# Sifting Property of the Impulse Function

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
Homework Helper
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

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
Homework Helper
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] Well p(z) will only have contibutions when z= x+y, or z-x-y=0, so you can write $$p_{Z}(z) = \int \int p_{X}(x)p_{Y}(y) \delta(z-x-y)dxdy$$ $$p_{Z}(z) = \int \int p_{X}(x)p_{Y}(z-x) dx$$ which shows the distribution of the sum of two RVs is given by the convolution of their distributions lanedance Homework Helper Under what circumstances does it make life simpler to write [itex]\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

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
Homework Helper
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.