Quant Job Interview Questions (2009) [pdf] • Hacker News Board

Re: Quant Job Interview Questions (2009) [pdf]

#151

Post by foldr » Mon, Nov 14, 2016, 5:32 PM UTC

murbard2 wrote: post #150

foldr wrote: post #148

Earlier quoted context omitted.

Your way of looking at it is perfectly sensible. But a sensible person might also say that a uniform prior over strategies is appropriate in the absence of any information about your opponent. Choice of priors is always going to be partly subjective, and reasonable people can make different choices. I would point out that in a real world scenario, a uniform prior is barely any less realistic than the a prior derived…

Do you think it's equally likely that a random opponent will redraw below 0.01 as it is that they will redraw below 0.5 or below phi? Most people do not give the right answer not because of philosophical issues around the Nash equilibrium but because they have a hard time understanding that looking to maximize the expected value is not optimal.

>Do you think it's equally likely that a random opponent will redraw below 0.01 as it is that they will redraw below 0.5 or below phi?

If random really means random, then that's presumably true by definition. If we're talking about an opponent randomly selected from a representative group of people, then no, but equally, I wouldn't assume that they're going to pick the golden ratio as their threshold either.

The thing is that maximizing the expected value is optimal in the absence of any information about the opponent's strategy. But sure, I can easily believe that many people would get the question wrong anyway.

Re: Quant Job Interview Questions (2009) [pdf]

#152

Post by murbard2 » Mon, Nov 14, 2016, 6:43 PM UTC

foldr wrote: post #151

murbard2 wrote: post #150

Earlier quoted context omitted.

Do you think it's equally likely that a random opponent will redraw below 0.01 as it is that they will redraw below 0.5 or below phi? Most people do not give the right answer not because of philosophical issues around the Nash equilibrium but because they have a hard time understanding that looking to maximize the expected value is not optimal.

>Do you think it's equally likely that a random opponent will redraw below 0.01 as it is that they will redraw below 0.5 or below phi? If random really means random, then that's presumably true by definition. If we're talking about an opponent randomly selected from a representative group of people, then no, but equally, I wouldn't assume that they're going to pick the golden ratio as their threshold either. The thin…

> If random really means random, then that's presumably true by definition.

No it's not. Randomness is always with respect to a given distribution. There is nothing special about the uniform distribution over a threshold. The space of strategies also include strategies where your opponent redraw only when they hit a number between 0.1 and 0.2, or where they redraw if the second decimal digit of the first draw is even.

> The thing is that maximizing the expected value is optimal in the absence of any information about the opponent's strategy.

Your statement is meaningless without a prior on the strategy. Under which prior is maximizing the expected value optimal? If you assume the other party randomly picks a threshold between 0 and 1, then the optimal answer is (sqrt(39)-3)/6 not 0.5. Not that this prior makes any sense. It's far more sensible to assume a strategic opponent.

Rethrowing below 0.5 is typically the optimal strategy when the opponent has no strategy at all, i.e. always or never rethrows.

Re: Quant Job Interview Questions (2009) [pdf]

#153

Post by foldr » Mon, Nov 14, 2016, 7:46 PM UTC

murbard2 wrote: post #152

foldr wrote: post #151

Earlier quoted context omitted.

>Do you think it's equally likely that a random opponent will redraw below 0.01 as it is that they will redraw below 0.5 or below phi? If random really means random, then that's presumably true by definition. If we're talking about an opponent randomly selected from a representative group of people, then no, but equally, I wouldn't assume that they're going to pick the golden ratio as their threshold either. The thin…

> If random really means random, then that's presumably true by definition. No it's not. Randomness is always with respect to a given distribution. There is nothing special about the uniform distribution over a threshold. The space of strategies also include strategies where your opponent redraw only when they hit a number between 0.1 and 0.2, or where they redraw if the second decimal digit of the first draw is even…

If a strategy is any function from the preceding game state to keep/redraw, then I don't have the mathematical chops to specify a uniform prior over all such strategies and compute the optimal strategy that way. (In fact it may not necessarily be possible to do this in a sensible way, as Bertrand's paradox suggests, as another poster pointed out.)

However, there is a much simpler argument suggesting that maximizing EV would be the right approach in the absence of any knowledge about the opponent's strategy. Say the opponent chooses to redraw at random. Then in effect they are drawing once. If we redraw when the value is Modeling a process that we know nothing about as a random process seems reasonable. And the reasoning above does not actually involve computing EVs, though it happens that the right answer is the answer that maximizes the EV.

To justify this a bit further, we can reason as follows. If we know nothing about the opponent's strategy, then their initial draw tells us nothing about whether or not they will choose to redraw. So our estimate of their probability of redrawing, p, must be independent of their initial draw. Then the probability of winning is 0.5ap + (1-a)(1-p)(a + 0.5) + (1-a)p(a + 0.5) + 0.5a(1-p), which is invariant with respect to p, and which is at its maximum when a = 0.5.

>It's far more sensible to assume a strategic opponent.

Well, this is going round in circles, but I would say that this is sensible only if you know something about the opponent! I don't see how it's sensible to assume anything about your opponent if you know nothing about them. And if we're talking about the "real world", then of course you do know something, but you certainly don't know that they're perfectly rational.

Re: Quant Job Interview Questions (2009) [pdf]

#154

Post by pps43 » Wed, Nov 16, 2016, 3:38 PM UTC

dreeves wrote: post #79

murbard2 wrote: post #54

Here's my favorite interview question (spent 10 years as a quant, interviewed a bunch of people, most do not do well on this) We're going to play a game. You draw a random number uniformly between 0 and 1. If you like it, you can keep it. If you don't, you can have a do-over and re-draw, but then you have to keep that final result. I do the same. You do not know whether I've re-drawn and I do not know whether you've…

Oh my god it's the golden ratio! That's so cool! EDIT: I'll show my work, rot13'd... Jr pna cnenzrgrevmr n fgengrtl ol n guerfubyq g: gur inyhr gung gur svefg qenj arrqf gb or yrff guna va beqre gb pubbfr gb qenj ntnva. Hfvat guerfubyq g, gur cebonovyvgl bs trggvat yrff guna g vf g^2 orpnhfr lbh unir gb qenj orybj gur guerfubyq gjvpr va n ebj. Gung chgf gur cebonovyvgl bs raqvat hc nobir gur guerfubyq ng 1-g^2. Gur c…

There is another way to do it, without integrals and convolutions. Say player A has threshold a and player B has threshold b, a After some simplification, total Player A win probabilty is (-ba^2 + (b^2-b+1)a +(b^2-b+1))/2. Find a that maximizes this probability for any given b by taking the derivative and setting it to 0, which would be a = (b^2-b+1)/2b.

The game is symmetric, so at Nash equilibrium a==b. Now solve a = (a^2-a+1)/2a to get the golden ratio.

The lift over naive 0.5 threshold is very small, though, less than 1%.