Recent content by Brad Warren

You can't do better than zero in expectation. You're playing a game at even odds with a fair coin. You have no advantage.

Congratulations to the winners!

Well if it's for a quant job, I would be surprised if they didn't ask any technical questions at some point. But it sounds like this call is just a brief interview for you to talk about your background. And think of some questions to ask too.

Another way from 2 to 3: x^-1, cos^-1, tan, x^2

I like your solution. But to get the odd k terms (for Player 1 winning), you should actually add the binomial powers, and then we need to divide by a factor of 2. The probability should be...

That formula works for n=2. But the game can go on more than 2 rolls. This one works for n <= 6: (\frac{79n^4+10n^3+5n^2+50n-24}{120n^4}) but I don't see a pattern for the general solution.

I interned in Technology at an investment bank last summer. I'm a junior too, so I'm trying to get a quant internship this summer. I'm not sure exactly what I want to do yet, which is another reason to get experience in a quant internship. This firms list is a good starting point. I've found a...

I'm in a similar situation as you. I would recommend self-study to try and improve on the areas you feel are lacking. I'm planning on applying for jobs as well as MFE, and then the decision depends on what job I can get.

I don't see anything negative about athletics. Especially at the collegiate level, it requires a significant amout of discipline and time. If someone has the same academic record and technical skills as other candidates, and they were able to do that while spending hours every day at practice...

You could do it either way. Starting with a vector of uncorrelated standard normals, using the covariance matrix is more direct since you have to factor in the individual volatilities anyway.

An alternative solution: For a Brownian motion, there is equal probability that Wt = a or Wt = -a. Taking the average, (1/2(N(a)+N(-a))=1/2)

For G(4,4), there are 2 groups of 4 such that all santas are friends with everyone in the other group. For G(4,3), there are 4 santas in GROUP 2 that each need to have 3 friends in GROUP 3, but they must also have only 2 mutual friends in GROUP 3 (the third mutual friend is the first santa)...

There are \(\frac{nk}{2}=\frac{k}{2}(1+k+\frac{k(k-1)}{l})\) edges. Either k or \(\frac{k(k-1)}{l}\) has to be even. I think those conditions are sufficient. Never mind, that doesn't explain why G(4, 3) doesn't exist. But k does not have to be even.

It hits both barries before reaching time T, not at the same time.

The process is Markov, so we can condition on the next car passing. The time to cross the road is 20 seconds if no cars pass for 20 seconds. Otherwise, the expected time to cross is the time the car passes plus the original expectation. The time a car passes is distributed as Exp(lambda = 1/10)...