A question encounted in my interview

Hinstings · February 17th, 2006, 9:36 pm

Suppose the current stock price is $60. The stock price follows a geometric brownian motion with parameter (mu, sigma). What is the expected time that the stock hits $150? I cannot work it out, because I don't know how to calculate the probablity that the stock won't go above $150 within time [0, t], where 0 is the current time. Somebody please help.

Claudio2323 · February 19th, 2006, 7:18 pm

HI,the answer should be E[t_b]=infinitywhere t_b is the stopping time the Brow. Mot. = 150

Paolos · February 20th, 2006, 4:47 pm

QuoteOriginally posted by: Claudio2323HI,the answer should be E[t_b]=infinitywhere t_b is the stopping time the Brow. Mot. = 150Only if: mu <= 0.5*Vol^2otherwise E[t_b]=ln(H/So)/(mu-0.5*vol^2)where H is the level of the barrier and Sò is the current price of the stockRegardsP.

Claudio2323 · February 20th, 2006, 10:47 pm

That's right Paolo,... I forgot the drift in my calc..

noexpert · August 10th, 2007, 3:23 am

Can someone explain this concept to me. Thanks

SPMars · August 12th, 2007, 4:16 pm

QuoteOriginally posted by: noexpertCan someone explain this concept to me. ThanksTake logs inS_t = level.There are standard formulae for the hitting times of levels by brownian motions with drift.Actually, if S_t has drift d and volatility v there are two distinct situations:i) d - 1/2 vol^2 and log(level/S_0) have the same sign (in which case the expected time to hit is given below).ii) When these two quantities have opposite sign (in which case the expected time to hit is infinite).

noexpert · August 14th, 2007, 12:57 am

Thanks SPMars...Ya I found this expression in some text books !