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Derivative of a std Normal CDF?

  1. Nov 24, 2011 #1
    I was wondering how I can find the derivative of a normal cdf with respect to a boundary parameter?
    I can get an answer with Mathematica or something but I have no idea how to actually do this. I don't know how fundamental theorem of calculus can be applied. (if it can be)


    [itex]\Phi[/itex]((log(S/K)+(r+σ^2)T)/(σ sqrt(T)))

    where phi is the std normal cdf.

    I'm doing option pricing analysis but I can't really figure out how to derive this with respect to r or sigma or any other parameter. I would appreciate any help.
     
  2. jcsd
  3. Nov 24, 2011 #2

    mathman

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    The standard normal distribution is usually expressed in terms of 2 parameters, the mean and variance. Your expression has one parameter, so I can't tell what you mean.
     
  4. Nov 24, 2011 #3
    standard normal has a mean 0 and variance 1.
     
  5. Nov 24, 2011 #4
    So basically I need to derive
    ∫[itex]\frac{1}{\sqrt{2\pi}}[/itex]exp([itex]\frac{-x^2}{2}[/itex]) dx

    with upper bound
    (log(S/K)+(r+σ^2)T)/(σ sqrt(T))
    lower bound -inf

    with respect to r, Sigma or any parameter so I can learn how to do this.
     
  6. Nov 25, 2011 #5

    mathman

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    The cdf of the normal distribution cannot be expressed analytically. To use it you need to calculate the upper bound explicitly and get the answer from a table.
     
  7. Nov 25, 2011 #6
    How do you think those values were found?
    The derivative with respect to r is this... according to Mathematica. But I don't know how.


    [itex]\frac{\sqrt{T}}{\sqrt{2\pi}σ}[/itex]exp(-([itex]\frac{(T (r + σ^2) + Log[S/K] )}{\sqrt{2 T σ}}[/itex])^2)

    with changing a few things ==

    T * Normal PDF(-log(S/K),Tσ^2) at point T*(r+^2)

    Technically this is suppose to be 0 as normal pdf is 0 at any point since it is continuous but something different can be acquired with deriving with respect to something else.
     
  8. Nov 25, 2011 #7

    Mute

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    Just use the chain rule. e.g., for r

    [tex]\frac{d\Phi(f(r))}{dr} = \frac{d\Phi(y)}{dy} \frac{dy(r)}{dr},[/tex]

    where y = f(r) is the argument of your cdf. Since it's a standard normal distribution, [itex]d\Phi(y)/dy = \exp(-y^2/2)/\sqrt{2\pi}[/itex], and then plug in y = f(r), of course.

    You can do a similar thing treating the argument of the cdf as a function of [itex]\sigma[/itex].
     
  9. Nov 26, 2011 #8
    Thank you. That was of great help!
     
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