is 0.5 at each step.

Interpretation B: We sample the fraction after each family is finished. Suppose N families have finished. Then there are N girls, and some random variable B of boys. The expected fraction of girls is E[N/(N+B)] > N/(N+E[B]) = 0.5 by Jensen's inequality. So at any finite step, the expected fraction is more than 0.5! By the law of large numbers, N/(N+B) converges almost surely to 0.5 as N tends to infinity.

Interpretation C: We start with a fixed population and we want to know what happens in the long run. In fact, the population will die out, so the fraction isn't even defined. Let p be the probability that a population with one man survives. Let b_i be the probability that a man has i boys. Then p = b_0 + b_1*p + b_2*p^2+... = . The only two solutions are p=0 and p=1 (the RHS is convex, so there can be at most two). Since there is positive probability the population dies out at the first step, p != 1, so we must have p = 0.

[Gaussian Process @GaussianProcess]

The answer depends on your interpretation.

Interpretation A: We sample the fraction after each child is born. Then we have a series where each element is independently boy or girl, so by symmetry the expected fraction of girls is 0.5 at each step.

Interpretation B: We sample the fraction after each family is finished. Suppose N families have finished. Then there are N girls, and some random variable B of boys. The expected fraction of girls is E[N/(N+B)] > N/(N+E[B]) = 0.5 by Jensen's inequality. So at any finite step, the expected fraction is more than 0.5! By the law of large numbers, N/(N+B) converges almost surely to 0.5 as N tends to infinity.

Interpretation C: We start with a fixed population and we want to know what happens in the long run. In fact, the population will die out, so the fraction isn't even defined. Let p be the probability that a population with one man survives. Let b_i be the probability that a man has i boys. Then p = b_0 + b_1*p + b_2*p^2+... = . The only two solutions are p=0 and p=1 (the RHS is convex, so there can be at most two). Since there is positive probability the population dies out at the first step, p != 1, so we must have p = 0.

[Gaussian Process @GaussianProcess]

@maxresnick I believe the rule is within each flashblock, you

1. Group by eoa, within each group order by nonce
2. Order groups by min(priority fee)

[Gaussian Process @GaussianProcess]

@angeris A similar proof, but written more combinatorially:

Pick a team of size at most pn uniformly from all possible teams. Let X_i be the indicator person i is selected.

P(X_i = 1) <= p, so we have

log(LHS) = H(X_1,…,X_n) <= H(X_1) + … + H(X_n) <= nH(p) = log(RHS)

[Gaussian Process @GaussianProcess]

@angeris In fact for small eps, you can do slightly better than the random construction via Reed-Solomon codes, see e.g. https://t.co/7jtBRMP197

So one angle for intuition here is good error correcting codes exist => you can find lots of nearly orthogonal vectors

[Gaussian Process @GaussianProcess]

@Galois_Capital Yep, and note that your recurrence telescopes.

Another way to phrase this is to let I_k = 1 if a loop is formed when k noodles remain, and 0 otherwise. As above, E[I_k] = 1 / (2k-1).

Then by linearity of expectation,

E[loops] = E[I_100 + ... + I_1] = 1/199 + ... + 1

[Gaussian Process @GaussianProcess]

@danrobinson @ksrini_ @ThogardPvP @phildaian If you're the only arbitrageur, 1s blocks are at least as good as 2s blocks because you can replicate any 2s block performance by just ignoring every second block

[Gaussian Process @GaussianProcess]

@danrobinson @ksrini_ @ThogardPvP @phildaian As an aside, someone's feelings on short block times should match their feelings on uniform block times, because f(1)+ f(1) vs f(2) is the same comparison as f(1) + f(1) vs f(2) + f(0)

[Gaussian Process @GaussianProcess]

@danrobinson @ksrini_ @ThogardPvP @phildaian In the screenshot I'm replying to a thread where the assumed model is that MEV is proportional to sqrt(block time) https://t.co/KNlNl5E36V

In reality I think it could be convex (like in academic LVR models), ~linear (CPMM with no fee), or concave (monopolist arbitrageur)

[Doug Colkitt @0xdoug]

1/ Toy model showing that MEV extraction from ordinary users will be slightly worse in post-Merge PoS compared to PoW...

[Gaussian Process @GaussianProcess]

@libevm @MetaMask Note that using logs in this way isn't very safe, it would be quite easy for an attacker to generate false positives, see e.g. https://t.co/gvopjH9xF3

[Gaussian Process @GaussianProcess]

@MevRefund I think mev blocker only keeps the signature private and reveals the complete unsigned tx. This stops an attacker from including a mev blocker tx in a bundle (e.g. sandwiching it), but nothing stops an attacker from sending their own competing tx, so no leak necessary here.

[Gaussian Process @GaussianProcess]

Yet another solution, starting from the following well known facts:

(1) If X ~ Gamma(k) and Y ~ Gamma(n), then X/(X+Y) ~ Beta(k, n) and X/(X+Y) is independent of X+Y

(2) Beta(k, n) has the same distribution as V_(k), the kth smallest of k+n-1 uniform r.v.s - (uniform r.v. order statistics)

(3) Gamma(k) ~ sum of k iid Exp(1)

(4) e^{-Exp(1)} ~ Unif(0, 1) - (probability integral transform)

(3) + (4) imply that e^-Gamma(m) ~ U_1...U_m for iid uniform U_i

(1) + (2) imply that X = (X+Y) * X/(X+Y) ~ Gamma(k+n) * V_(k)

Combining, U_1...U_k ~ (U_1...U_{k+n})^{V_(k)}

Setting k=n=1, we recover the original problem U_1 ~ (U_1U_2)^(V_1)

[Gaussian Process @GaussianProcess]

A nice generalization that makes what's going on clearer:

Imagine radioactive sludge decaying (i.e. a Poisson process).

Repeat the following: pick an undecayed particle i, and wait until it decays at time T_i. We retain a uniform r.v. fraction of particles at each step, so the fraction of particles undecayed after n steps is U_1...U_n.

We can also work 'backwards'. If we know that the nth particle decayed at time T_n, then we can generate T_1,...,T_{n-1} by picking n-1 uniform r.v.s V_i and setting T_i = V_(i) * T_n, where V_(i) is the ith smallest of the V_i.

If the fraction of undecayed particles at time t is k, then the fraction of undecayed particles at time ct is k^c.

Then the fraction of undecayed particles at time T_m is (U_1...U_n)^V_(m).

So we must have that (U_1....U_n)^V_(m) ~ U_1...U_m

And in particular for n=2 and m=1 we get (U_1 * U_2)^V_1 ~ U_1

As an aside, V_(m) has distribution Beta(m, n-m), and -log(U_1...U_n) has distribution Gamma(n)

So this is equivalent to Beta(m, n-m) * Gamma(n) ~ Gamma(m), a more well known fact

[Gaussian Process @GaussianProcess]

@MihaiCNica @3blue1brown I addressed this in my last few sentences above.

Consider X, Y, Z which are i.i.d. uniform random variables, independent of U,V,W.

XY and UV have the same distribution and are both independent of W, so (UV)^W and (XY)^W have the same distribution, and X,Y,W are i.i.d.

[Gaussian Process @GaussianProcess]

@danrobinson Infinite money glitch:

If you think E[EUR per USD tomorrow] = EUR per USD today, then E[USD per EUR tomorrow] > USD per EUR today

[Gaussian Process @GaussianProcess]

@ksrini_ fun problem