Let $X$ be a random variable and $Y=g(X)$

Define $$\tag{1} \chi = \{x: f_X(x)>0\}\quad \text{and}\quad \mathcal{Y} = \{y:y=g(x) \text{ for some } x \in \chi\} $$

Define $g^{-1}(y) = \{x\in \chi:g(x) = y\}$

Define: A random variable $X$ is continuous if $F_X(x)$ is a continuous function of $x$.

My question is: how come, in the theorem below, the statement in (b) requires X to be a continuous random variable but the statement in (a) does not

The relevant theorem is (Theorem 2.1.3 in Casella and Berger 2nd Edition)

Let $X$ have cdf $F_X(x)$, let $Y=g(X)$, and let $\chi$ and $\mathcal{Y}$ be defined as in (1)

  • (a) If $g$ is an increasing function on $\chi$, $F_Y(y) = F_X(g^{-1}(y))$ for $y\in \mathcal{Y}$

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

Another way of stating what I am asking is that, prior to stating this theorem, Casella and Berger state

if $g(x)$ is an increasing function, then using the fact that $F_Y(y) = \int_{x\in\chi : g(x)\leq y} f_X(x)dx$, we can write $$ F_Y(y) = \int_{x\in\chi : g(x)\leq y} f_X(x) \, dx = \int_{-\infty}^{g^{-1}(y)} f_X(x) \, dx = F_X(g^{-1}(y)) $$

If $g(x)$ is decreasing, then we have

$$ F_Y(y) = \int_{g^{-1}(y)}^\infty f_X(x) \, dx = 1-F_X(g^{-1}(y)) $$ "The continuity of $X$ is used to obtain the second equality

My question(restated) is in yellow box below:

My question (restated) is: How come, when $g(x)$ is an increasing function we do not need to use continuity of $X$, but we do for the case when $g(x)$ is decreasing?

  • (A side question, I will accept answer so long as answers the above question): this is continuity of the random variable, but the integral uses the PDF. what is the relation between continuity of $X$ and it's pdf? (specifically, I think there may be some strangeness if $F_X$, the CDF of $X$ is continuous but not differentiable)?

What came to my mind was Fundamental theorem of calculus maybe, but there is a version of it that doesn't require continuity of $f$ I think? Plus, here we have $X$ is continuous, if that matters -- I'm not sure.

  • $\begingroup$ How is the function $g^{-1}$ defined? $\endgroup$
    – drhab
    May 26, 2018 at 9:15
  • 2
    $\begingroup$ If $X$ has a density function $f_X$ then it is a continuous random variable (in the sense its distribution function is continuous) The question does not make much sense to me. $\endgroup$ May 26, 2018 at 12:21
  • $\begingroup$ @drhab It is the inverse function. I added a definition to the post. $\endgroup$
    – user106860
    May 26, 2018 at 17:27
  • 1
    $\begingroup$ @user106860 As is pointed out, proposition (a) involves $\mathcal Y$, the definition of which requires the existence of $f_X$, and that implies $X$ is continuous. $\endgroup$
    – Ѕааԁ
    Jun 15, 2018 at 5:05
  • 1
    $\begingroup$ Where you wrote $Y=g(x),$ I'm guessing you meant $Y=g(X). \qquad$ $\endgroup$ Jun 15, 2018 at 22:22

2 Answers 2


Recall that we have $$F_Y(y) = P(Y \leq y)$$ If $g$ is an increasing function, then the event $Y \leq y$ is equivalent to the event $X \leq g^{-1}(y)$, so we have $$F_Y(y) = P(Y \leq y) = P(X \leq g^{-1}(y)) = F_X(g^{-1}(y))$$

If $g$ is a decreasing function, then the event $Y \leq y$ is instead equivalent to the event $X \geq g^{-1}(y)$, so we have $$F_Y(y) = P(Y \leq y) = P(X \geq g^{-1}(y)) = 1-P(X < g^{-1}(y))$$

Note that when we took complements we moved to strict inequality. If $X$ is a continuous variable, this isn't an issue -- the event that $X=g^{-1}(y)$ has probability $0$, so the probability of strict inequality is the same as the cumulative distribution value.

But if $X$ is discrete and the event $X=g^{-1}(y)$ has positive probability, you'd need to add an extra term to account for this.


$$ \int_{g^{-1}(y)}^\infty f_X(x) \, dx = 1-F_X\left(g^{-1}(y)\right) \text{ ?} $$ We have: $$ 1-F_X(g^{-1}(y)) = 1 - \Pr(X\le g^{-1}(y)) = \Pr\left(X>g^{-1}(y)\right) $$ We may consider continuity of $F$ at $g^{-1}(y)$ or continuity of $F$ at points greater than $g^{-1}(y).$ Nothing about continuity at points less than $g^{-1}(y)$ can matter here.

In the first place $$\Pr(a< X < b) = \int_a^b f_X(x)\,dx\tag 1$$ only if $X$ has a density function $f_X,$ and that in itself requires continuity of $F_X$ (and in fact requires something more than just continuity). If $\Pr(x = c)>0,$ where $c$ is some number between $a$ and $b,$ then line $(1)$ above is not true of any function in the role of $f.$

However, statement $(b)$ of the theorem does not mention integration of any density function. The statement is in effect $\Pr(Y\le y) = 1- \Pr(X>g^{-1}(y))$ if $F_X$ is continuous.

Cumulative distribution functions are non-decreasing. The only kind of discontinuity that a non-decreasing function can have is a jump. A jump in $F_X$ at $g^{-1}(y)$ would mean $\Pr(X = g^{-1}(y))>0.$ If that happens then \begin{align} & \Pr(Y\le y) = \Pr(Y=y) + \Pr(Y<y) \\[10pt] = {} & \Pr(X=g^{-1}(y)) + \Pr(X>g^{-1}(y)) \\[10pt] = {} & \Pr(X=g^{-1}(y)) + \int_{g^{-1}(y)}^\infty f_X(x)\,dx. \end{align} If the first term in the last line is positive rather than zero, then equality between the second term in the last line and $\Pr(Y\le y)$ is not true.

But now suppose it had said $\Pr(Y\ge y).$ Then we would have $$ \Pr(Y\ge y) = \Pr(X\le g^{-1}(y)) = F_X(g^{-1}(y)). $$ The difference results from the difference between $\text{“}<\text{''}$ and $\text{“} \le \text{''}$ in the definition of the c.d.f., which says $F_X(x) = \Pr(X\le x)$ and not $F_X(x) = \Pr(X<x).$

As for the relationship between continuity and density functions, that is more involved. The Cantor distribution is a standard example, defined like this: A random variable $X$ will be in the interval $[0,1/3]$ or $[2/3,1]$ according to the result of a coin toss; then it will be in the upper or lower third of the chosen interval according to a second coin toss; then in the upper or lower third of that according to a third coin toss, and so on.

The c.d.f. of this distribution is continuous because there is no individual point between $0$ and $1$ that gets assigned positive probability.

But notice that there is probability $1$ assigned to a union of two intervals of total length $2/3,$ then probability $1$ assigned to a union of intervals that take up $2/3$ of that union of intervals, thus $4/9$ of $[0,1],$ then there is probability $1$ assigned to a set taking up $2/3$ of that space, thus $(2/3)^3 = 8/27,$ and so on. Thus there is probability $1$ that the random variable lies within a certain set whose measure is $\le (2/3)^n,$ no matter how big an integer $n$ is. The measure of that set must therefore be $0.$ If you integrate any function over a set whose measure is $0,$ you get $0.$ Hence there can be no function $f$ such that for every measurable set $A\subseteq[0,1]$ we have $$ \Pr(X\in A) = \int_A f(x)\,dx, $$ i.e. there can be no density function.

Thus the Cantor distribution has no point masses and also no probabilities that can be found by integrating a density function.

Thus existence of a density function is a stronger condition on than mere continuity of the c.d.f.

  • $\begingroup$ Thank you very much! (just one thing: when you say "If $Pr(x=c)>0$, where $c$ is some number between $a$ and $b$ then line (1) above is not true of any function in the role of f", you mean weakly between?) $\endgroup$
    – user106860
    Jun 22, 2018 at 22:05
  • $\begingroup$ Also, a part of your answer that I am slightly confused about: You say that \begin{align} & \Pr(Y\le y) = \Pr(Y=y) + \Pr(Y<y) \\ = {} & \Pr(X=g^{-1}(y)) + \Pr(X>g^{-1}(y)) \\ = {} & \Pr(X=g^{-1}(y)) + \int_{g^{-1}(y)}^\infty f_X(x)\,dx. \end{align} But if there is a jump in $F_X$ then $F_X$ is not continuous, so is this last line valid? That is, you said that $$\Pr(a< X < b) = \int_a^b f_X(x)\,dx\tag 1$$ requires continuity of $F_X$, so if $F_X$ is not continuous can we still say that $$\Pr(X>g^{-1}(y))=\int_{g^{-1}(y)}^\infty f_X(x)$$?? (I get that argument holds without last line too) $\endgroup$
    – user106860
    Jun 22, 2018 at 22:31
  • 1
    $\begingroup$ @user106860 : That line is valid only if $F_X$ has no jumps in the open interval $(g^{-1},+\infty).$ A jump at $g^{-1}(y)$ won't upset it. $\endgroup$ Jun 23, 2018 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.