Consider a probability space $(\Omega,\Sigma,P)$.

We say that a real random variable $X\colon \Omega \to \mathbb{R}$ is absolutely continuous when there exists a function $f_X\colon \mathbb{R} \to [0,+\infty)$ such that $\mu_X((-\infty,x])=\int_{-\infty}^x f_X(t)\,\text{d}t$ for all $x \in \mathbb{R}$, where $\mu_X\colon \mathcal{B}(\mathbb{R}) \to [0,1] \mid \mu_X(A)=P(X \in A)$, namely $\mu_X$ is the Borel pushforward measure (or image measure) of $X$. $f_X$ is said to be the probability density function (PDF) of $X$.

$ $

Now consider Radon-Nikodym derivative:

Consider the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu_X)$. The Radon-Nikodym derivative $\frac{\text{d}\mu_X}{\text{d}\lambda}$ is a measurable function $f\colon \mathbb{R} \to [0,+\infty)$ such that for all $A \in \mathcal{B}(\mathbb{R}) \quad \mu_X(A)=\int_A f\,\text{d}\lambda$ where $\lambda$ is the Lebesgue measure.

$ $

So my question: is it true that $f_X=f$?

I have foound on wikipedia that: "The probability density function of a random variable is the Radon–Nikodym derivative of the induced measure with respect to some base measure (usually the Lebesgue measure for continuous random variables)". So I suppose that the answer to my question is "yes".

$ $

We know that $\forall \,x \in \mathbb{R} \quad \mu_X((-\infty,x])=\int_{-\infty}^x f_X(t)\,\text{d}t=\int_{(-\infty,x]} f_X\,\text{d}\lambda$, where obviously $(-\infty,x] \in \mathcal{B}(\mathbb{R})$. In order to show that $f_X=f$ we should prove that the above relation holds for every $A \in \mathcal{B}(\mathbb{R})$.

Thank you for your time!


1 Answer 1


The answer to your question is yes.

If $f:\mathbb R\to\mathbb R$ is a nonnegative measurable map such that $\int_{\mathbb R}f(x)\,dx=1$, then there exists a unique probability measure $\mu$ on $\mathcal B(\mathbb R)$ such that for all $x\in\mathbb R$, $\mu((-\infty,x])=\int_{-\infty}^xf(t)\,dt$. This is the very well known fact that the cumulative distribution function characterises a probability measure. So this measure $\mu$ is the one defined for all $A\in\mathcal B(\mathbb R)$ by $\mu(A)=\int_Af(x)\,dx$. We deduce that a real-valued random variable $X$ is absolutely continuous iff its distribution $\mu_X$ is absolutely continuous with respect to the Lebesgue measure $\lambda$, in which case $X$ admits the probability density function $f_X=\frac{d\mu_X}{d\lambda}$.

You could also check that if $X:\Omega\to\mathbb N$ is a discrete random variable, then the distribution $\mu_X$ of $X$ admits a density with respect to the counting measure $\nu=\sum_{n\in\mathbb N}\delta_n$. More precisely, the density $\frac{d\mu_X}{d\nu}=f_X:\mathbb N\to\mathbb R$ is defined for all $n\in\mathbb N$ by $f(n)=\mathbb P(X=n)$. Indeed we have for any $A\subset\mathbb N$ and measurable bounded map $h:\mathbb N\to\mathbb R$ that $$ \mathbb E[h(X)]=\sum_{n\in\mathbb N}h(n)\mathbb P(X=n)=\int_{\mathbb N}h(n)f_X(n)\,\nu(dn). $$

  • $\begingroup$ Does the Radon Nikodym theorem also hold in the case of discrete r.v. for which there exists only a pmf ? Can we also write: $\mathbb P(X=n)=\int_{\mathbb N}f_X(n)\,\nu(dn)$ ? Why you needed the expectation ? $\endgroup$
    – user996159
    Commented Jan 26 at 0:10
  • 1
    $\begingroup$ Yes the Radon Nikodym theorem holds. And I write the expectation because it characterises the probability distribution, but it is true that if $X$ is integer-valued then the family $(\mathbb P(X=n))_{n\in\mathbb N}$ characterises the distribution as well. $\endgroup$
    – Will
    Commented Jan 26 at 20:27
  • $\begingroup$ Thanks. Let $P, Q$ be two measures with $P<<Q$, the Radon Nikodym derivative $f$ and the corresponding pdf's, if they exist, are $f_1,f_2$ in respect to the Lebesgue measure. Then, $dP=f_1dx, dQ=f_2dx,dP/dQ=f_1/f_2, f_2\neq 0.$ Can one state that $f=f_1/f_2$ is the Radon Nikodym derivative of $P$ in respect to $Q$ without further notice of their absolute continuity? In particular, if either one of $P, Q$ has no pdf or both of them have no pdf's, and given that the cdf's always exist, can one state that $f=cdf_P/cdf_Q, cdf_Q\neq 0,$ is the Radon Nikodym derivative of $P$ in respect to $Q$? $\endgroup$
    – user996159
    Commented Jan 27 at 7:39
  • $\begingroup$ I think you should post it as a question. $\endgroup$
    – Will
    Commented Jan 27 at 7:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .