Variance of a point chosen at random on the circumference of a circle

  • #1
Master1022
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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.
 
Last edited:

Answers and Replies

  • #2
PeroK
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What about considering ##E(x^2 + y^2)##?
 
  • #3
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What about considering ##E(x^2 + y^2)##?
Thanks @PeroK ! I hadn't considered this at all. Would we hope to exploit a symmetry between ##x## and ##y## with this method?

The reason I ask is because I am thinking along these lines - although I really ought to give this more thought:
[tex] E(x^2 + y^2) = E(r^2) = r^2 [/tex]

Then, this is where my hand-waiving starts and I can't yet fully convince myself of the mathematics. If we assume ##x^2## and ##y^2## are independent (not sure how fair this is), then:
[tex] E(x^2 + y^2) = E(x^2) + E(y^2) = r^2 [/tex]
Then, if we exploit a symmetry between ##x## and ##y##, then we can let (again, very hand wavy) ##E(x^2) = E(y^2) ##. This doesn't feel right.

Nonetheless, ##Var(X) = E(X^2) - [E(X)]^2 = E(X^2) ## becase ##E(X) = 0##. Which leads us to ##Var(X) = \frac{r^2}{2}##. Now this method makes the independence and symmetry assumptions, which I am not truly convinced hold.

I'll keep thinking about this though.
 
  • #4
PeroK
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Looks good to me!
 
  • #5
Master1022
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Looks good to me!
Oh wow! Thanks @PeroK . Just a few follow up questions I had:

1) What made you think about considering ##E(x^2 + y^2)## to start off with?
2) Why do the independence and symmetry properties hold for this question?

3) For part (b), I am guessing we could do a similar method with ##E(x^2 + y^2 + z^2)## and, after making the same assumptions, end up with ##\frac{r^2}{3}##?
 
  • #6
Stephen Tashi
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Homework Statement: Let us randomly generate points on the circumference of a circle (two dimensions).

[tex] E(x^2 + y^2) = E(r^2) = r^2 [/tex]

Are we assuming "a circle" must be a circle with center (0,0) ?
 
  • #7
Master1022
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Are we assuming "a circle" must be a circle with center (0,0) ?
I believe so. The problem doesn't say, but I just assumed that we were.
 
  • #8
PeroK
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To check that ##E(x^2+y^2) = E(x^2)+E(y^2)## you could pick a few random points and check it out. It's not that ##x## and ##y## are independent, it's that expectation is additive. This should be easy to prove.

There is an obvious symmetry in ##x## and ##y## values round the circle.

You just need to remove the false assumption of independence.
 
  • #9
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To check that ##E(x^2+y^2) = E(x^2)+E(y^2)## you could pick a few random points and check it out. It's not that ##x## and ##y## are independent, it's that expectation is additive. This should be easy to prove.
Sure thing, the additive property is clear to me. I think I was getting confused with variance, for which there is the extra covariance term for sums of variables.

There is an obvious symmetry in ##x## and ##y## values round the circle.
Agreed

You just need to remove the false assumption of independence.
Okay many thanks - this makes more sense
 
  • #10
haruspex
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to ensure that the point picked is truly random
That doesn't really mean anything. You can have a 'random' variable the distribution of which only allows one value.
What I presume you intend is that the distribution is uniform along the arc of the circle.
 

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