How do they acquire this cumulative distribution function?

Question

How do they acquire the CDF in this problem? I don't even see $f_{X}$ and its integral anywhere. They simply jump straight to the answer. Can someone enlighten me? TIA!

Answer 1

The set of equations for $F_Y$ (and it really is a set of equations, because there are two $=$ signs between the formulas) gives you two options of how to calculate $F_Y$. One option is to use the formula from the right side of the second equation: $$ F_Y(y) = \int_{\{x\mid g(x) \leq y\}} f_X(x)\,dx. $$ But the other option is to just use the first equation: $$ F_Y(y) = P(g(X) \leq y). $$

In example 4.1, the text uses the first equation: $$ F_Y(y) = P(\sqrt X \leq y). $$ Then we observe that $X$ is guaranteed to be non-negative (since its support is $[0,1]$, and a good thing too because otherwise $\sqrt X$ could be undefined), so $\sqrt X \leq y$ if and only if $X \leq y^2.$ That is, the event "$\sqrt X \leq y$" is the exact same event as the event "$X \leq y^2$". Therefore $$ P(\sqrt X \leq y) = P(X \leq y^2). $$

Finally, you could evaluate $P(X \leq y^2)$ by integrating $f_X(x) = 0$ from $x=-\infty$ to $x=0$ and $f_X(x) = 1$ from $x=0$ to $x=y^2.$ But do you really need to write out the integral every time when some one gives you a variable, in this case $X$, with uniform distribution on $[0,1]$, and some number $z$, and asks you what is the probability $P(X \leq z)$? The author didn't think so, so they didn't write the integral. The probability is always $P(X \leq z) = z$ if $z \in [0,1].$ And in this case $z = y^2.$