Ito's Lemma

Discussion in 'CT8' started by James123, Dec 2, 2015.

  1. James123

    James123 Member

    I am getting confused as to the final term in Ito's Lemma:
    sigma(df/dx(t))d(B(t))

    My confusion is with regards to the d(B(t)) part, as this is not consistent in the questions I am doing in the Q&A Bank - sometimes it is not dB(t), but is for example dx(t).

    eg.2.7(d)
    f = 100 + 10Xt
    The final term in Itos lemma is: (df/dx(t))d(B(t))

    eg. 2.13
    Y(t) = log(((xt)^-1) - 1)
    [Are also given the SDE (dX(t) in the Question, which contains B(t)'s if this has an impact?].
    The final term in Itos lemma to find dY(t) is: (df/dx(t))d(X(t))

    eg. 2.15
    S(t)= e^(x(t))
    [Are also given that S(t)=geometric bm =e^(mu(t)+sigma(w(t))]
    The final term in Itos lemma to find dS(t) is: (df/dx(t))d(W(t))

    Apologies is this comes across as confusing, but I just cant see any consistency between the examples and as a result am struggling to understand how you would know which to use in different cases - such as in that final example, how do we know to use d(W(t)) rather than d(x(t)) ?

    [Also, apologies if my Q numbers are incorrect - I am using an old version of the q&a bank]

    Thanks in advance !
     
    Last edited by a moderator: Dec 3, 2015
    James123, Dec 2, 2015
    #1
  2. Steve Hales

    Steve Hales ActEd Tutor Staff Member

    Okay, the first thing to notice is that notation isn't consistent across all questions (either in the Notes or in the exams). For instance, whilst a standard Brownian motion is usually denoted by \(B(t)\) it also appears as \(X(t)\) and sometimes \(W(t)\). There's nothing mysterious about it, you just have to read the question carefully to known how its referring to standard Brownian motion.

    Secondly, I'm not sure you've quite understood how Ito's lemma works. Let's take 2.7(ii)(d) for example (although this is 2.12(ii)(d) in the 2016 version of the Notes).

    Remember that the SDE for \(dX_t\) depends on \(dB_t\), and that \(f\) is a function of \(X_t\). Then we have: \(f'(X_t)=10\) and \(f''(X_t)=0\). Ito's lemma then says: \[df(X_{t})=\frac{\partial f}{\partial X_t}dX_t+\dfrac{1}{2}\frac{\partial^2 f}{\partial X_t^2}(dX_t)^2\]Substituting in the values for the derivatives we have:
    \[df(X_{t})=10dX_t+0\]And part (i) of the questions tells us what \(X_t\) is. So the Brownian motion \(B_t\) creates the noise in the process \(X_t\), which in turn drives the function \(f\).

    Let me know how you get on with the other parts of this question :)
     
    Steve Hales, Dec 7, 2015
    #2
  3. James123

    James123 Member

    Hi Steve, yes I understand the first part of the notation for Brownian motion not being consistent.

    I now understand how you have solved the question in your response, but am struggling to apply that methodology to other questions. Can you please explain how you would go about solving another part of that question where the function is log(Bt), using the same approach you outlined in your response?

    We have f'(x)=(Bt)^-1 and f''(x)=-(Bt)^-2

    If subbing these into Itos lemma as you described below I get:

    df(Xt) = ((Bt)^-1)dXt + 0.5(-(Bt)^-2)((dXt)^2)

    But this is clearly not the solution so what would be the next step?
     
    Last edited by a moderator: Dec 11, 2015
    James123, Dec 11, 2015
    #3
  4. Steve Hales

    Steve Hales ActEd Tutor Staff Member

    Hi

    It looks as though you're very nearly there. The problem is that the function in the question is a function of \(B_t\) and not \(X_t\) explicitly.

    Notice that \(dX_t=\sigma_t dB_t +\mu_tdt\) and that \(dX_t=dB_t\) when \(\sigma_t=1\) and \(\mu_t=0\). If you make this substitution for \(dX_t\) in your solution (and recall that \((dB_t)^2=dt\)) then you'll be done!

    Let me know if this still doesn't work out for you.
     
    Steve Hales, Dec 14, 2015
    #4
  5. James123

    James123 Member

    Hi Steve, thanks for your response, I can see what you've done but just to clarify:

    Here We have used that sigma=1 and mu=0 . Therefore using the substitution dXt = dBt

    However in the previous example we left mu and sigma as they are, to use the substitution dXt = (sigma)dBt + (mu)dt

    How do you know when you can sub in sigma=1, mu=0 as opposed to keeping mu and sigma as they are?...so in that first example could I have subbed in mu=0 and sigma=1 into the answer to get 10dBt as an alternative solution?

    Or am I over complicating this and is it just as simple as saying:
    - if the function is an explicit function of Bt we can use sigma=1, mu=0 and thus dXt=dBt.
    -Whereas if it's a function of X only we keep mu and sigma as they are and thus use the substitution dXt = (sigma)dBt + (mu)dt


    Sorry to be a pain just trying to wrap me head around this!
     
    James123, Dec 14, 2015
    #5
  6. Steve Hales

    Steve Hales ActEd Tutor Staff Member

    Perfect :)
     
    Steve Hales, Dec 14, 2015
    #6

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