Goldman Sachs Interview Question
AnalystsI think the interviewer asked you to use the normal distribution probability function and interpolate it over 4 times the length of a quarter (whose cumulative probability is 0.16 around the mean) and calculate the probability of losing all the money. It's a formula I don't remember ....
en.wikipedia.org/wiki/Normal_distribution
normal distribution is characterised by 2 variables mean and variance, so you need 2 pieces of information to determine the distribution. We only have one piece of information so can't figure this out.
The normal distribution is a red herring and
1-0.84^4 ~ 50% is the correct answer I believe.
There are two ways we can lose money:
3 quarters of losses or 4 quarters of losses. Otherwise we end up with gains or break even (since probability stays constant).
As such, for 3 quarters of losses we have: .16*.16*.16*.84 (and there are 4 ways to arrange this, so multiply by 4).
And for 4 quarters of losses we have: .16^4. As such, the probability of losing money at the end of the year is 4 * [.16^3*.84] + .16^4
I assume the problem means log returns of the investment during each quarter are i.i.d. Normal ( otherwise doesn't make sense financially ). We let it being a random variable X ~ Normal(mu, sigma)
Prob(X<0) = 0.16, which means sigma = mu, because Prob(X<mu-sigma) ~ 0.16.
Then the end of year log return is 4X ~ Normal(4*mu, 2*sigma).
So Prob(4X<0) = Prob(4X<4*mu - 2*(2*sigma)), i.e. the probability of a normal random variable being smaller than mean by more than twice standard deviation, which is about 2%.
I do not quite understand the problem, for example, if losing money in Q1, and winning money in Q2-Q4, then does that count as losing too?
- screwedupagain March 20, 2010(The interviewer hinted that one should connect the figure 16% with the mean and deviation of normal distribution, blahblahblah)