Determine marginal distribution of random vector on unit sphere

[psie]

Homework Statement

Let $(X,Y,Z)$ be a point chosen uniformly within the three-dimensional unit sphere. Determine the marginal distributions of $(X,Y)$ and $X$.

Relevant Equations

The marginal distributions of say $X$ is obtained by integrating out the other variables. The surface area of the unit sphere is $4\pi$.

I'm tempted to write the joint density $f_{XYZ}$ as $$f_{XYZ}(x,y,z)=\begin{cases}\frac1{4\pi}&\text{if }x^2+y^2+z^2=1, \\ 0&\text{otherwise.}\end{cases}$$However, from other sources, I've read that a uniform distribution on the unit sphere does not have a density in three variables. If this is true, I'd be grateful if someone could explain this. Moreover, if there is no $f_{XYZ}$, then how does one tackle the problem?

 

[Orodruin]

I'm tempted to write the joint density $f_{XYZ}$ as $$f_{XYZ}(x,y,z)=\begin{cases}\frac1{4\pi}&\text{if }x^2+y^2+z^2=1, \\ 0&\text{otherwise.}\end{cases}$$However, from other sources, I've read that a uniform distribution on the unit sphere does not have a density in three variables. If this is true, I'd be grateful if someone could explain this. Moreover, if there is no $f_{XYZ}$, then how does one tackle the problem?

A surface distribution indeed does not have a regular function as a three-dimensional density. A three-dimensional density is a function that, integrated over a volume, gives the total within that volume. However, the volume of the sphere's surface is zero (it is two-dimensional) so integrating anything like the function you have given over it is going to return zero.

There are two things you may do:

1. Only consider the surface distribution. Define a function $\sigma$ only on the 2-dimensional surface such that $\sigma\, dA$ is the charge on a small area $dA$.
2. Use a generalised function - a distribution - to describe the 3-dimensional density. The relevant distribution here would involve a delta-distribution $\delta(r - \sqrt{x^2 + y^2 + z^2})$.

 

[psie]

I think I understand why $W=(X,Y,Z)$ doesn't have a density. Just to reiterate. We have $$P(W\in S^2)=P_W(S^2)=\int_{S^2} f\, dx=0,$$ for any nonnegative Borel measurable function$f$ and where $P_W$ is the law of $W$. The last equality follows from the fact that $S^2$ has zero $3$-dimensional Lebesgue measure, but we should have $P(W\in S^2)=1$.

Anyway, I'm somewhat lost on how to continue. I'm attaching a partial solution to the problem. I have several question marks.

1. Overall, I don't understand what they are doing. Why express $\sin^2\theta$ as $x^2+y^2$? What is the approach they are taking?
2. I'm not very familiar with wedge products, and I think they use these. However, isn't an area element of the unit sphere simply $\sin\theta d\varphi d\theta$? Why write $\sin\theta d\varphi\land d\theta$?
3. In the third line of the displayed equations, they have divided by $2$ and $\cos\theta$. I believe they express $\cos\theta$ as $\sqrt{1-x^2-y^2}$. But we have $\cos^2\theta=1-\sin^2\theta=1-x^2-y^2$, so $|\cos\theta|=\sqrt{1-x^2-y^2}$. Note the absolute values. Hence I do not understand what happened in the third equation.
4. Lastly, I don't really understand why $(X,Y)$ has a density? Or $X$?

Grateful if someone could comment on this, or help in some way with the problem.

 

[Orodruin]

1. They are expressing $\sin^2\theta$ in terms of $x$ and $y$. The aim is to differentiate it to obtain the differential of the coordinate function $\theta$. As it (happily) turns out, the differential of $\sin^2\theta$ contains the product of $\sin\theta$ and $d\theta$, which appears directly in the area form. Once they have divided by $\cos\theta = \sqrt{1-x^2-y^2}$ they have $\sin\theta \, d\theta$.
2. $d\theta \wedge d\varphi$ is an area form - a differential 2-form (a $p$-form is a fully antisymmetric (0,p) tensor). You can think of it as a directed area element. It takes two vectors (on the sphere) as argument and returns the area spanned by those vectors. In the theory of integration of differential forms, the vectors plugged into the differential forms are the tangent vectors of the coordinate lines corresponding to your integration coordinates. This is why, when you use coordinates $\theta$ and $\varphi$, the scalar area element becomes $\sin(\theta) d\theta\, d\varphi$. The mathematics of the wedge product of differentials will directly give you the Jacobian determinant when you perform a coordinate change. What you need to remember is essentially that the wedge product is fully antisymmetric. (You will need this when taking the wedge product $\sin\theta \, d\theta \wedge d\varphi$)
3. You can restrict yourself to the upper half of the sphere for which $z \geq 0$ (i.e., $\cos\theta \geq 0$). By symmetry the lower half of the sphere will give the same density in the x-y-projection. You will anyway normalise the distribution in the end.
4. They have changed to parametrise the sphere by the $x$ and $y$ coordinates. The area form in the $x$-$y$ plane is just $dx\wedge dy$ so whatever multiplies this is the density in $xy$. To find the density in $x$ only you must marginalise $y$.

Edit: Note their typo $d\varphi \wedge d\theta$ instead of $d\theta \wedge d\varphi$. Normally I would suspect the typical mathematician case of using the opposite convention for which angle to call what, but this is contradicted by the $\theta$ in the sine argument. It is also easy to check that their wedge product would have the opposite sign if they actually had $d\varphi \wedge d\theta$.

 

[psie]

Ok, thanks a lot.

After having computed this myself, I think there's a typo in the solution I attached, or? They write $$\sin\theta d\varphi\land d\theta=\frac1{\sqrt{1-x^2-y^2}}dx\wedge dy,$$ however I claim it should be $$\sin\theta d\varphi\land d\theta=\frac1{\sqrt{1-x^2-y^2}} dy\wedge dx,$$i.e. instead of $dx\wedge dy$ we have $dy\wedge dx$. We have\begin{align}\sin\theta d\theta&=\frac{xdx+ydy}{\sqrt{1-x^2-y^2}} \nonumber \\ d\varphi&=\frac1{1+\frac{y^2}{x^2}}\left(\frac1xdy-\frac{y}{x^2}dx\right)=\frac{x^2}{x^2+y^2}\left(\frac1xdy-\frac{y}{x^2}dx\right) \nonumber \end{align} So, \begin{align}\sin\theta d\varphi\wedge d\theta&=d\varphi\wedge\sin\theta d\theta \nonumber \\ &=\frac{x dy}{x^2+y^2}-\frac{y dx}{x^2+y^2}\bigwedge \frac{xdx}{\sqrt{1-x^2-y^2}}+\frac{ydy}{\sqrt{1-x^2-y^2}}\nonumber \\ &=\frac{x^2}{x^2+y^2}\frac{1}{\sqrt{1-x^2-y^2}}dy\wedge dx-\frac{y^2}{x^2+y^2}\frac{1}{\sqrt{1-x^2-y^2}}dx\wedge dy\nonumber \\ &=\frac{1}{\sqrt{1-x^2-y^2}} dy\wedge dx\nonumber.\end{align}

 

[Orodruin]

Ok, thanks a lot.

After having computed this myself, I think there's a typo in the solution I attached, or? They write $$\sin\theta d\varphi\land d\theta=\frac1{\sqrt{1-x^2-y^2}}dx\wedge dy,$$ however I claim it should be $$\sin\theta d\varphi\land d\theta=\frac1{\sqrt{1-x^2-y^2}} dy\wedge dx,$$i.e. instead of $dx\wedge dy$ we have $dy\wedge dx$. We have\begin{align}\sin\theta d\theta&=\frac{xdx+ydy}{\sqrt{1-x^2-y^2}} \nonumber \\ d\varphi&=\frac1{1+\frac{y^2}{x^2}}\left(\frac1xdy-\frac{y}{x^2}dx\right)=\frac{x^2}{x^2+y^2}\left(\frac1xdy-\frac{y}{x^2}dx\right) \nonumber \end{align} So, \begin{align}\sin\theta d\varphi\wedge d\theta&=d\varphi\wedge\sin\theta d\theta \nonumber \\ &=\frac{x dy}{x^2+y^2}-\frac{y dx}{x^2+y^2}\bigwedge \frac{xdx}{\sqrt{1-x^2-y^2}}+\frac{ydy}{\sqrt{1-x^2-y^2}}\nonumber \\ &=\frac{x^2}{x^2+y^2}\frac{1}{\sqrt{1-x^2-y^2}}dy\wedge dx-\frac{y^2}{x^2+y^2}\frac{1}{\sqrt{1-x^2-y^2}}dx\wedge dy\nonumber \\ &=\frac{1}{\sqrt{1-x^2-y^2}} dy\wedge dx\nonumber.\end{align}

Like I said previously:

Note their typo $d\varphi \wedge d\theta$ instead of $d\theta \wedge d\varphi$. Normally I would suspect the typical mathematician case of using the opposite convention for which angle to call what, but this is contradicted by the $\theta$ in the sine argument. It is also easy to check that their wedge product would have the opposite sign if they actually had $d\varphi \wedge d\theta$.

 

[psie]

Ok, thanks @Orodruin, I missed your edit.

I think I'm starting to understand this more and more (although I must confess, I really believe the problem is meant to be for the unit ball, but whatever, this is also an interesting problem!). I'm trying to formulate the problem into an equation where we want to find the unknown density; I'm tempted to write $$\iint f(x,y)\,dxdy=\iint \sin\theta \,d\theta d\varphi =1,$$where the coordinates in the latter integral are the spherical coordinates, and say we want to find $f$, which then should be the Jacobian of that transformation. But is this correct? After all, spherical coordinates are a 3-dimensional thing, so I'm not sure where $\theta$ and $\varphi$ "live", or how to write down the relationship between $x,y$ and $\theta,\varphi$.

 

[Orodruin]

Spherical coordinates are generally three-dimensional. However, restricting r to be fixed, you have a two-dimensional coordinate system for the sphere of the fixed radius.

Yes, you can do it like that with the Jacobian, but using the wedge product does that essentially for you.

 

[psie]

I'm tempted to write $$\iint f(x,y)\,dxdy=\iint \sin\theta \,d\theta d\varphi =1,$$where the coordinates in the latter integral are the spherical coordinates, and say we want to find $f$, which then should be the Jacobian of that transformation.

This was partly not right, since $\iint \sin\theta \,d\theta d\varphi=4\pi$, so if anything we'd want $\iint f(x,y)\,dxdy=\iint \frac{\sin\theta}{4\pi} \,d\theta d\varphi =1$.

The area form in the $x$-$y$ plane is just $dx\wedge dy$ so whatever multiplies this is the density in $xy$.

I have one more question. How do we know that whatever multiplies it is unique? I'm not sure uniqueness is a requirement for the density, but it seems like it should be.

 

[Orodruin]

I have one more question. How do we know that whatever multiplies it is unique? I'm not sure uniqueness is a requirement for the density, but it seems like it should be.

I am not sure I understand the question. You have a particular distribution on the sphere and changing coordinates is not going to change that you have that particular distribution, just how that distribution is expressed.