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If $P$ is a random point on the circle $x^2+y^2=1$, what are the mean and variance $P$'s distance to $(0,1)$?

The random variable has the function $$X=\sqrt{-2x+2}$$ can't quite figure out what to do next...

Parcly Taxel
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pi47
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  • Hello, what is your take on this? You need to add what you have done yourself. – Cardinal Dec 01 '17 at 04:58
  • Hi. I took the distance between point (0,1) and any (x,y) as $\sqrt ((x-1)^2 + y^2)$ then substituted $x^2 + y^2 = 1$ to get X = $\sqrt (-2x+2)$ can't quite figure out what to do next.. – pi47 Dec 01 '17 at 05:08
  • Considering half of the circle, I think since you choose randomly, then the ditance will be uniformly distributed between the lowest value and the highest value. – Cardinal Dec 01 '17 at 05:23
  • Considering half of the circle, the distance will be uniformly distributed between $0$, the lowest value, and $2$, the highest value.

    Thus, the mean will be $\mu = \frac{2+0}{2} = 1$ and the variance will be $\frac{(2-0)^2}{12}=\frac{1}{3}$.

    For the complete circle, the probabilities will be doubled and we should find a way to relate them. However, I am not sure if it's right.

     – Cardinal Dec 01 '17 at 05:35
  • You have given me a place to start. thanks! – pi47 Dec 01 '17 at 05:42
  • @Cardinal There is no reason to believe from the offset that the distance is uniformly distributed. What makes you say that? – Isaac Browne Dec 01 '17 at 06:03
  • @Isaac Browne. I suppose the probability of of each point on the circle is the same. And each point on the half-circle translates to a unique distance. – Cardinal Dec 01 '17 at 06:40
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    @Cardinal Yes, unfortunately, this does not imply that the distance is uniformly distributed. There are bijective functions which are not uniformly distributed, such as $y=x^3$, which has a higher density of $x$ values towards $y=0$ – Isaac Browne Dec 01 '17 at 06:42
  • @Issac Browne I see. Thanks for making it clear. I was thinking discretely while distance is continuous !!! :-) – Cardinal Dec 01 '17 at 06:48

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Since we are supposed to choose a random point on the circle, I think it is best to write our circle as $(\cos(t),\sin(t))$ in order that we go through each point at the same speed ($1$), and thus count them equally. Then we can write our definitions of mean and variance as follows (I only go from $t=0$ to $t=\pi$ because of symmetry. $$E(X) = \frac{\int_0^\pi \sqrt{(1-\cos(t))^2+\sin^2(t)}dt}{\pi}$$ $$E(X^2) = \frac{\int_0^\pi (1-\cos(t))^2+\sin^2(t)dt}{\pi}$$ $$\mu(X) = E(X^2)-E(X)^2$$ Now it's just evaluating integrals. If you need more help, just ask!

Isaac Browne
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