Feb 12th, 2021
Learn what it means to talk about conditional probabilities with random variables. Be able to calculate conditional expectations. Be able to use the law of total expectation.
Q: Would you mind defining obfuscate again?
A1: here, obfuscate means obscure X_i half of the time with a random coin flip. Y_i is still mathematically defined in terms of X_i, but we introduce an unpredictable fair coin flip so you’re not fully revealing the private data that might be associated with X_i.
A2: However, the statistics of X_i (expectation, for instance) can still be derived from Y_i, and if you’re only interested in the statistics, you can still get them without disclosing the private data 100% of the time.
Q: So we could also use the fact that one table sums to one accross columns and the other does across rows?
A1: absolutely.. that’s what I did :)
A2: And that’s just what Chris did too. :)
Q: do you need the parentheses in #8 or is it just there for fun?
A1: Just for fun... not needed.
Q: Is there a difference between E[X|Y] and E[X|Y=y]?
A1: nope, just two different notations.
Q: How did we do the first step?
A1: E[X|Y] is a function of Y, so the average value of that function, where each value of that function is weighted by the probability that Y = y.
Q: so we could find the expectation of S by summing over diffrent values for D_2?
A1: You certainly can. But just to clarify, do you mean computing E[S] from E[S|D_2]? That’s how I’m reading your question.
Q: Can you explain again why E[X,Y] doesn't make sense?
A1: You can only compute the expectation of a single random variable, like X, or Y, or Z = X + Y. X, Y isn’t a single random variable, because it’s not clear how X and Y are being combined.
Q: sorry I missed E[X, Y]. Is this asking E[X and Y], so a number?
A1: No, that particular one doesn’t make sense.
A2: You can only compute the expectation of a single random variable, like X, or Y, or Z = X + Y. X, Y isn’t a single random variable, because it’s not clear how X and Y are being combined.
Q: I mean we could find the expectation of S (overall) by inputing all valid numbers for D_2 (1,2,3, …) into E[E[S|D_S]]
A1: Absolutley. E[S| D_2] (which is what I think you mean) is a function of D_2, so you can now compute the weighted average of that function for all valid D_2 outcomes.
Q: how is the position of the second best engineer figured into k/i+1
A1: Technicaly, all engineers better than any in the first k are ones that stop the process. We basically want to maximize the likelihood that the best engineer isn’t among the first k engineers, and that instead is the first engineer of all better-than-first-k engineers to be interviewed once you can actually hire someone.