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Quant Job Interview Questions (2009) [pdf]
51–60 of 154 posts
Re: Quant Job Interview Questions (2009) [pdf]
#52Earlier quoted context omitted.
Is ito calculus really the extent of the math you need to know to go into quant??
It's probably like any coding interview. You'll get asked a bunch of things that won't be a problem when you're on the job. Basically, it's a "have you heard of this" type thing.
Oh well that makes sense, if stochastic calculus is all you need to know, i was half considering becoming a Quant there for a second ;)
Re: Quant Job Interview Questions (2009) [pdf]
#53did it start snowing at 7:30am?
Re: Quant Job Interview Questions (2009) [pdf]
#54Here'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 re-drawn. Decisions are made independently. We compare our numbers and the highest one wins $1.
What strategy do you use?
EDIT: The answer is not 0.5, it is not 0.7 either.
Re: Quant Job Interview Questions (2009) [pdf]
#55Here'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…
This sounds like a variant of the two envelope problem: https://en.m.wikipedia.org/wiki/Two_envelopes_problem
So is this one of those questions that's not possible to answer you're just trying to see how they handle the stress of the unsolvable?
Re: Quant Job Interview Questions (2009) [pdf]
#56Here'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…
This sounds like a variant of the two envelope problem: https://en.m.wikipedia.org/wiki/Two_envelopes_problem So is this one of those questions that's not possible to answer you're just trying to see how they handle the stress of the unsolvable?
It isn't. There's a clear, well defined answer.
Re: Quant Job Interview Questions (2009) [pdf]
#57I wrote about this a long time ago here: https://news.ycombinator.com/item?id=8698986#8699260 I think most of what I wrote is still valid. I believe the below reference: http://www.decal.org/file/2945 and the book Heard on teh street are also great refresher's for the type of material you'll need. https://www.amazon.ca/Heard-Street-Quantitative-Questions-In... To be honest most of this won't matter, the number one qu…
You stress the importance of having a Phd - How much of that is because it's really needed and how much of it is a filter / credibility signal?
You need to learn the maths and for that there is nothing close to sitting down in a master/phd classroom for years, doing hard exercises for the whole duration of it, that could get you fired of the school if you ever get a single one wrong at an exam.
Re: Quant Job Interview Questions (2009) [pdf]
#58Here'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…
This sounds like a variant of the two envelope problem: https://en.m.wikipedia.org/wiki/Two_envelopes_problem So is this one of those questions that's not possible to answer you're just trying to see how they handle the stress of the unsolvable?
But here you know the content of the "envelope" before you decide whether to switch or not.
If the adversary was only allowed one pick I guess I'd only re-draw if I had less than 0.5 since statistically you'd be more likely to have a greater number than a weaker one. I'm not sure what the maths are but a simple implementation gave me an average value of about 0.625 using this strategy, so definitely an improvement over a single pick.
But I have the feeling that since the adversary can chose the same strategy maybe you should be a little more aggressive and redraw if you are at 0.5? But then the adversary would do the same so I guess the solution would eventually converge to a value a little above 0.6?
That'd be my guess at least.
Re: Quant Job Interview Questions (2009) [pdf]
#59Earlier quoted context omitted.
This sounds like a variant of the two envelope problem: https://en.m.wikipedia.org/wiki/Two_envelopes_problem So is this one of those questions that's not possible to answer you're just trying to see how they handle the stress of the unsolvable?
It isn't. There's a clear, well defined answer.
I'm just a programmer, but this seems really easy to me. What am I missing?
Re: Quant Job Interview Questions (2009) [pdf]
#60Here'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…
If first draw is >= .7 don't redraw since chance of being larger than .7 is 50% for two draws. 50% probability is chosen since if you pick any number lower than .7, the probability increases favoring your opponent drawing a higher number.
calc: p(a union b) = p(a) + p(b) - (p(a)*p(b)); p(>=.7)=3/10; .51 = (3/10) + (3/10) - (.3)(.3)