 #1
Master1022
 611
 116
 Homework Statement:

Let us randomly generate points ##(x,y)## on the circumference of a circle (two dimensions).
(a) What is Var(x)?
(b) What if you randomly generate points on the surface of a sphere instead?
 Relevant Equations:
 Variance
Hi,
I was looking at this problem and just having a go at it.
Question:
Let us randomly generate points ##(x,y)## on the circumference of a circle (two dimensions).
(a) What is ##\text{Var}(x)##?
(b) What if you randomly generate points on the surface of a sphere instead?
Attempt:
In terms of understanding the question, I think I have to understand whether we pick ##X## or are picking ##\theta##. In my mind, if we are picking ##X## from a uniform distribution, then the variance will just be ## \frac{(b  a)^2}{12} = \frac{r^2}{3} ##, which isn't particularly difficult (so I don't think it would justify being asked in an interview). Also, I feel like then the choice of ##Y## isn't random as it is restricted to two points for a given choice of ##X##. Therefore, to ensure that the point picked is truly random, it made more sense to me (perhaps I am wrong) to pick a ##\theta## measured from axis and then ## x = r cos(\theta) ## and ## y = r sin(\theta) ##.
Proceeding with the ##\theta## argument: ## \theta## will be from a uniform distribution between ##0## and ##2\pi ## with probability ## \frac{1}{2\pi} ##. Then we would need to transform the distribution of ## \theta ## to the distribution for ## X ##
[tex] X = r cos(\theta) \rightarrow p(x) dx = p(\theta) d\theta \rightarrow p(x) = p(\theta) \frac{d\theta}{dx} [/tex]
[tex] p(x) = \frac{1}{2\pi} \cdot \frac{1}{\frac{dx}{d\theta}} = \frac{1}{2\pi} \cdot \frac{1}{r sin(\theta)} = \frac{1}{2\pi} \cdot \frac{1}{\sqrt{1  x^2}} [/tex]
and then proceed with some integration to find ##Var(X)##... However, I feel like there ought to be a more elegant method of reaching the solution?
Would appreciate any advice/tips outlining where I have gone astray. Thanks in advance.
[EDIT]: changed the pdf of ##\theta## to ##\frac{1}{2\pi}## as it should be. Apologies.
I was looking at this problem and just having a go at it.
Question:
Let us randomly generate points ##(x,y)## on the circumference of a circle (two dimensions).
(a) What is ##\text{Var}(x)##?
(b) What if you randomly generate points on the surface of a sphere instead?
Attempt:
In terms of understanding the question, I think I have to understand whether we pick ##X## or are picking ##\theta##. In my mind, if we are picking ##X## from a uniform distribution, then the variance will just be ## \frac{(b  a)^2}{12} = \frac{r^2}{3} ##, which isn't particularly difficult (so I don't think it would justify being asked in an interview). Also, I feel like then the choice of ##Y## isn't random as it is restricted to two points for a given choice of ##X##. Therefore, to ensure that the point picked is truly random, it made more sense to me (perhaps I am wrong) to pick a ##\theta## measured from axis and then ## x = r cos(\theta) ## and ## y = r sin(\theta) ##.
Proceeding with the ##\theta## argument: ## \theta## will be from a uniform distribution between ##0## and ##2\pi ## with probability ## \frac{1}{2\pi} ##. Then we would need to transform the distribution of ## \theta ## to the distribution for ## X ##
[tex] X = r cos(\theta) \rightarrow p(x) dx = p(\theta) d\theta \rightarrow p(x) = p(\theta) \frac{d\theta}{dx} [/tex]
[tex] p(x) = \frac{1}{2\pi} \cdot \frac{1}{\frac{dx}{d\theta}} = \frac{1}{2\pi} \cdot \frac{1}{r sin(\theta)} = \frac{1}{2\pi} \cdot \frac{1}{\sqrt{1  x^2}} [/tex]
and then proceed with some integration to find ##Var(X)##... However, I feel like there ought to be a more elegant method of reaching the solution?
Would appreciate any advice/tips outlining where I have gone astray. Thanks in advance.
[EDIT]: changed the pdf of ##\theta## to ##\frac{1}{2\pi}## as it should be. Apologies.
Last edited: