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A question encounted in my interview
Posted: February 17th, 2006, 9:36 pm
by Hinstings
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.
A question encounted in my interview
Posted: February 19th, 2006, 7:18 pm
by Claudio2323
HI,the answer should be E[t_b]=infinitywhere t_b is the stopping time the Brow. Mot. = 150
A question encounted in my interview
Posted: February 20th, 2006, 4:47 pm
by Paolos
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.
A question encounted in my interview
Posted: February 20th, 2006, 10:47 pm
by Claudio2323
That's right Paolo,... I forgot the drift in my calc..
A question encounted in my interview
Posted: August 10th, 2007, 3:23 am
by noexpert
Can someone explain this concept to me. Thanks
A question encounted in my interview
Posted: August 12th, 2007, 4:16 pm
by SPMars
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).
A question encounted in my interview
Posted: August 14th, 2007, 12:57 am
by noexpert
Thanks SPMars...Ya I found this expression in some text books !