Understanding the Use of the Delta Function in Probability Density Functions

[Mathsey]

Homework Statement

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

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

where I know the relationship between r and R is given by $$R=r\sin\theta$$

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

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

so that it becomes

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

etc...

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

 

[andrewkirk]

Homework Statement

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

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

where I know the relationship between r and R is given by $$R=r\sin\theta$$

There seems to be something wrong with the way this is expressed.

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

$$\displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(r\sin\theta-r\sin\theta)d\theta d\phi dr<br /> =\int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(0)d\theta d\phi dr$$

which will be infinite.

 

[Mathsey]

There seems to be something wrong with the way this is expressed.

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

$$\displaystyle \int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(r\sin\theta-r\sin\theta)d\theta d\phi dr<br /> =\int_{0}^{r_{max}}\int_0^{2\pi}\int_0^{\pi}\sin\theta\ \delta(0)d\theta d\phi dr$$

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.

 

[Ray Vickson]

Homework Statement

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

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

where I know the relationship between r and R is given by $$R=r\sin\theta$$

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

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

so that it becomes

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

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, $\delta(f(x) - a) \neq \delta(x - f^{-1}(a))$.

 

[andrewkirk]

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.