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Let $X$ be the random variable corresponding to the radial distance of a point chosen randomly and uniformly in the unit disc.Then the pdf of $X$ is given by: $$ f(x)= 2x, \leq x \leq 1$$ How can we find the dsitribution of $Y=X^2?$ I willbe grateful for any help/hints.

AgnostMystic
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  • Presumably, $X$ is given in terms of Cartesian or polar coordinates...in either case two parameters would be needed to specify its support. More to the point, the density is certainly not identically zero. Please clarify the question. – Math1000 Jul 16 '23 at 05:58

2 Answers2

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I think you might have a typo in the above?

Either way, I would recommend trying to compute the CDF of $Y$. In particular, you can compute $$\mathbb{P}(X^2 \leq a) = \mathbb{P}(|X| \leq \sqrt{a}).$$ From there, you can obtain the density by differentiating with respect to $a$.

Alan Chung
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I am posting this answer for seeking a clarification in the final answer . For the sake avoiding confusion with the x-coordinate ,let us denote what we called $X$ by $W$ and $V=W^2$. We note that the $f(w)$ is the pdf of the radial distance of a point chosen randomly and uniformly in the unit circle. So its CDf is $F_W(w)=w^2$So we are looking for the distribution of $W^2.$ As pointed out above in the (partial) answer,we can write the CDF of $V$ as : \begin{align}F_V(v)&=\mathbb{P}(|W| \leq \sqrt{a})\\ &=\mathbb{P}( -\sqrt{a}\leq W \leq \sqrt{a})\\ &=F_W(\sqrt{a} )-F_W(-\sqrt{a} )\\ &=a-\mathbb P (W \leq -\sqrt{a} )\\ &=a \end{align} so that the pdf of V is given by $$ f_V(v)=1,0 \leq v \leq 1,$$ which is the pdf of a uniform random variate on $[0,1].$ Is my final calculaiton correct or is square of radial distance of a point chosen radomly and uniformly over the unit disc indeed a uniform random variable ?

AgnostMystic
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