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I am new to this site. I would like to ask when can we invert the function in the inequality? Question: Let X be a continuous random variable with pdf . Suppose () is strictly monotonic, differentiable function of x. The random variable =() has the PDF.

$$ {f_Y} (y)= \begin{cases} {f_X}[g^{-1}(y)] \left|\dfrac{d}{dy}g^{-1}(y)\right|, & \text{if y = g(x) for some x}\\[2ex] 0, &\text{if y $\ne$g(x) for all x} \end{cases} $$ Proof. Suppose that y=g(x) for some x. Then, with Y=g(X),

\begin{aligned} F_Y(y) & =P\{g(X)\le y\}\\ & = P\{X\le g^{-1}(y)\}\\ &=F_X(g^{-1}(y)) \end{aligned}

I am not sure why function $g(x)$ can be converted to $g^{-1}{(x)}$ over the inequality sign. Is it because $g(x)$ is strictly monotonic so it definitely has the inverse function?

Many thanks for your help.

Question and proof

hmakholm left over Monica
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user590977
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What we need is that $g$ is a strictly increasing function. This implies that for any numbers $x_1$ and $x_2$ in the domain of $g,$ we will have $g(x_1) \leq g(x_2)$ if and only if $x_1 \leq x_2.$

Now put $X$ for $x_1$ and $g^{-1}(y)$ for $x_2.$ So we have $g(X) \leq g(g^{-1}(y))$ if and only if $X \leq g^{-1}(y).$ That is, the event that $g(X) \leq y$ is exactly the same event as the event that $X \leq g^{-1}(y)$ (you cannot have an outcome that satisfies one of those inequalities and not the other), and it has the same probability.

If $g$ were strictly monotonic but decreasing rather than increasing, then we would have $g(x_1) \leq g(x_2)$ if and only if $x_1 \geq x_2,$ and when making the substitutions $X$ for $x_1$ and $g^{-1}(y)$ for $x_2$ we end up with $g(X) \leq y$ if and only if $X \geq g^{-1}(y),$ which is not what we need for the theorem.

According to the way I learned it (something like this), monotonic does not imply increasing. But perhaps the author of the theorem believed it does; or perhaps they had implied earlier that they would use the term "monotonic" to describe only monotonically increasing functions, never decreasing functions.

David K
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  • +1. A minor followup for the OP: as written here, you need $g$ to be strictly increasing and surjective onto the domain of $F_Y$. This in particular implies that $g$ is continuous. When $g$ is not surjective onto the domain of $F_Y$, there is an appropriate generalization; when $g$ is a CDF $F_Z$, this generalization is called the quantile function of $Z$. https://en.wikipedia.org/wiki/Quantile_function – Ian Sep 07 '18 at 19:31