probabilityprobability distributionsuniform distribution
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?
All you need to care about is if $X = c$, then $Y = 0$, and if $X = d$, then $Y = 1$. So in particular, we want the solution to the system $$0 = ac + b, \quad 1 = ad + b$$ for $a, b$ given $c, d$. This gives us $$a = \frac{1}{d-c}, \quad b = -\frac{c}{d-c}.$$ No solution is possible if $c = d$, but this is assumed to not be possible.
Of course, this is not a unique solution because there is no stipulation that the transformation $Y = aX + b$ must also be order-preserving. That is to say, you could also solve the system $$1 = ac + b, \quad 0 = ad + b$$ and the resulting transformation would also give $Y \sim \operatorname{Uniform}(0,1)$.
The joint density looks like this:
So, the marginal distribution of $Y$ is given by the following integral
$$\int_y^1 2 \ dx=2(1-y)$$
if $0\leq y\leq 1$.
Best Answer
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 $$ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}\frac{y^2}{3} =\frac{2}{3}y. $$