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Thread: stochastic calc and expected value

  1. #1
    Senior Member
    Nov 2010
    Hong Kong

    stochastic calc and expected value

    This is rather an unfamiliar territory for me, but I managed to do the first part of the question, so there is no reason why I cannot do the second, especially with some help ))) Besides, I really want to understand how I can use MGF here.


    If v_t=e^{-w_t} and w_t={\gamma}z_t+\frac{1}{2}\gamma^2t, ie dw_t=\frac{1}{2}\gamma^2dt+{\gamma}dz_t,

    where z_t is standard Brownian motion, show that


    Show also that E(e^{-w_t})=1. (Hint: consider the moment generating function of the standard normal variable \frac{z_t}{t}.


    I use Ito's lemma to prove the first:

    first and second derivatives of v_t: f'=(e^{-w_t})'=-e^{-w_t}; f''=e^{-w_t}

    dv_t=(\frac{1}{2}\gamma^2(-e^{-w_t})+\frac{1}{2}e^{-w_t}\gamma^2)dt+\gamma(-e^{-w_t})dz_t=-{\gamma}v_tdz_t (as the first bracket is zero.)

    Now, for the second part, I could try substitution the formula for w_t :

    E(e^{-w_t})=E(e^{-{\gamma}z_t-\frac{1}{2}\gamma^2t})=e^{-1/2\gamma^2t}E(e^{-{\gamma}Z_t})=e^{-1/2{\gamma}^2t}M_{Z_t}(-\gamma)=e^{-1/2\gamma^2t}e^{1/2{\gamma}^2} - somehow I feel I should get e^0=1 but I don't see it yet. Perhaps I am neglecting properties of Brownian motion?

    (in my MGF for Zt I assumed variance of Zt as 1. If, say, std deviation of z_t=\sqrt{t} then I get my result 1.


    Got it! Checked again the definiton of Standard Brownian motion and the variance is t!!

    So, e^{-1/2{\gamma}^2t}M_{Z_t}(-\gamma)=e^{-1/2\gamma^2t}e^{1/2{\gamma}^2t}=e^0=1

    Now tell me if I was wrong!

    PS Who would have thought studying maths could be such an emotional roller-coaster
    Last edited by Volga; Feb 24th 2011 at 12:48 AM.
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