A pdf is usually written as $f(x|\theta)$, where the lowercase $x$ is treated as a realization or outcome of the random variable $X$ which has that pdf. Similarly, a cdf is written as $F_X(x)$, which has the meaning $P(X<x)$. However, in some circumstances, such as the definition of the score function and this derivation that the cdf is uniformly distributed, it appears that the random variable $X$ is being plugged into its own pdf/cdf; by doing so, we get a new random variable $Y=f(X|\theta)$ or $Z=F_X(X)$. I don't think we can call this a pdf or cdf anymore since it is now a random variable itself, and in the latter case, the "interpretation" $F_X(X)=P(X<X)$ seems like nonsense to me.

Additionally, in the latter case above, I am not sure I understand the statement "the cdf of a random variable follows a uniform distribution". The cdf is a function, not a random variable, and therefore doesn't have a distribution. Rather, what has a uniform distribution is the random variable transformed using the function that represents its own cdf, but I don't see why this transformation is meaningful. The same goes for the score function, where we are plugging a random variable into the function that represents its own log-likelihood.

I have been wracking my brain for weeks trying to come up an intuitive meaning behind these transformations, but I am stuck. Any insight would be greatly appreciated!

  • 4
    $\begingroup$ The notation may be confusing you. E.g., $F_X(X)$ is exactly as meaningful as applying any measurable function to $X$ would be. For a correct interpretation you will need to be very clear about what a random variable is. For any random variable $X:\Omega\to\mathbb{R},$ the function $$Y:\omega\to F_X(X(\omega))$$ for $\omega\in\Omega$ clearly is a random variable and therefore has a distribution $F_Y.$ (Note the two distinct meanings of the symbol "$X$" in "$F_X(X)$.") $F_Y$ is uniform if and only if $X$ has a continuous distribution. $\endgroup$
    – whuber
    Feb 27, 2018 at 19:17
  • 1
    $\begingroup$ This isn't really a measure-theoretic issue: to understand it, you may safely ignore all references to "measurability." You might benefit from studying a little set theory early in your graduate career: that's where most people learn what this basic (and ubiquitous) mathematical terminology and notation really mean, so it's best not to put off learning it. $\endgroup$
    – whuber
    Feb 27, 2018 at 21:01
  • $\begingroup$ Maybe a word on why one should do a crazy thing like this: inserting a RV into its own density!!?! One example: say you want to estimate the density of X then you could measure how good you are by integrating over $f(x)-f_X(x)$ but this is “unfair”: you will never achieve good approximation when you do not have much data examples (I.e. the true density is small). Hence, a “fair” evaluation would be to weight the term by the true density. This is more or less the effect of inserting RV into their own densities... $\endgroup$ Feb 27, 2018 at 21:31
  • $\begingroup$ See also stats.stackexchange.com/questions/324768/… $\endgroup$ Feb 27, 2018 at 21:31

2 Answers 2


Like you say, any (measurable) function of a random variable is itself a random variable. It is easier to just think of $f(x)$ and $F(x)$ as "any old function". They just happen to have some nice properties. For instance, if $X$ is a standard exponential RV, then there's nothing particularly strange about the random variable $$Y = 1 - e^{-X}$$ It just so happens that $Y=F_X(X)$. The fact that $Y$ has an Uniform distribution (given that $X$ is a continuous RV) can be seen for the general case by deriving the CDF of $Y$.

\begin{align*} F_Y(y) &= P(Y \leq y) \\ &= P(F_X(X) \leq y) \\ &= P(X \leq F^{-1}_X(y)) \\ &= F_X(F^{-1}_X(y)) \\ &= y \end{align*}

Which is clearly the CDF of a $U(0,1)$ random variable. Note: This version of the proof assumes that $F_X(x)$ is strictly increasing and continuous, but it's not too much harder to show a more general version.

  • 1
    $\begingroup$ Your conclusion is incorrect for most strictly increasing $F_X$: you have assumed $F_X\circ F_X^{-1}$ is the identity--but that's not always the case. $\endgroup$
    – whuber
    Feb 27, 2018 at 19:19
  • $\begingroup$ Yes, thank you. The random variable $X$ clearly must be continuous. Am I missing anything now? $\endgroup$
    – knrumsey
    Feb 27, 2018 at 19:32
  • 1
    $\begingroup$ $F_X$ does not need to be bijective. Take, for example, the case where $X$ itself has a uniform distribution! The closure of the image of $F_X$ needs to be the entire interval $[0,1].$ That's essentially the definition of a continuous distribution. $\endgroup$
    – whuber
    Feb 27, 2018 at 19:45

A transform of a random variable $X$ by a measurable function $T:\mathcal{X}\longrightarrow\mathcal{Y}$ is another random variable $Y=T(X)$ which distribution is given by the inverse probability transform $$\mathbb{P}(Y\in A) = \mathbb{P}(X\in\{x;\,T(x)\in A\})\stackrel{\text{def}}{=} \mathbb{P}(X\in T^{-1}(A))$$ for all sets $A$ such that $\{x;\,T(x)\in A\}$ is measurable under the distribution of $X$.

This property applies to the special case when $F_X:\mathcal{X}\longrightarrow[0,1]$ is the cdf of the random variable $X$: $Y=F_X(X)$ is a new random variable taking its realisations in $[0,1]$. As it happens, $Y$ is distributed as a Uniform $\mathcal{U}([0,1])$ when $F_X$ is continuous. (If $F_X$ is discontinuous, the range of $Y=F_X(X)$ is no longer $[0,1]$. What is always the case is that when $U$ is a Uniform $\mathcal{U}([0,1])$, then $F_X^{-}(U)$ has the same distribution as $X$, where $F_X^{-}$ denotes the generalised inverse of $F_X$. Which is a formal way to (a) understand random variables as measurable transforms of a fundamental $\omega\in\Omega$ since $X(\omega)=F_X^{-}(\omega)$ is a random variable with cdf $F_X$ and (b) generate random variables from a given distribution with cdf $F_X$.)

To understand the paradox of $\mathbb{P}(X\le X)$, take the representation $$F_X(x)=\mathbb{P}(X\le x)=\int_0^x \text{d}F_X(x) = \int_0^x f_X(x)\,\text{d}\lambda(x)$$if $\text{d}\lambda$ is the dominating measure and $f_X$ the corresponding density. Then $$F_X(X)=\int_0^X \text{d}F_X(x) = \int_0^X f_X(x)\,\text{d}\lambda(x)$$ is a random variable since the upper bound of the integral is random. (This is the only random part of the expression.) The apparent contradiction in $\mathbb{P}(X\le X)$ is due to a confusion in notations. To be properly defined, one needs two independent versions of the random variable $X$, $X_1$ and $X_2$, in which case the random variable $F_X(X_1)$ is defined by$$F_X(X_1)=\mathbb{P}^{X_2}(X_2\le X_1)$$the probability being computed for the distribution of $X_2$.

The same remark applies to the transform by the density (pdf), $f_X(X)$, which is a new random variable, except that it has no fixed distribution when $f_X$ varies. It is nonetheless useful for statistical purposes when considering for instance a likelihood ratio $f_X(X|\hat{\theta}(X))/f_X(X|\theta_0)$ which 2 x logarithm is approximately a $\chi^2$ random variable under some conditions.

And the same holds for the score function$$\dfrac{\partial \log f_X(X|\theta)}{\partial \theta}$$which is a random variable such that its expectation is zero when taken at the true value of the parameter $\theta$, i.e.,$$\mathbb{E}_{\theta_0}\left[ \dfrac{\partial \log f_X(X|\theta_0)}{\partial \theta}\right]=\int \dfrac{\partial \log f_X(x|\theta_0)}{\partial \theta}f_X(x|\theta_0)\,\text{d}\lambda(x)=0$$

[Answer typed while @whuber and @knrumsey were typing their respective answers!]

  • $\begingroup$ Could you explain in words what is the meaning/interpretation of the statement $F_X(X_1)=P(X_2 \leq X_1)$? It still seems to me that saying "the cdf of a r.v. has a uniform distribution" does not make any sense. $\endgroup$
    – mai
    Feb 27, 2018 at 20:35
  • $\begingroup$ The cdf of a rv $F_X$ is not the same thing as the transform of a rv $X$ by the cdf of this rv, namely $F_X(X)$. $\endgroup$
    – Xi'an
    Feb 27, 2018 at 20:43
  • $\begingroup$ Yes, I agree that they are not the same thing. In the first instance it is not a r.v., while in the second case it is a r.v. Am I correct? $\endgroup$
    – mai
    Feb 27, 2018 at 21:34
  • $\begingroup$ Yes, which relates to the different meanings of $X$ in $F_X(X)$ $\endgroup$
    – Xi'an
    Feb 27, 2018 at 21:35
  • $\begingroup$ Could you explain what you mean by "expectation is zero when taken at the true value of the parameter $\theta$? It seems like $\theta$ is being treated as a variable here. What changes if $\theta$ is not at its "true value"? $\endgroup$
    – mai
    Mar 1, 2018 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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