Ito process

Discussion in 'CT8' started by hatton02, Apr 7, 2014.

  1. hatton02

    hatton02 Member

    I need help with either using Taylor's formula or Ito lemma. Taylor's expression on page 2ish of the tables don't seem to match what we need. In the core notes we have a question that says consider the SDE

    DSt = aSt dt + sig St dBt

    It then says divide by St to separate the variables

    Then it says we can use itos lemma to calculate d log St.

    D log St = 1/St dSt + 1/2 (-1/St^2)(dSt)^2

    How do we get to this?
     
  2. DaddyCruel

    DaddyCruel Member

    I think they use the separation of variables to give you an intuition of what the likely solution is going to be.

    Of course you can use regular calculus with stochastic processes so you get d[Ln(St)] by applying Ito's lemma.
     
  3. Graham Aylott

    Graham Aylott Member

    Let f(St) = log(St), then using a Taylor Series in one variable, we have:

    df(St) = d(log(St) = df/dSt * dSt + 0.5 * d^2f/dSt^2 * (dSt^2)

    Here:

    df/dSt = 1/St and d^2f/dSt^2 = -1/St^2

    So, we have:

    d(log(St) = 1/St * dSt + 0.5 * -1/St^2 * (dSt^2)

    Finally, as:

    dSt= a*St*dt + sig*St*dBt

    and:

    dSt^2 = (a*St*dt)^2 + (sig*St*dBt)^2 + 2*a*St*dt*sig*St*dBt

    = 0 + sig^2*St^2*dt + 0 = sig^2*St^2*dt

    we have:

    d(log(St) = 1/St *(a*St*dt + sig*St*dBt) + 0.5 * -1/St^2 * sig^2*St^2*dt

    So, cancelling all the St terms gives:

    d(log(St) = a*dt + sig*dBt - 0.5 *sig^2*dt

    ie:

    d(log(St) = (a - 0.5 *sig^2)*dt + sig*dBt

    I hope this helps. :)
     
  4. hatton02

    hatton02 Member

    Thanks that helps. The answer to my question I think is "using Taylor series" but I don't really understand what Taylor series is! I mean in your derivation below you've said what df(St) is but I don't remember doing just df.....we did df/dx so in my eyes it's missing a denominator and then I'm screwed! I can't see how Taylor series in one variable is derived as surely things would cancel out to give:

    df(St) = df + 0.5 d^2f.

    Or is this not how differentiation works?! I've genuinely forgotten.

    With practice I've managed to work out how to use itos lemma and a previous thread on here explains there's two starting points depending on whether you have geometric brownian motion or ormsbeck

     
  5. Graham Aylott

    Graham Aylott Member

    In general, df(x) is the differential of f(x) and represents the change in the value of function f in response to a small change in x.

    Conversely, df/dx is the rate of change of f with respect to x.

    The Taylor Series on page 3 in the Tables show us how the value of f(x+h), where "h" is a small increment in x, is related to f(x) itself and the partial derivative terms in x, df/dx, d^2f/dx^2 etc. (The higher-order partial deriviatives allow for the fact that df/dx itself is generally not constant, ie the relationship between f and x is generally not linear.) Remember that a partial derivative tells us how f changes in response to a small change in x, all else being equal (ie nothing else changing).

    So, df(x) = f(x+h) - f(x) and from page 3:

    df(x) = df/dx *dx + 0.5*df^2/dx^2 + ...

    ie the change in f (df) is equal to the change in x (dx) times the rate rate of change of with respct to x.

    In practice, if x is a well behaved deterministic function then we can ignore the higher-order terms (which are "small") and just write:

    df(x) = df/dx *dx

    In CT8, f is typically a derivative price whose value depends on two variables, usually deterministic time t and the stochastic asset price x, ie f(t, x). So here the overall change in the value of f will depend on the changes in both t and x over a short dt instant of time. So, we allow for this by using the Taylor Series in two variables (see page 3 in the Tables again) and including terms in the partial derivatives of both t and x and also the cross terms.

    Generally in CT8, as t is deterministic and x is stochastic, the partial derivatives of order two or higher in t and the partial derivatives of order three or higher in x and all of the cross term in both t and x are sufficiently "small" that we can ignore them and so we have the overall change in f(t, x) as:

    df(t,x) = df/dt* dt + df/dx*dx + 0.5*d^2f/dx^2 * (dx)^2

    So, we just need to work out the three partial derivatives and substitute them into this formula, plus we substitute in the formula for dx and also square the formula for dx (using the multiplication table on page 19 of Chapter 9) and substitute that in.

    Ito's Lemma is just a special case of this Taylor Series approach, in which dx has the form:

    dx = a(t,x)*dt +b(t,x)*dZt

    where a(t,x) is the drift function and b(t,x) the volatility function and the substitution of dx and dx^2 into the Taylor Series formula has already been done for us (using the multiplication table). :)
     

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