Sifting Property of the Impulse Function

[El Moriana]

1. The problem
I have a problem grasping what the point of the sifting property of the Dirac function is.
It isolates the value at a point in a function, right?
Doesn't just substituting that point into the function do exactly the same thing?

Homework Equations

Sifting poperty:
if f(t) is continuous at t=a then

\int_{-∞}^∞f(t)δ(t-a)dt= f(a)

from: Glyn James, Advanced Modern Engineering Mathematics (3rd Ed), Section 2.5, p.155

 

[lanedance]

so based on the properties of the delta function you know
\int \delta (x) dx = 1

A handwaving explanation is that if f is continuous and if you zoom in on a small enough region (x-\epsilon, x+\epsilon), then f(x) will be close to constant on this region.

The delta function zero everywhere except at x=a and the integral evaluates to exactly the value of the function at the point x=a
\int f(x)\delta (x-a) dx = f(a)

It is the same as substituting into the function for f(a), and this is exactly what the inequality tells you. It is a useful propperty of the delta function for various mathematical manipulations

 

[El Moriana]

First of all, thanks for the reply.

I understand how the shifting property works and how it equates to the function at point x=a.

What I don't see are the "various mathematical manipulations", its usefulness.
Under what circumstances does it make life simpler to write \int f(t)δ(t-a)dt than simply f(a)?

 

[lanedance]

Hows your probability? one that springs to mind is as follows...

say you have two independent random variables X & Y, with joint pdf p_{X,Y}(x,y) = p_{X}(x)p_{Y}(y)

And say you want to find the probability distirbution for Z= X+Y [\itex]<br /> <br /> Well p(z) will only have contibutions when z= x+y, or z-x-y=0, so you can write <br /> p_{Z}(z) = \int \int p_{X}(x)p_{Y}(y) \delta(z-x-y)dxdy<br /> p_{Z}(z) = \int \int p_{X}(x)p_{Y}(z-x) dx<br /> <br /> which shows the distribution of the sum of two RVs is given by the convolution of their distributions

 

[lanedance]

Under what circumstances does it make life simpler to write \int f(t)δ(t-a)dt than simply f(a)?

In fact its almost the opposite, in that if you end up with an integral with a delta function in it, you can use the above fact to simplify the expression

 

[El Moriana]

I'm starting to get it. Though, looking at it, would your example not result in:

p_{Z}(z) = \int p_{X}(x)p_{Y}(x-z) dx

due to the sign of the y and the one integral being cancelled?

I take it you would also be able to use this to eliminate the integral w.r.t x if it turns out to be more convenient?

p_{Z}(z) = \int p_{Y}(y)p_{X}(y-z) dy

 

[lanedance]

Not quite you can only use it in this case because you only want contributions from where z=x+y. When integrating over y, this occurs when y=z-x in the delta function.