Interview Question

Strategist Interview

-London, England

Goldman Sachs

1) Manufacture and option from plain vanillas that makes money in case of low volatility and does not lose too much in the opposite scenario. 2) We are in a junction. p is the probability that at least one car will pass through it within the next 20 minutes. Give me the probability that no car will pass within the next 5 minutes.

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5 Answers

20

Let q be the probability of seeing no car in any given 5 minutes. Then the prob of not seeing a car in 20 mins is q^4 (or 1 - p). So q = (1-p)^0.25

Denys on

2

For question 2, one needs to use the Poisson distribution. The number of cars passing through within 20 mins distributes according to a Poisson distribution with parameter lambda. 1-p is the probability that no car will pass through within the next 20 minutes. Therefore 1-p=e^(-lambda). The number of cars passing through within 5 mins distributes according to a Poisson distribution with parameter lambda/4. Therefore, the probability that no car will pass within the next 5 minutes is e^(-lambda/4)=(1-p)^(1/4)

Hugh on

2

1) Butterfly: Call(S0 - K) -2*Call(S0) + Call(S0 + K), K = Sigma*Sqrt(T), T: option expiration.

FMP on

0

Standard and fair interview

Anonymous on

2

[2] 1 - p/4 for no cars in first 5 min

Manish on

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