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

[Master1022]

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 $X$

X = r cos(\theta) \rightarrow p(x) dx = p(\theta) d\theta \rightarrow p(x) = p(\theta) |\frac{d\theta}{dx}|
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}}

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.

 

[PeroK]

What about considering $E(x^2 + y^2)$?

 

[Master1022]

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:
E(x^2 + y^2) = E(r^2) = r^2

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:
E(x^2 + y^2) = E(x^2) + E(y^2) = r^2
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.

 

[PeroK]

Looks good to me!

 

[Master1022]

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 $\frac{r^2}{3}$?

 

[Stephen Tashi]

Homework Statement: Let us randomly generate points on the circumference of a circle (two dimensions).

E(x^2 + y^2) = E(r^2) = r^2

Are we assuming "a circle" must be a circle with center (0,0) ?

 

[Master1022]

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.

 

[PeroK]

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.

 

[Master1022]

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

 

[haruspex]

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.