Derivative of a std Normal CDF?

[yamdizzle]

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)


$\Phi$((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.

 

[mathman]

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.

 

[yamdizzle]

standard normal has a mean 0 and variance 1.

 

[yamdizzle]

So basically I need to derive
∫$\frac{1}{\sqrt{2\pi}}$exp($\frac{-x^2}{2}$) 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.

 

[mathman]

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.

 

[yamdizzle]

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.


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

 

[Mute]

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

$$\frac{d\Phi(f(r))}{dr} = \frac{d\Phi(y)}{dy} \frac{dy(r)}{dr},$$

where y = f(r) is the argument of your cdf. Since it's a standard normal distribution, $d\Phi(y)/dy = \exp(-y^2/2)/\sqrt{2\pi}$, 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 $\sigma$.

 

[yamdizzle]

Thank you. That was of great help!