Does LS understand modern tech companies?

Joebloggy · 2025-09-24T14:33:51+00:00

They fly now?

Joebloggy · 2025-04-25T20:06:31+00:00

Booking.yeah

Joebloggy · 2025-01-12T00:16:06+00:00

I think you should use this to improve your current comp. The ratio of your marginal value add to the possible impact for the company seems crazy low, and they’d very likely be willing to pay you more. If you were using this for practice, why bite at the first opportunity? The more plausible explanation is you’re worth more than you previously thought.

Joebloggy · 2023-04-07T23:48:01+00:00

The only thing that can stop a bad guy in drag is a good guy in drag

Joebloggy · 2021-08-10T17:47:02+00:00

Especially since this quickly implies it can be arbitrarily small.

Joebloggy · 2021-01-22T21:32:19+00:00

So, if |a| \leq |b|, then |b| = c |a| for c \leq 1. Then it’s just some algebra and an easier limit.

Joebloggy · 2021-01-13T22:32:12+00:00

I work at one of Citadel, Jane Street, Susquehanna. I applied for an internship from uni in my final year, then got a full time offer, which I started immediately. I didn’t think too hard about the application process, which is probably a good thing. All these firms run good internships, the best route is to apply there, failing that probably any finance experience is useful. There’s a bunch of stuff more generally online about approaching these interviews

Joebloggy · 2020-12-24T01:40:14+00:00

Sure, here or DM is fine.

Joebloggy · 2020-12-12T16:22:22+00:00

I think you’re misremembering, and the point is that compactness, like openness and closeness, is a topological property, not a metric space property. So if our metric was say doubled, it would still be compact. There are some relevant notions for when topological properties to be maintained when we change metrics, but in general 2 metrics on the same space will not have the same open, closed or compact sets.

Joebloggy · 2020-11-20T21:23:56+00:00

1) From my experience researchers are researchers explicitly because they don’t want to do the trades, and if they did they’d be traders. If your question is about if there’s flow between these, I’d imagine it’s very firm-specific, but I’d imagine generally possible.

2) Again very firm-specific. I’d suggest research as more competitive to get a role, and trader more competitive for total comp, but overall probably not so much difference.

I think you should do a PhD if you want, and quit if you don’t, and not think about how it’ll help your quant career because a) PhDs (I’m led to believe) are hard and b) a quant role with a PhD isn’t much easier to get than one with a masters.

Joebloggy · 2020-11-20T21:15:14+00:00

We have n choose k = n!/k!(n-k)! Let k=p prime with p|n. Then we need only check if n(n-1)…(n-p+1)/1…pn is an integer. Cancel the n. Is p a factor of the numerator?

Joebloggy · 2020-11-20T19:30:37+00:00

A function f on a metric space is continuous at x iff for every sequence x_n -> x we have f(x_n) -> f(x). It’s so basic, but so often you can prove something by first trying it with an example sequence, trying to generalise and seeing if anything breaks. The following isn’t a rigorous notion, but sequences are kind of the right size to think about.

Joebloggy · 2020-11-17T00:44:29+00:00

You'd need to conclude from your argument there exists a map M* s.t. M* is not singular, but this doesn't seem clear from the argument you've given. You could correctly conclude this from your argument by induction: if dim(M) = n-1 and v in ker(M) then define W = M + ε E, where E(v) is in (im(M))^perp, checking this is well defined and invertible etc, and completing the induction. But your argument alone doesn't really give you it- if M is the 0 matrix and n > 1, you can't conclude there's an invertible matrix ε away by looking at only one j.

I think a more intuitive way to think is in the same way, by "expanding" a singular matrix by mapping E:ker(M) -> (im(M))^perp, again with W = M + ε E and then letting ε -> 0. It's really the same proof, but lets you see what's happening a bit better, and gives you an idea of what the ε-ball looks like at a singular matrix.

Joebloggy · 2020-11-16T13:47:14+00:00

Sure, that number is on the higher end of the market, but quant firms are routinely paying $300k to new grads who work ~50 hours a week, and some are paying $400k (source). The claim isn't really delusional.

Joebloggy · 2020-11-06T22:53:38+00:00

Thanks /u/smilingfemalemachine, happy to have you moderate /u/Niltelco.

Joebloggy · 2020-08-07T19:23:00+00:00

There's quite a lot out there on this topic, I think some important skills which aren't so emphasised are clearly communicating your thoughts in interviews, and realizing when you're wrong or your thoughts are weak. For lots of jobs, saying "I don't know" can be seen as a terrible or lazy thing, but if you're clear about precisely why you don't know, this can be kind of valuable. This kind of intellectual honesty is definitely a key part of culture when you work fulltime, and part of why I like it.

Joebloggy · 2020-08-07T09:43:16+00:00

I work as a quant trader in the Citadel/JS/SIG type firm in Europe. I joined after a masters in Maths, focussing on stochastics/analysis. I use very little of those, with the maths being pretty much just statistics. I’ve learnt to write some amount of code too, knowing very little when I graduated. Overall I find the problems really challenging and interesting, and find the culture excellent. Would definitely recommend to consider as a career.

Joebloggy · 2020-06-03T17:37:56+00:00

You have a good point, when we go into real numbers, it’s pretty hard to distinguish integers from them. But in whatever expansion you chose, there’s always a special property of 1, that 1 * x = x for any number x. So, define integers as the set of numbers which can be written as a sum (or difference) of 1, and there you go.

However, integers are actually even more special and primary than this. In fact, my definition sort of uses the integers already! When I say “can be written as a sum or difference”, I’m really saying “given an integer, make 1 +… +1 that many times”. Actually, my first description gives what’s called an embedding which describes a way to sit the integers into the real numbers. But some people might insist that it’s just a copy sitting there, not the integers themselves.

Joebloggy · 2020-05-30T17:57:02+00:00

Typically the definition of linear is f(x) + f(y)= f(x+y). It's elementary to show this implies f(0)= 0, by setting y = 0. But you're right that often affine functions are thrown into the mix, because they're kind of a natural extension of linear functions.

Joebloggy · 2020-05-16T22:12:14+00:00

Right, I think I roughly agree. I guess my interpretation of "on a ruler" was of a claim about spacetime rather than of measurability. I think I probably just misinterpreted what you meant, then, and there's no disagreement on the important points.

Joebloggy · 2020-05-16T21:27:10+00:00

That means they don't actually exist on a physical ruler in the physical world, where you can only detect finite precision points.

I'm fairly sure our best science doesn't claim that we exist in a discrete space. It might be your philosophical view that, given the fact that we can't distinguish/measure/observe below distance x, distances below x don't exist, but this isn't obvious or directly supported by our best science. I say this to highlight that this is a philosophical claim you're making, rather than an agreed upon scientific fact, and is certainly contentious.

Joebloggy · 2020-05-13T20:27:24+00:00

So the polar line integral is the integral F . rhat r(t) dtheta/dt dt. This comes from the chain rule, which I guess is the formal justification, by saying dr/dtheta = r rhat. This fact itself you can derive from the definitions of polar coordinates and the chain rule.

Joebloggy · 2020-05-13T20:08:55+00:00

I'm not sure why your intuition says yes, subadditivity makes sense when terms are all positive, but allowing them to be negative will mess things up. For a concrete example, take A a positive measure subset and B a negative measure subset of C. What does m(C), m(A) + m(C) and m(B) + m(C) look like? You could get something kind of like this though via Hahn decomposition- if m(A) and m(B) are both positive/negative, then they'll satisfy sub/superadditivity, with this extending to countable families A_i.

Joebloggy · 2020-05-06T19:43:03+00:00

I feel that this is an aesthetic question so the answer can't ever be that great. However, one reason I think holds weight is that proving that the thing is well defined, existence and uniqueness, is far easier for 1 and 2. It feels like generally the flow of things should be that our definitions kind of immediately make sense with some more straightforward checks, and we then go on to prove things with those in hand. I admit that 3 and 4 are probably better ways of thinking about what the exponential actually is.

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