Understanding the Use of the Delta Function in Probability Density Functions

In summary, there seems to be an issue with the way the integral is expressed in the problem, and the use of the Dirac delta function may not be appropriate. There may be an error in the derivation that led to this expression.
  • #1
Mathsey
2
0

Homework Statement


So I have an issue evaluating the integral for a joint probability distribution given by:

[tex]Pr(R) = \displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta \delta(R-r\sin\theta)d\theta d\phi dr[/tex]

where I know the relationship between r and R is given by [tex]R=r\sin\theta[/tex]

Are there any special properties of the delta function I should be aware of besides it's sifting property?

Homework Equations

The Attempt at a Solution



I have tried evaluating this by re-writing the integral as

[tex]\int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta \delta(\sin^{-1}\left(\frac{R}{r}\right) - \theta)d\theta d\phi dr[/tex]

so that it becomes

[tex]\int_{0}^{r_{max}}\int_0^{2\pi} \sin[\sin^{-1}\left(\frac{R}{r}\right)] d\phi dr[/tex]

etc...

but this cannot be evaluated so I think this is just wrong, any help would be great.
 
Last edited:
Physics news on Phys.org
  • #2
Mathsey said:

Homework Statement


So I have an issue evaluating the integral for a joint distribution given by:

[tex]\displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta \delta(R-r\sin\theta)d\theta d\phi dr[/tex]

where I know the relationship between r and R is given by [tex]R=r\sin\theta[/tex]
There seems to be something wrong with the way this is expressed.

If ##R=r\sin\theta## then the integral is

[tex]\displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(r\sin\theta-r\sin\theta)d\theta d\phi dr
=\int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(0)d\theta d\phi dr[/tex]

which will be infinite.
 
  • #3
andrewkirk said:
There seems to be something wrong with the way this is expressed.

If ##R=r\sin\theta## then the integral is

[tex]\displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(r\sin\theta-r\sin\theta)d\theta d\phi dr
=\int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(0)d\theta d\phi dr[/tex]

which will be infinite.

So I don't know if this helps but I think this is a fairly standard way of expressing a probability density function of multiple variables, I should probably have made that more clear, see https://en.wikipedia.org/wiki/Probability_density_function, under multiple variables.
 
  • #4
Mathsey said:

Homework Statement


So I have an issue evaluating the integral for a joint probability distribution given by:

[tex]Pr(R) = \displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta \delta(R-r\sin\theta)d\theta d\phi dr[/tex]

where I know the relationship between r and R is given by [tex]R=r\sin\theta[/tex]

Are there any special properties of the delta function I should be aware of besides it's sifting property?

Homework Equations

The Attempt at a Solution



I have tried evaluating this by re-writing the integral as

[tex]\int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta \delta(\sin^{-1}\left(\frac{R}{r}\right) - \theta)d\theta d\phi dr[/tex]

so that it becomes

[tex]\int_{0}^{r_{max}}\int_0^{2\pi} \sin[\sin^{-1}\left(\frac{R}{r}\right)] d\phi dr[/tex]

etc...

but this cannot be evaluated so I think this is just wrong, any help would be great.

Why can't it be evaluated? ##\sin( \arcsin( y)) = y##.

However, a more serious criticism is that ##\delta(R - r \sin \theta) \neq \delta (\arcsin(R/r) - \theta)##. In general, ##\delta(f(x) - a) \neq \delta(x - f^{-1}(a))##.
 
  • #5
Mathsey said:
So I don't know if this helps but I think this is a fairly standard way of expressing a probability density function of multiple variables, I should probably have made that more clear, see https://en.wikipedia.org/wiki/Probability_density_function, under multiple variables.
The problem with the expression in the OP relates to the use of the Dirac delta function. The section of that wiki article on multiple variables does not contain any uses of the Dirac delta.

It might help if you posted the original problem. What you have posted looks like something derived, and my guess is that the derivation, which was not posted, contains an error.
 

Related to Understanding the Use of the Delta Function in Probability Density Functions

What is an integral with delta function?

An integral with delta function is a type of mathematical integral that involves the use of the Dirac delta function, which is a mathematical function that is zero everywhere except at one point, where it is infinite. This type of integral is often used in physics and engineering to model point-like sources or to represent impulses in a system.

How is an integral with delta function calculated?

To calculate an integral with delta function, you first need to define the limits of integration and then substitute the Dirac delta function into the integral. The resulting integral will have a value of either 0 or the value of the function at the point where the Dirac delta function is located.

What are some applications of integrals with delta function?

Integrals with delta function have many applications in physics and engineering, such as modeling point-like sources of electromagnetic fields, analyzing impulse responses in control systems, and solving differential equations with discontinuous boundary conditions.

What is the relationship between integrals with delta function and the Heaviside step function?

The Heaviside step function is closely related to the Dirac delta function and is often used in conjunction with integrals with delta function. The Heaviside step function is defined as 0 for negative values and 1 for positive values, and when multiplied by the Dirac delta function, it can help to define the limits of integration for an integral with delta function.

What are some properties of integrals with delta function?

Some properties of integrals with delta function include linearity, time shifting, and scaling. Linearity means that the integral with delta function can be split into multiple integrals, time shifting allows the function to be shifted along the x-axis, and scaling allows the amplitude of the function to be multiplied by a constant.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
649
  • Calculus and Beyond Homework Help
Replies
9
Views
351
  • Calculus and Beyond Homework Help
Replies
3
Views
940
  • Calculus and Beyond Homework Help
Replies
4
Views
365
  • Calculus and Beyond Homework Help
Replies
5
Views
765
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
934
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
998
  • Calculus and Beyond Homework Help
Replies
3
Views
453
Back
Top