Stock price probability question

Discussion in 'SP6' started by barbados, Jul 3, 2011.

  1. barbados

    barbados Member

    Hi,

    I came upon the following question:

    Stock price is currently 50. Its expected return and volatility are 12% and 30%, respectively. What is the probability that the stock price will be greater than 80 in 2 years.

    Obviously, the answer can be obtained by calculating P[S_2 > 80] = P[ln S_2 > ln80] = ... = 1 - Phi(0.7542)

    However, why is the following reasoning not working:

    We know that the change in the stock price is given by:
    dS = mu S dt + sigma S dz -> dS ~ N(mu S dt, sigma^2 S^2 dt)

    So given the question above, I need to calculate P[dS > 80 - 50] =

    P[ N(0,1) > (30 - 0.12 x 50 x 2)/(0.30 x 50 x sqrt(2))] = 1 - Phi(0.8485)

    This is not equal to the first method. Where am I going wrong in the second method.

    PS - the second method I took from Hull Q12.12 which is a similar question.

    Thanks,
    Barbados.
     
    barbados, Jul 3, 2011
    #1
  2. Mike Lewry

    Mike Lewry Member

    Stock price probabilities

    Hi barbados,

    The stock price process does not have independent increments and so we need to condition on the current value, ie assuming the figures you give are annualised, we want to calculate:

    P[S2>80 | S0=50] = P[ln(S2/S0)>ln1.6]

    Now: ln(S2/S0) ~ N(mu.t , sigma^2.t) = N(0.24 , 0.18)

    So: Prob = 1-Phi[(ln1.6-0.24)/sqrt(0.18)] = 1-Phi[0.5421]

    (There might be something else to sort out as well as I can't see where your figure of 0.7542 comes from.)

    The Hull question asks about the stock price the next day, which is very different from asking about two years into the future.

    The expression you give for dS is correct for infinitesimal time intervals, but can't be used for intervals of two years. The normal distribution you state has a mean and variance that both depend on S, but S will vary considerably over two years, so you can't just put in the starting value of 50. The distribution of dS when S=50 is very different from its distribution when S=75, say - we'd expect bigger movements for larger values of S, due to its exponential nature.

    This is the reason that the Greeks can be usfeul for estimating the effect of changes in variables and parameters over the next day, but are fairly worthless over much longer periods.

    The probability you get with the second method is much too small (about 20% instead of about 30%), which is what we'd expect, since it doesn't allow for bigger possible movements as the stock price rises above 50.
     
    Mike Lewry, Jul 4, 2011
    #2

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