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A dart player always hits the dartboard (with a radius of 20cm), but has such a poor aim that the distribution of darts is uniform across the entire board. Let $R$ be the distance in cm between the dart and the center. Evaluate the probability density function for $R$ at $10$.


As the distribution is uniform, the PDF would be $\frac{1}{b-a}$ where $b$ and $a$ are 2 extremes.

But here we have the entire area. So for $R=10$,

PDF would be $\frac{1}{\pi * 10^2} = \frac{1}{100*\pi}$.

But the above answer is wrong.

Can someone please help me in understanding this concept?

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The pdf is $\frac{1}{400 \pi}=f(x,y)$, now if you want $f(r)$ you should first write $f$ in polar coordinates and then integrate over $\theta$.

$f(r)= \int_{0}^{2 \pi}\frac{r}{400 \pi}d\theta=\frac{r}{200}$. Now if you integrate over $r$ you obtain $\int_{0}^{20}\frac{r}{200}dr=1$ which make sense since the total probability must be one. So we have $f(10)=\frac{1}{20}$. Tell me if it's right.

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  • $\begingroup$ Hey, can you please explain how did you convert the original pdf to polar coordinate? $\endgroup$
    – Ramesh
    Commented Nov 19, 2017 at 14:05
  • $\begingroup$ @RajeshR since the pdf is constant it remains constant in every coordinates, the $r$ is just the absolute value of the determinant of the Jacobian matrix of the transformation: $|det(J)|=r$. $\endgroup$
    – chak
    Commented Nov 19, 2017 at 14:11
  • $\begingroup$ Please explain it in simple words. I don't know what is Jacobian matrix etc. $\endgroup$
    – Ramesh
    Commented Nov 19, 2017 at 14:14
  • $\begingroup$ $J= \left( \begin{matrix} \frac{\partial x(\theta,r)}{\partial \theta} & \frac{\partial x(\theta,r)}{\partial r} \\ \frac{\partial y(\theta,r)}{\partial \theta} & \frac{\partial y(\theta,r)}{\partial r} \end{matrix}\right)$. You have to write the change of cordinates $x=x(\theta,r),y=y(\theta,r)$. $\endgroup$
    – chak
    Commented Nov 19, 2017 at 15:03
  • $\begingroup$ @chak thanks for the answer. But can you write the answer in terms of x and y only ? Without converting to polar coordinates ? Maybe it will require double integration. $\endgroup$
    – Zephyr
    Commented Nov 19, 2017 at 16:29

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