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

In summary, The conversation discusses randomly generating points on the circumference of a circle and the variance of the x-coordinate. The attempt at solving the problem involves considering the distribution of theta and transforming it to the distribution of x. However, a simpler method is proposed by considering E(x^2 + y^2) and exploiting the symmetry between x and y. The assumptions of independence and symmetry are questioned and it is suggested to remove the false assumption of independence. The conversation also briefly touches on generating points on the surface of a sphere.
  • #1
Master1022
611
117
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.
 
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  • #2
What about considering ##E(x^2 + y^2)##?
 
  • #3
PeroK said:
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.
 
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  • #4
Looks good to me!
 
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Likes Master1022
  • #5
PeroK said:
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
Master1022 said:
Homework Statement: Let us randomly generate points on the circumference of a circle (two dimensions).

Master1022 said:
[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
Stephen Tashi said:
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
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
PeroK said:
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.

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

PeroK said:
You just need to remove the false assumption of independence.
Okay many thanks - this makes more sense
 
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  • #10
Master1022 said:
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.
 

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

1. What is the formula for calculating the variance of a point chosen at random on the circumference of a circle?

The formula for calculating the variance of a point chosen at random on the circumference of a circle is Var = (π/2)^2 - (π/4)^2 = π^2/4 - π^2/16 = 3π^2/16.

2. How is the variance of a point chosen at random on the circumference of a circle related to the radius of the circle?

The variance of a point chosen at random on the circumference of a circle is directly proportional to the square of the radius of the circle. This means that as the radius increases, the variance also increases.

3. Can the variance of a point chosen at random on the circumference of a circle be negative?

No, the variance of a point chosen at random on the circumference of a circle cannot be negative. It is always a positive value.

4. How does the number of points on the circumference of a circle affect the variance?

The number of points on the circumference of a circle does not have a direct effect on the variance. However, as the number of points increases, the variance becomes a more accurate measure of the spread of the points.

5. Is the variance of a point chosen at random on the circumference of a circle affected by the shape of the circle?

Yes, the variance of a point chosen at random on the circumference of a circle is affected by the shape of the circle. For example, a circle with a larger radius will have a higher variance compared to a circle with a smaller radius, even if they have the same number of points on the circumference.

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