# Probability density function and Radon-Nikodym derivative

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 $$X$$.



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 $$A \in \mathcal{B}(\mathbb{R})$$.

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 $$\mathbb E[h(X)]=\sum_{n\in\mathbb N}h(n)\mathbb P(X=n)=\int_{\mathbb N}h(n)f_X(n)\,\nu(dn).$$
• 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 ? Commented Jan 26 at 0:10
• 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.
• 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 $Q$? Commented Jan 27 at 7:39