61–70 of 154 posts
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…
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…
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)
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Serious question... do people really enjoy working on this stuff? Do most quants go to work thinking it is fun to work on mathematical models to optimize financial transaction. Or is it just a job that pays really well towards long term financial independence to do what they really want?
There's some interesting things about this work: - You get to know something that isn't published. You find a strategy, you don't tell anyone. Sure, you can talk about it in broad strokes (we're trend following...) but it's pretty unlikely you'll ever tell anyone enough for them to be able to replicate it. Except people you trust, of course. - You're always thinking about philosophy of science. How do I know that thi…
Earlier quoted context omitted.
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)
Why p=.5 for > .7 w/ 2 draws?
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…
Decisions and draws are independent - we can ignore the other guy and just go for the highest value. Draw the first number. If it's below 0.5, draw again, since the odds then are that the next draw will be higher.
Is there more?
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…
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…
This seems so simple. Decisions and draws are independent - we can ignore the other guy and just go for the highest value. Draw the first number. If it's below 0.5, draw again, since the odds then are that the next draw will be higher. Is there more?
Earlier quoted context omitted.
This seems so simple. Decisions and draws are independent - we can ignore the other guy and just go for the highest value. Draw the first number. If it's below 0.5, draw again, since the odds then are that the next draw will be higher. Is there more?
Consider heads or tails: you get two flips, what are your chances of getting heads at least once? 50% for the first time then theres 50% chance you didn't, of that you now have a 50% chance to still get it .5*.5=.25. so for the two chances p(1)=.5, p(2)=.25, therefore overall p1+p2=.75 for redrawing at <50% or getting heads once with two tries.
Though in this problem it seems the "average" final number would be 0.625, not 0.75 - since half the time you have an 0.75 expected outcome and the other half a 0.5 expected.
Earlier quoted context omitted.
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.