# Distributions of functions and random variables

I am reading the book statistical inference by Casella and Berger. I am having trouble in understanding the following theorem:

Let X have cdf $F_X$(x), let Y = g(X), and let $S_1$ and $S_2$ be the sample spaces,

a. If g is an increasing function on $S_1$, $F_Y$(y) = $F_X$($g^{-1}$(y)) for y $\in$ $S_2$.

b. If g is a decreasing function on $S_1$ and X is a continuous random variable, $F_Y$(y) = 1 - $F_X$($g^{-1}$(y)) for y $\in$ $S_2$.

Thanks.

• Can you think of why the result might be true if $g(\cdot)$ is a strictly monotone increasing function, that is, if $a < b$, then $g(a) < g(b)$ and vice versa. Then, can you figure out what, if anything, goes wrong if $a<b$ merely implies that $g(a) \leq g(b)$ (which some people include in the term increasing while others, more properly, call nondecreasing ? – Dilip Sarwate Aug 4 '16 at 16:06

a. Using that $g$ is increasing we have that $g(x_1)<g(x_2)$ if and only if $x_1<x_2$:

$$F_Y(y) = P[Y\leq y] = P[g(X)<y] = P[X \leq g^{-1}(y)] = F_X(g^{-1}(y)).$$

b. Using that $g$ is decreasing we have that $g(x_1)>g(x_2)$ if and only if $x_1<x_2$:

\eqalign{ F_Y(y) &= P[Y\leq y] = P[g(X)\leq y] = P[X>g^{-1}(y)] =1 - P[X\leq g^{-1}(y)] \\ &= 1- F_X(g^{-1}(y)). }

Note that $g(X)\leq y$ if and only if $g^{-1}(g(X)) \geq g^{-1}(y)$. This is $X \geq g^{-1}(y)$, since $g$ is decreasing.

• @Silverfish I am not allowed to comment at such a young age. – Selfie Aug 4 '16 at 16:10
• @Selfie why change of signs, when taking inverses? – Prerit Aug 4 '16 at 16:13
• @Prerit Because $g$ is decreasing , then, if $x_1<x_2$ it follows that $g(x_1)>g(x_2)$. Just picture a decreasing function in your mind, such as $f(x)=-x$ – Selfie Aug 4 '16 at 16:17
• @JuhoKokkala well spotted. Copy paste typo. Thanks. – Selfie Aug 4 '16 at 16:19