Average shortest distance between a circle and a random point lying in it

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If the point lies in a ring radius $x$, thickness $\delta x$, centred at the centre of the circle, then the distance is $r-x$. The ring has area $2\pi x\ \delta x$ and the circle has area $\pi r^2$. So assuming the distribution for the point $P$ is uniform over the disk, the expected distance is $\frac{1}{\pi r^2}\int_0^r(r-x)2\pi x\ dx=\frac{r}{3}$.

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Sophie Agnesi
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Sophie Agnesi

Updated on June 11, 2020

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  • Sophie Agnesi
    Sophie Agnesi over 3 years

    What is the average shortest distance between the circle $(x-a)^2+(y-b)^2=r^2$ and a random point lying in it?

    This question is just idle curiosity. Basically, it's the same as finding the difference between its radius and the average distance between the random point and its center. Let $D$ denote the shortest distance between the circle $(x-a)^2+(y-b)^2=r^2$ and the random point $P(X,Y)$, then

    \begin{equation} D=r-\sqrt{(X-a)^2+(Y-b)^2} \end{equation}

    We may assume $X$ and $Y$ are independently uniformly distributed in $(0,a)$ and $(0,b)$, respectively. Then its joint pdf is

    \begin{equation} f_{X,Y}(x,y)=f_X(x)\cdot f_Y(y)=\frac{1}{ab} \end{equation}

    Hence the average of $D$ is

    \begin{equation} E[D]=\int_0^b\int_0^a d\ f_{X,Y}(x,y)\ dx\ dy=r-\frac{1}{ab}\int_0^b\int_0^a \sqrt{(x-a)^2+(y-b)^2}\ dx\ dy \end{equation}

    Is my approach correct? If not, how does one find the correct $E[D]$?

    • almagest
      almagest over 7 years
      Your ranges for $X,Y$ are wrong. Even making them $(a-r,a+r)$ and $(b-r,b+r)$ does not work, because you do not want a rectangular range. You want the distribution for $P$ to be uniform over the disk.
    • Sophie Agnesi
      Sophie Agnesi over 7 years
      @almagest Be that as it may, I think to denote the random point in your answer you should use 2 random variables, namely $R\sim U(0,r)$ and $\Theta\sim U(0,2\pi)$.
    • almagest
      almagest over 7 years
      There is no need for $\theta$ because of the symmetry.
    • Sophie Agnesi
      Sophie Agnesi over 7 years
      Are you implying the answer to my question is the same as the average distance between 2 random points on a line with length $r$?
    • almagest
      almagest over 7 years
      Yes, if you do that calculation, it also comes to $\frac{r}{3}$.
  • Sophie Agnesi
    Sophie Agnesi over 7 years
    Can you elaborate your answer a bit? I just need a clear & a detail explanation of how it must be true? Thanks
  • almagest
    almagest over 7 years
    This is just a straightforward integration. All points in the shaded ring have the same distance $r-x$. The ring has circumference $2\pi x$ and thickness $\delta x$ hence area $2\pi x\ \delta x$.
  • Jack D'Aurizio
    Jack D'Aurizio over 7 years
    This works if we pick a random point with respect to the uniform distribution over the circle, but that is not the OP's assumption, apparently.
  • Sophie Agnesi
    Sophie Agnesi over 7 years
    Sorry, if it's so obvious but how do you explain that the pdf is $\frac{2x}{r^2}$? I'm still confused?
  • almagest
    almagest over 7 years
    The assumption is that the prob of $P$ lying in an area $A$ is 0 if the area is outside the disk and proportional to the area if it lies inside it. So the prob that the point lies inside the shaded ring is $\frac{2\pi x\ \delta x}{\pi r^2}$.