## Interview Question

Strategist Interview

-London, England

# 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.

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