# Finding the pdf of $\sqrt X$ when $X\sim U(0,3)$

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?

• Where did you stuck? – kludg Aug 3 '17 at 8:47
• Found. Now what should I do with it? – Did Aug 3 '17 at 8:47
• Sorry for the vagueness! Just updated the question @Did – Alexander Aug 3 '17 at 8:58
• Modulo some annoying typos and an annoying lack of conditions on $y$, yes. – Did Aug 3 '17 at 9:02
• 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 – Henry Aug 3 '17 at 9:30

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