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Let $X$ be a uniformly distributed random variable over $(0, 3)$, find the probability density function of $Y = \sqrt{X}$ , $f_y(y)$, in explicit formula.

From here What I got so far:

$F_y(y) = P(Y \le y) = P(\sqrt x \le y) = P(x \le y^2) = F_x(y^2) = \int^{y^2}_0 \frac{1}{3} dt = \frac{1}{3}t |^{y^2}_0 = \frac{1}{3}y^2$

Is this correct?

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  • $\begingroup$ Where did you stuck? $\endgroup$ – kludg Aug 3 '17 at 8:47
  • $\begingroup$ Found. Now what should I do with it? $\endgroup$ – Did Aug 3 '17 at 8:47
  • $\begingroup$ Sorry for the vagueness! Just updated the question @Did $\endgroup$ – Alexander Aug 3 '17 at 8:58
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    $\begingroup$ Modulo some annoying typos and an annoying lack of conditions on $y$, yes. $\endgroup$ – Did Aug 3 '17 at 9:02
  • $\begingroup$ Your $F_y=\frac13y^2$ with $0 \lt y \lt \sqrt{3}$ is the cumulative distribution function. You need to take the derivative to get the density function $\endgroup$ – Henry Aug 3 '17 at 9:30
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Yes, and you can simply use the transformation formula to find directly the density of $Y$. $g(X) = \sqrt{X}$ is monotone (increasing) function on $[0,3]$, and $g^{-1}(Y) = Y^2$, thus $$ f_Y(y) = f_Y(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right| = \frac{2}{3}y, \quad y\in (0,\sqrt{3}). $$

Recall that the density function is the derivative of the cumulative distribution function, i.e., $$ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}P(Y\le y), $$ in your case you can verify that $$ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}\frac{y^2}{3} =\frac{2}{3}y. $$

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  • $\begingroup$ So your answer is $\frac{2}{3} y$ does that mean my answer $\frac{1}{3} y^2$ is incorrect? $\endgroup$ – Alexander Aug 3 '17 at 9:11
  • $\begingroup$ No, you have the cumulative distribution function, i.e., $F'_Y(y) = (y^2/3)' = 2y/3=f_Y(y)$. $\endgroup$ – V. Vancak Aug 3 '17 at 9:13
  • $\begingroup$ Oh I get it. So should the $f_y(y) = 2y/3$ and $F_y(y) = 1/3 y^2$? $\endgroup$ – Alexander Aug 3 '17 at 9:17
  • $\begingroup$ I've edited the answer $\endgroup$ – V. Vancak Aug 3 '17 at 9:20
  • $\begingroup$ @Alexander That should be $f_Y(y)$ and $F_Y(y)$. Case sensitivity is important. $\endgroup$ – Graham Kemp Aug 3 '17 at 9:55

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