Integral of the Dirac Delta Function

[Quasar Chunawala]

Hi guys,

In the introductory chapter in Shreve II, he talks about a sequence of functions [imath](f_n)_{n=0}^{\infty}[/imath] converging to [imath]f[/imath] almost everywhere. He further gives an example of the sequence of functions [imath](f_n)[/imath], where [imath]f_n[/imath] is the density of Gaussian [imath]N(0,\frac{1}{n})[/imath] random variable:

[math] f_n(x) = \sqrt{\frac{n}{2\pi}}e^{-\frac{nx^2}{2}} [/math]
If we fix [imath]x = x_0, x_0 \neq 0[/imath], its easy to see that the limit [imath]\lim_{n \to \infty} \frac{\sqrt{n}}{e^{n{x_0}^{2}/2}}[/imath] by the L'Hopital's rule is equal to [imath]\lim \frac{1}{(x_0^2/2)} \cdot \frac{1}{2\sqrt{n}} \cdot e^{-n\frac{x_0^2}{2}}[/imath]. Since, the exponential term is bounded, the limit approaches zero.

At [imath]x = 0[/imath], [imath]f_n(0)[/imath] is unbounded. So, [imath](f_n)[/imath] converges pointwise to :

[math] f(x) = \begin{cases}0 & \text{ if } x \neq 0 \\ \infty & \text{ if } x = 0\end{cases} [/math]
I recognize that this is the Dirac Delta function. If we were to try and find the Riemann integral of [imath]f[/imath], the lower integral is [imath]0[/imath], but the upper integral is unbounded, regardless of the partition [imath]P[/imath]. But, I read that, [imath]\int \delta(x) dx = 1[/imath], somehow, magically; so how does this come about..

Thanks,
Quasar

 

[Daniel Duffy]

The above analysis is wrong, in essence. Classical maths breaks down because Dirac is a generalised function aka distribution.

Read this background first and we can talk further

Distributions: What Exactly is the Dirac Delta “Function”?

The theory of distributions , also known as generalized functions, often remains something of an enigma…

www.cantorsparadise.com

L'Hopital's rule
Riemann
no.

 

[Quasar Chunawala]

That was the theory of distributions 101 for me.

So, a generalized function or distribution [imath]T[/imath] is any scattering of unit mass over the real line. And to describe the distribution, we are interested in the density/mass over small intervals, so we use measurement devices aka test functions [imath]\phi \in C^{\infty}_c[/imath] such as the bump function. And we describe the distribution by its action on the test functions; how it maps [imath]\phi[/imath] to a real number [imath]\int_{-\infty}^{\infty} f(x) \phi(x) dx[/imath].

The definition of differentiability of classical functions is too strong, so we define a weak(generalized) derivative, by the rule [imath]T'(\phi) =T(-\phi')[/imath].

Dirac distribution concentrates all mass at a singular point.

 

[Quasar Chunawala]

I didn't realize this, until I actually read it on Wilmott. The delta of a European call with time to expiration [imath]0[/imath] is just the heavyside function, so the gamma of the call is the Dirac distribution.

 

[Daniel Duffy]

Have you been reading my (@Cuchulainn) posts, Quasar?

https://forum.wilmott.com/viewtopic.php?f=34&t=102822

This thesis => compute gamma as a PDE

https://www.datasim.nl/application/files/5915/7045/5027/Matt_Robinson_Thesis_.pdf

 

[Quasar Chunawala]

It's nice, inspiring to read Wilmott, before winding up the day; I especially like some of the math brain-teasers, you ask and the general conversation there. I always learn something new.

I am hoping to learn and explore some of the methods, in the thesis paper you shared, during the PDE/FDM course (hope to start in October).