# Simple probability question about a 2-sphere

• binbagsss
In summary: Well, it's not clear to me whether it should have said##\rho(\theta,\phi)=\frac 1{4\pi}##, ##\int\int \rho \sin(\theta).d\theta d\phi=\int\int \frac 1{4\pi} \sin(\theta).d\theta d\phi##, or##\rho(\theta,\phi)=\frac 1{4\pi}\sin(\theta)##, ##\int\int \rho \sin(\theta).d\theta d\phi=\int\int \frac 1{4\pi} \sin^2(\theta).d\theta d\phi##,but you
binbagsss

## Homework Statement

I have probabilty density ##\rho(\theta,\phi)=\frac{1}{4\pi} sin \theta d\theta d\phi##
The question is let ##\theta## be measured with respect to the z-axis, find the probability ##p_{z}(z) dz ## that the particle lies between ##z## and ##z+dz##?

see above

## The Attempt at a Solution

My method:

##z=cos \theta ##

##\frac{d\theta}{dz}= \frac{-1}{sin\theta}##

##\int^{2\pi}_{0} d\phi \int \frac{1}{4\pi} sin \theta dz . \frac{d\theta}{dz} ##

= ##2\pi \frac{-1}{4\pi} = - \frac{1}{2} ##

This is wrong by a minus sign and the solutions instead do:

##P_{z}(z)=...=p(\theta) |\frac{d\theta}{dz}|= 1/2##?

(the '...' being the integration over ##\phi##. )

MY QUESTION:

I don't understand where or why the modulus signs come from in the transformation variables : ##|\frac{d\theta}{dz}|##?

Many thanks

binbagsss said:
This is wrong by a minus sign
it also is not of the form ##
p_{z}(z) dz## at all.
With ##z = \cos\theta## you seem to be working on a unit sphere. Is that a correct assumption ?

You correctly have ##dz = \sin\theta\;d\theta = -d\cos\theta##
The minus sign is only because the directions of ##z## and ##\theta ## are opposite.
You are looking at a probability density function and the probability (area between ##z## and ##z+dz##) is probability density times ##|dz|##
With no phi dependence you get the ##1\over 2## so the correct answer to the question should be ##{1\over 2} dz##, not just ##1\over 2##

BvU said:
You are looking at a probability density function and the probability (area between ##z## and ##z+dz##) is probability density times ##|dz|##
With no phi dependence you get the ##1\over 2## so the correct answer to the question should be ##{1\over 2} dz##, not just ##1\over 2##

okay thanks I think that is making more sense.

However from what i understand you've just said , that it is probability density times ##|dz|##, I get ## - 1 \over 2 ## ##|dz|##, instead
Because the probability density comes with the minus sign, and then you say that rather than multiplying by just ##dz## we should multiply by ##|dz|##?

Many thanks

binbagsss said:

## Homework Statement

I have probabilty density ##\rho(\theta,\phi)=\frac{1}{4\pi} sin \theta d\theta d\phi##
The question is let ##\theta## be measured with respect to the z-axis, find the probability ##p_{z}(z) dz ## that the particle lies between ##z## and ##z+dz##?

see above

## The Attempt at a Solution

My method:

##z=cos \theta ##

##\frac{d\theta}{dz}= \frac{-1}{sin\theta}##

##\int^{2\pi}_{0} d\phi \int \frac{1}{4\pi} sin \theta dz . \frac{d\theta}{dz} ##

= ##2\pi \frac{-1}{4\pi} = - \frac{1}{2} ##

This is wrong by a minus sign and the solutions instead do:

##P_{z}(z)=...=p(\theta) |\frac{d\theta}{dz}|= 1/2##?

(the '...' being the integration over ##\phi##. )

MY QUESTION:

I don't understand where or why the modulus signs come from in the transformation variables : ##|\frac{d\theta}{dz}|##?

Many thanks

Before I start:
(1) the probability density is just ##\frac{1}{4\pi} \sin \theta##; it does not include ##d \theta## and ##d \phi##, which refer to integration of the density.
(2) do not write ##sin \theta## and ##cos \theta##; they look ugly and are hard to read. Instead, write ##\sin \theta## and ##\cos \theta##, which you do by putting a "\" in front of the sin and cos (so write "\sin" instead of "sin", etc). That holds as well for all the other trig functions and things like "arcsin", etc., as well as for "exp", "log", "ln", "max", "min", "lim", and all the hyperbolic functions like "sinh", etc.

Rather than using canned formulas, I, personally, prefer to work first with the cumulative distribution, then differentiate it to get the density. So, if ##Z## is your ##z##-random variable, we have
$$F_Z(z) = P(Z \leq z) = P\{(\theta, \phi) : \cos \theta \leq z\} = \int_{\phi=0}^{2\pi} \int_{\theta=\arccos(z)}^{\pi} \rho(\theta,\phi)\, d \phi \, d \theta.$$
The ##\theta## limits go from ##\arccos(z)## to ##\pi## because we are looking at points on the sphere that lie below ##z##.

Now the probability density of ##Z## is ##f_Z(z) = dF_Z(z)/dz##, which is easily calculated using standard rules for differentiation of integrals with respect to their upper or lower limits.

The correct answer for the density function of ##Z## is, indeed, ##f_Z(z) = 1/2##; it should not have a ##dz## in it, because the ##dz## comes into play only when you integrate the density; it is not part of the density itself. See, eg.,
https://en.wikipedia.org/wiki/Probability_density_function or
https://onlinecourses.science.psu.edu/stat414/node/97 .

Last edited:
Ray Vickson said:
$$\int_{\phi=0}^{2\pi} \int_{\theta=\arccos(z)}^{\pi} \rho(\theta,\phi)\, d \phi \, d \theta.$$
For polar integration, shouldn't there be a ##\sin \theta ## factor in the integrand?

haruspex said:
For polar integration, shouldn't there be a ##\sin \theta ## factor in the integrand?

That factor is already embedded in ##\rho(\theta, \phi)##.

Ray Vickson said:
That factor is already embedded in ##\rho(\theta, \phi)##.
Is it? If ρ is the surface density then the integral is ∫∫ρ sin(θ)dθdφ. The factor sin(θ) in the function ρ is extra, no?

Edit: the question is, is the density ##\frac 1{4\pi}## or, as you wrote in post #4, ##\frac 1{4\pi}\sin(\theta)##?

Last edited:
haruspex said:
Is it? If ρ is the surface density then the integral is ∫∫ρ sin(θ)dθdφ. The factor sin(θ) in the function ρ is extra, no?

Edit: the question is, is the density ##\frac 1{4\pi}## or, as you wrote in post #4, ##\frac 1{4\pi}\sin(\theta)##?

In post #1 the OP defined ##\rho(\theta,\phi) = \sin(\theta)/4\pi## (but mistakenly including the differentials ##d\theta## and ##d\phi##.

Ray Vickson said:
In post #1 the OP defined ##\rho(\theta,\phi) = \sin(\theta)/4\pi## (but mistakenly including the differentials ##d\theta## and ##d\phi##.
Well, it's not clear to me whether it should have said
##\rho(\theta,\phi)=\frac 1{4\pi}##, ##\int\int \rho \sin(\theta).d\theta d\phi=\int\int \frac 1{4\pi} \sin(\theta).d\theta d\phi##, or
##\rho(\theta,\phi)=\frac 1{4\pi}\sin(\theta)##, ##\int\int \rho \sin(\theta).d\theta d\phi=\int\int \frac 1{4\pi} \sin^2(\theta).d\theta d\phi##,
but you cannot have it both ways.

haruspex said:
Well, it's not clear to me whether it should have said
##\rho(\theta,\phi)=\frac 1{4\pi}##, ##\int\int \rho \sin(\theta).d\theta d\phi=\int\int \frac 1{4\pi} \sin(\theta).d\theta d\phi##, or
##\rho(\theta,\phi)=\frac 1{4\pi}\sin(\theta)##, ##\int\int \rho \sin(\theta).d\theta d\phi=\int\int \frac 1{4\pi} \sin^2(\theta).d\theta d\phi##,
but you cannot have it both ways.

If (as I think the person setting the problem intended) we regard the angles ##\Theta## and ##\Phi## as random variables, then for a uniform distribution on the sphere's surface we have
$$P(\theta < \Theta < \theta + d \theta, \phi < \Phi < \phi + d \phi) = c \sin(\theta) d\theta d\phi,$$
so the density of ##(\Theta, \Phi)## is ##f_{\Theta, \Phi} (\theta, \phi) = c \sin (\theta)##, because by definition of a bivariate probability density we have
$$\text{probability} = \text{density} \times d \text{(variable 1)}\; \times d \text{(variable 2)}.$$
The density includes the Jacobian, etc. See, eg.,
https://www.cs.ubc.ca/~murphyk/Teaching/Stat406-Spring08/homework/changeOfVariablesHandout.pdf
or
http://math.arizona.edu/~jwatkins/n-bivariate.pdf

Ray Vickson said:
If (as I think the person setting the problem intended) we regard the angles ##\Theta## and ##\Phi## as random variables, then for a uniform distribution on the sphere's surface we have
$$P(\theta < \Theta < \theta + d \theta, \phi < \Phi < \phi + d \phi) = c \sin(\theta) d\theta d\phi,$$
so the density of ##(\Theta, \Phi)## is ##f_{\Theta, \Phi} (\theta, \phi) = c \sin (\theta)##, because by definition of a bivariate probability density we have
$$\text{probability} = \text{density} \times d \text{(variable 1)}\; \times d \text{(variable 2)}.$$
The density includes the Jacobian, etc. See, eg.,
https://www.cs.ubc.ca/~murphyk/Teaching/Stat406-Spring08/homework/changeOfVariablesHandout.pdf
or
http://math.arizona.edu/~jwatkins/n-bivariate.pdf
Ok, so ρ is not the surface density. Haven't come across that source of confusion before.
Thanks for explaining.

## 1. What is a 2-sphere?

A 2-sphere, also known as a 2-dimensional sphere, is a geometric shape that is formed when a circle is rotated around its diameter in 3-dimensional space. It is a closed surface with no edges or corners, and is often used in mathematics and physics to represent objects such as planets or atoms.

## 2. How is simple probability calculated for a 2-sphere?

Simple probability for a 2-sphere can be calculated by dividing the surface area of the desired event by the total surface area of the sphere. This can be represented by the formula P = A/E, where P is the probability, A is the area of the event, and E is the total area of the sphere.

## 3. What is the difference between simple probability and conditional probability for a 2-sphere?

Simple probability is the likelihood of a specific event occurring on a 2-sphere, while conditional probability takes into account additional information or conditions that may affect the probability of the event. For example, the probability of rolling a 6 on a fair 2-sphere is 1/6, but the probability of rolling a 6 given that the sphere is weighted on one side may be higher or lower.

## 4. Can simple probability be applied to non-uniformly distributed events on a 2-sphere?

Yes, simple probability can still be applied to non-uniformly distributed events on a 2-sphere. The total surface area of the sphere can still be used as the denominator, but the area of the event will vary based on the distribution. For example, if a certain area of the sphere has a higher concentration of the event, then the probability will be higher in that area.

## 5. How is simple probability used in real-world applications involving 2-spheres?

Simple probability is commonly used in fields such as physics, astronomy, and biology to make predictions and analyze data. For example, in astronomy, simple probability can be used to calculate the likelihood of a planet orbiting within a certain distance from its star. In biology, simple probability can be used to determine the probability of certain genetic traits being passed down through generations.

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