This article contains very little substance. Show me the math!

AgentOrange1234 7 hours ago | parent [-]

Yes I found this very hard to follow. I appreciate expressing ideas in math like E_a[X] as much as the next guy, but there is no definition or even description of what the heck E or E_a or Var(x) even mean, so how is anyone supposed to understand the reasoning here? All I get from this is a claim that experienced latency is different than the mean, which sounds important, but I still have no intuition as to why this is. Which is sad, because Booker's blog is often deeply amazing.

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NightMKoder 7 hours ago | parent | next [-]

This is standard statistics terminology - E(X) is https://en.wikipedia.org/wiki/Expected_value . E_a is presumably Alice's perceived expected value. Var(X) is https://en.wikipedia.org/wiki/Variance . The law of large numbers says the arithmetic average of observations becomes E(X) with enough samples.

I'm pretty sure what the author is saying is:

E(X) =:= \sum_t(t * P(X = t)) is the definition

another important note is P(X^2 = t^2) = P(X = t) - because it's the same distribution.

E_a(X) is a bit sloppy, but consider X_a aka Alice's latency "experience" distribution. The argument is:

P(X_a = t) = t * P(X = t) / \sum_u(u * P(X = u)) - i.e. scale the probability up by t but make it sum to 1.

Then

E(X_a) = \sum_t(t * P(X_a = t)) = \sum_t(t * t * P(X = t) / \sum_u(u * P(X = u))

aka

E(X^2) / E(X)

Then (from wikipedia)

Var(X) = E(X^2) - (E(X))^2

And we get

E(X_a) = (Var(X) + (E(X))^2) / E(X) = E(X) + Var(X) / E(X)

	▲	gruberben 4 hours ago \| parent [-]
		Thanks!

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canjobear 5 hours ago | parent | prev [-]

I was clicking around in response to this article and found this video that explains the inspection paradox nicely. https://youtu.be/Jd1wNizPjoE