Quantitative analyst Interview Questions

2011 is prime

More details in addition to the last post. 1) subtracting the 1st equation from the 2nd one gives: (b-a)*(b+a-2c)＝2011 2) 2011 is prime, so there are only two possibilities: b-a=1, b+a-2c=2011; or b-a=2011, b+a-2c=1 3) subtracting the 1st equation from the 3rd equation gives: (c-a)(c+a-2b)=x-1 4) plugging c-a=0.5*[(b-a)-(b+a-2c)], c+a-2b=(c-a)-2(b-a) into 3) gives you x

In 2) there should be four rather than two possibilities. Both terms could be negative.

A is larger than B. Maybe use Jensen's inequality and sqrt() is a concave function

A = B = sqrt(Variance). Variance is not a random variable, it is a constant. Then we get E[Variance] = Variance and E[ sqrt(Variance) ] = sqrt(Variance) .

Francesco, have you ever heard of stochastic volatility models ? Do you still think that "Variance is not a random variable, it is a constant." ? I agree with Jolene using Jensen inequality on a concave function implies that sqrt(E(...)) > E(sqrt(...))

If you can weigh 1,...,n-1 for n > 2, then you can weigh n by using 1 and n-1 (this does not hold for weighing 2). So you just need 1 and 2.

Not sure I fully follow the question, can't you just get everything in terms of the lightest, so just weigh the lightest to get your unit? e.g. number the weights w1, w2, ... ,w40 in increasing order (we can determine this without external knowledge, right?). Then we have w1 + w2 = w3 w1 + w3 = w4 w1 + w4 = w5 and also stuff like w2 + w3 = w5 substitute come stuff in and you find that w2 = 2w1, hence wn = nw1 So just 1?

delta^2*(x'x)^(-1)

I think you meant: sigma^2*(x'x)^-1

Both of these are wrong. Beta_hat = (X'X)^(-1)(X'Y)

exp(-r-q)

A typical change of numaraire. Change measure into that of S2 and find the new drift of S1 under it, then get the expectation of a GeoBrownian. A more common method is to decompose the two brownian motions W1, W2 of stocks S1, S2 to W1 and W2 = pW1 + sqrt(1-p^2)W, and the distribution of S1/S2 can be represented by a single brownian motion, then the coefficients of s1/s2 can be found, then similar integral bounds yield the answer in terms of CDF of Standard Normal and some coefficient.

In fact you don't need any CDF here because you are not asked to price any form of option, just a forward and therefore you are just interested in computing an expectation. There are 2 ways to do this: - If you assume a single brownian motion for both stocks then using S2 as a numeraire is straightforward and will give you a pretty simple result since S1/S2 is a martingale under S2 and you know the radon-Nikodym density for dP(S2)/dQ as well as how to write a brownian motion under S2 as a brownian motion under Q - If you assume 2 different brownian motions (correlation with rho) then you can not directly use the usual numeraire change result but you can still derive an ito calculus to write the risk-neutral dynamics of 1/S1, S2 and then S1/S2. When it is done will will notice that you get a lognormal distribution for S1/S2 and you can write the expectation as the exponential of something with sigma1^2, sigma2^2 and roh*sigma1*sigma2

P: The 5th fish has been caught, P=P1+P2+P2, where, P1: The 5th fish been caught as the first one P2: ... second one... P3: ....third one (or the last one). P1 = 0.5^5 P2 = 4*0.5^5 P3 = 6*0.5^5 The fifth fish will not be caught Pc = 1-P = 1-P1-P2-P3

I got a different answer :/ There's 2 ways the fish can survive: either the bear leaves before the 5th fish, or the bear fails to catch it. The probability the bear leaves is the probability the bear gets 3 fish in 4 attempts, or 4C3/16 = 1/4 The probability the bear tries * fails to catch the fish is 3/4 (the probability of trying) * 1/2 (the probability of failing) = 3/8. The total probability is thus 5/8? Not sure what I did differently

I could not answer it.

Idk, maybe I'm also wrong but it looks extremely simple. If the first lap is 20 km/h, then to get an average of 25 for two laps the second one should be 30 km/h. (20+30)/2 = 25.

There is another option. By definition, average speed is the total length of the route taken divided by the total time it took to cover that distance. Hence in this case it'll be V_avg = (V1*t1 + V2*t2)/(t1 + t2), where t1 and t2 are the times taken to cover the laps and V1*t1 and V2*t2 are the distances. Obviously, as it's the same lap the distances are equal: V1*t1 = V2*t2 = S, and t1 = S/V1, T2 = S/V2 Thus, equation for V_avg becomes V_avg = 2S/(S/V1 + S/V2). S is canceled and V_avg = 2/(1/V1 + 1/V2) = 2V1*V2/(V1 + V2). Solve for V2: V2 = V_avg*V1/(2V1 - Vavg) = 33.3 km/h