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Quant Job Interview Questions (2009) [pdf]

math.kent.edu

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

#141

Post by [deleted] »

Xcelerate wrote: post #134
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…

The problem seems ill-defined to me in the same sense that Bertrand's paradox is ( https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) ). Your optimal strategy involves maximizing an objective function that depends upon your opponent's strategy. "Assume nothing" about your opponent is an incomplete problem specification, because you must assume something in order to determine which function to maximize. For…

[deleted]

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

#142

Post by hkhall »

justanewguy wrote: post #140
SonOfLilit wrote: post #112

Earlier quoted context omitted.

All the ideas I gave were things that happened (or at least, that I read on the internet from reliable sources.) Chartering fast ships is from 1790, see e.g.: http://dewinforex.com/forex-basics/high-frequency-trading-fi... Can't find the link for the donut investor, but I'm pretty sure it was somewhere in Matt Levine's (highly recommended) Money Stuff column. Sat imagery of parking lots appears all the time in Matt L…

I got most of the references, but the donut one just seems very interesting. Please tell me if anyone here manages to find it. I'll be looking too.

I was working for the leading print marketing company that was starting to have serious competitors and post-IPO we were really concerned that people were going to order business cards from us all the time (since they were almost free) to track our daily internal sales via order number. I can't say that we ever had hard evidence that this was happening but we thought we had suspicious activity once so we changed it.

An aside, changing order number to a hash value can be a huge pain in a manufacturing environment when people have (mis)used it to for a variety of purposes beyond order identification.

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

#143

Post by murbard2 »

foldr wrote: post #132
CalChris wrote: post #119

Earlier quoted context omitted.

Agreed. I think the other 'more complex' answers assume there's communication or that you know your opponent's strategy. That's not in the problem statement.

Yeah, this came up the last time murbard2 posted this: https://news.ycombinator.com/item?id=8699033 The idea is to find a Nash equilibrium, but the question is stated in the wrong way, since it gives no indication that you're in a position to make any assumptions about your opponent's strategy.

It's an interview question, you're free to ask for clarifications.

That said, finding the Nash equilibrium is precisely the natural thing to do if you can't make assumptions about your opponent's strategy. It's a worst case scenario analysis. You can't statistically lose out to your opponent if you pick it. This does not mean that you know your opponent strategy.

Consider the context. This is algorithmic trading. If your strategy can be gamed, it will be gamed by someone else. Picking anything but the Nash equilibrium means you can be gamed.

A good candidate will immediately gravitates towards this solution, there is no ambiguity that this is the expected answer.

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

#144

Post by boromi »

SonOfLilit wrote: post #103
SonOfLilit wrote: post #101

Earlier quoted context omitted.

Saying "It's intuitive to me that this answer someone worked out and told me is correct" is cheating. But generally... The process is something like: If you choose 0.5, then 0.625 beats you because that is the expected average of someone redrawing below 0.5. If you choose 0.625, then 0.617[2] beats you because that is the expected average of someone redrawing below 0.625. If you choose 0.618, then 0.6182 beats you...…

OK I almost got it (edit: rot13 isn't very good at scrambling one-variable equations, so read at your peril, spoilers ahead): Jr'er ybbxvat sbe n svkrq cbvag bs "vs gur bccbarag pubbfrf gb erqenj orybj k, jr erqenj orybj uvf rkcrpgrq inyhr". V gevrq gb qrevir gur sbezhyn sbe uvf rkcrpgrq inyhr n srj gvzrf naq nyjnlf tbg pbashfrq ba jurer V fubhyq hfr k, jurer (1-k) naq jurer 0.5, fb riraghnyyl V gevrq qrevivat gur fc…

why do you seek the fixed point and not the k value that maximizes s(k) = (1 - k)(1 + k) / 2 + k / 2 ?

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

#145

Post by foldr »

murbard2 wrote: post #143
foldr wrote: post #132

Earlier quoted context omitted.

Yeah, this came up the last time murbard2 posted this: https://news.ycombinator.com/item?id=8699033 The idea is to find a Nash equilibrium, but the question is stated in the wrong way, since it gives no indication that you're in a position to make any assumptions about your opponent's strategy.

It's an interview question, you're free to ask for clarifications. That said, finding the Nash equilibrium is precisely the natural thing to do if you can't make assumptions about your opponent's strategy. It's a worst case scenario analysis. You can't statistically lose out to your opponent if you pick it. This does not mean that you know your opponent strategy. Consider the context. This is algorithmic trading. If…

In the absence of any information about your opponent, you should assume that they're equally likely to be using any strategy.

>You can't statistically lose out to your opponent if you pick it.

Right, but you can miss out on an opportunity to fully exploit the fact that your opponent is using a really bad strategy. So if you want to maximize your probability of winning against an opponent with a randomly chosen strategy, it's not the right answer. Nothing in the statement of the question indicates that you are entitled to make any assumptions about your opponent's strategy. And it's not as if determining the best strategy in a context where you lack perfect knowledge of your opponent's strategy is an unknown concept in game theory:

https://www.economics.utoronto.ca/workingPapers/UT-ECIPA-ECP...

I feel like the way you state the question at present may end up selecting for candidates who think exactly like you do. Why not just add the statement that the players have mutual knowledge of their rationality? That removes the ambiguity, and doesn't make the question fundamentally any easier.

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

#146

Post by SonOfLilit »

boromi wrote: post #144
SonOfLilit wrote: post #103

Earlier quoted context omitted.

OK I almost got it (edit: rot13 isn't very good at scrambling one-variable equations, so read at your peril, spoilers ahead): Jr'er ybbxvat sbe n svkrq cbvag bs "vs gur bccbarag pubbfrf gb erqenj orybj k, jr erqenj orybj uvf rkcrpgrq inyhr". V gevrq gb qrevir gur sbezhyn sbe uvf rkcrpgrq inyhr n srj gvzrf naq nyjnlf tbg pbashfrq ba jurer V fubhyq hfr k, jurer (1-k) naq jurer 0.5, fb riraghnyyl V gevrq qrevivat gur fc…

why do you seek the fixed point and not the k value that maximizes s(k) = (1 - k)(1 + k) / 2 + k / 2 ?

s(k) is the function for "The number I should choose if my opponent chooses k" [1]. We arrived at the formula by asking "what is the best number for us to choose assuming our opponent chooses k?".

Now, obviously a smart opponent would not choose a number k where k != s(k), because s(k) would be a better choice. So the only numbers it makes sense for him to choose would be fixed points of s(k), points where k = s(k). And the smartest thing for me to do in that case would be to choose s(k), which just so happens to be equal to k.

[1] which may be more or less than s(k), e.g. if my opponent chooses to always redraw, i.e. k=1, it should be the same as if he chooses to never redraw, k=0, and we expect for both cases s(k)=0.5 < 1 (and it indeed is)

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

#147

Post by murbard2 »

foldr wrote: post #145
murbard2 wrote: post #143

Earlier quoted context omitted.

It's an interview question, you're free to ask for clarifications. That said, finding the Nash equilibrium is precisely the natural thing to do if you can't make assumptions about your opponent's strategy. It's a worst case scenario analysis. You can't statistically lose out to your opponent if you pick it. This does not mean that you know your opponent strategy. Consider the context. This is algorithmic trading. If…

In the absence of any information about your opponent, you should assume that they're equally likely to be using any strategy. >You can't statistically lose out to your opponent if you pick it. Right, but you can miss out on an opportunity to fully exploit the fact that your opponent is using a really bad strategy. So if you want to maximize your probability of winning against an opponent with a randomly chosen strat…

1) There's no reason why you should assume every strategy is equally likely. It's absolutely not the case. When you play poker, do you assume your opponent play their cards randomly just because you don't know their strategy? It's not a realistic prior at all.

2) Playing the Nash equilibrium means that you may not be fully exploiting the opponent, true. It's also possible that you're opponent's strategy is to give you a million dollars if you say "abracadabra". So how about that? Check-mate interviewer.

3) If you want to make the problem harder on yourself, you can be my guest and try and assume a realistic distribution of strategies for your opponent. That distribution will have a peak around the nash equilibrium, so you'll have to end up calculating it.

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

#148

Post by foldr »

murbard2 wrote: post #147
foldr wrote: post #145

Earlier quoted context omitted.

In the absence of any information about your opponent, you should assume that they're equally likely to be using any strategy. >You can't statistically lose out to your opponent if you pick it. Right, but you can miss out on an opportunity to fully exploit the fact that your opponent is using a really bad strategy. So if you want to maximize your probability of winning against an opponent with a randomly chosen strat…

1) There's no reason why you should assume every strategy is equally likely. It's absolutely not the case. When you play poker, do you assume your opponent play their cards randomly just because you don't know their strategy? It's not a realistic prior at all. 2) Playing the Nash equilibrium means that you may not be fully exploiting the opponent, true. It's also possible that you're opponent's strategy is to give yo…

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 from the assumption of mutual knowledge of perfect rationality. Is your poker opponent equally likely to be doing their utmost to lose as to win? No. But on the other hand, they're not likely to be tracking 100% of available information and making the mathematically correct choice all the time.

In general, the question of what you should assume about an unknown opponent's strategy is a tricky philosophical question, and reasonable people can gravitate to different points of view on this issue. I think that might be why so few people give the answer you want. I was just suggesting that explicitly mentioning mutual knowledge of perfect rationality might help.

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

#149

Post by murbard2 »

SonOfLilit wrote: post #146
boromi wrote: post #144

Earlier quoted context omitted.

why do you seek the fixed point and not the k value that maximizes s(k) = (1 - k)(1 + k) / 2 + k / 2 ?

s(k) is the function for "The number I should choose if my opponent chooses k" [1]. We arrived at the formula by asking "what is the best number for us to choose assuming our opponent chooses k?". Now, obviously a smart opponent would not choose a number k where k != s(k), because s(k) would be a better choice. So the only numbers it makes sense for him to choose would be fixed points of s(k), points where k = s(k).…

Why do you bring up expected value in the calculation. You may end up with the right result, but there's no good justification for why expected value should come into play.

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

#150

Post by murbard2 »

foldr wrote: post #148
murbard2 wrote: post #147

Earlier quoted context omitted.

1) There's no reason why you should assume every strategy is equally likely. It's absolutely not the case. When you play poker, do you assume your opponent play their cards randomly just because you don't know their strategy? It's not a realistic prior at all. 2) Playing the Nash equilibrium means that you may not be fully exploiting the opponent, true. It's also possible that you're opponent's strategy is to give yo…

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.

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