8 Densities
8.1 Properties
8.1.1 Sufficient conditions for existence
The Radon Nikodym theorem provides a set of sufficient conditions (domination and sigma-finiteness) for the existence of a density. Let \mu, \nu be measures on (E, \mathcal E).
- \mu is \sigma finite.
- \mu dominates \nu (i,.e. \mu(A) = 0 \implies \nu(A) = 0 \forall A \in \mathcal E).
then there exists a non-negative measurable function \frac{d\nu}{d\mu} (x) such that \mu[f \frac{d\nu}{d\mu}] = \nu[f] for all positive measurable f.
8.1.2 Existence implies domination
\frac{d\nu}{d\mu} exists \implies \nu \ll \mu
Proof
Assume \mu(A) = 0 for some measurable set A. \nu(A) = \int \mathbb 1_A \frac{d\nu}{d\mu}(x) d\mu = 0
since the Lebesgue integral wrt \mu of any function over a \mu-negligible set is zero.
8.1.3 Uniqueness
If p and q are densities of \nu wrt \mu, p = q \mu-a.e.
Proof
Since they’re densities, for all positive measurable f, \nu[f] = \mu[fp] = \mu[fq]. Thus, \mu [f(p - q)] = 0. This implies f(p - q) is zero \mu-a.e. for all f, and thus p-q = 0 \mu-a.e.8.1.4 Condition for finiteness
\frac{d\nu}{d\mu} < \infty \quad \mu \text{-a.e. } \iff \nu \quad \sigma-finite
Assuming \frac{d\nu}{d\mu} exists and \mu is \sigma-finite.
Proof
For the first direction: Assume density \frac{d\nu}{d\mu} is finite almost everywhere in \mu. Let the set of infinite density points be E_0. \mu(E_0) = 0 so by domination \nu(E_0) = 0 < \infty.
By sigma finiteness of \mu, there exists a countable partition (E_i)_{i=1}^\infty such that \mu(E_i) < \infty and \cup_i E_i = E. Now, for m \in \mathbb N take E_{im} = {x \in E_i: \frac{d\nu}{d\mu} (x) \in [m, m+1) }. By the diagonal argument, E_{im} is a countable sequence of sets, and each one is measurable since \frac{d\nu}{d\mu} is a measurable function.
For each i, m, \nu(E_{i,m}) = \nu[\mathbb I_{E_{i,m}}] = \mu[\mathbb I_{E_{i,m}} \cdot \frac{d\nu}{d\mu}] \leq (m+1) \mu(E_{i,m}) \leq (m+1) \mu(E_{i}) < \infty. Thus, we have a countable partition \{E_0\} \cup (E_{im})_{i=1:\infty,m=0:\infty}
For the second direction, Assume \nu is \sigma-finite. Then there exists a countable partition (A_i) such that \nu(A_i) < \infty for each i. Now, consider B_i = \{x \in A_i: \frac{d\nu}{d\mu}(x) = \infty \}.
Now, if we prove that for each i, \mu(B_i) = 0, this would imply that the set of all infinite density points \cup_i B_i also has measure zero under \mu by countable additivity.
Assume \mu(B_i) > 0. Then, \nu(A_i) \geq \nu(B_i) = \mu[\frac{d\nu}{d\mu} \mathbb I_{B_i}] = \infty \cdot \mu(B_i) = \infty, which implies \mu(A_i) = \infty, leading to a contradiction. Thus, \mu(B_i) = 0.
8.1.5 Condition for strict positivity
\frac{d\nu}{d\mu} > 0 \quad \mu \text{-a.e. } \iff \mu \ll \nu
Proof
Assume $\frac{d\nu}{d\mu} > 0 \mu-a.e. Let \nu(B) = 0 = \int_B \frac{d\nu}{d\mu} d\mu. Since the density is strictly positive almost everywhere in \mu, this implies \mu(B) = 0.
For the converse, assume \nu >> \mu. Let A be set of zero density. \nu(A) = \int_A \frac{d\nu}{d\mu} d\mu = 0. Thus by domination, \mu(A) = 0.8.2 Manipulating densities
8.2.1 Linearity
Let \mu, \nu, \lambda be measures on (E, \mathcal E) and let a, b be positive measurable functions. Define:
\xi(A) = (a\mu + b\nu)(A) = \int_A a(x) \mu(dx) + \int_A b \nu(dx) . If the densities \frac{d\mu}{d\lambda} and \frac{d\nu}{d\lambda} exist, then the density \frac{d\xi}{d\lambda} also exists and:
\frac{d\xi}{d\lambda}(x) = a(x) \frac{d\mu}{d\lambda}(x) + b (x) \frac{d\nu}{d\lambda}(x)
8.2.2 Chain rule
If \frac{d\nu}{d\mu} and \frac{d\mu}{d\lambda} exist then \frac{d\nu}{d\lambda} exists and \frac{d\nu}{d\lambda}(x) = \frac{d\nu}{d\mu}(x) \cdot \frac{d\mu}{d\lambda}(x) \quad \lambda-\text{a.e.}
8.2.3 Reciprocal
If \frac{d\nu}{d\mu} and \frac{d\mu}{d\nu} both exist, then
\frac{d\mu}{d\nu}(x) = \frac{1}{\frac{d\nu}{d\mu}(x)} \quad \mu-\text{a.e. and } \nu-\text{a.e.}
8.2.4 Transformation
Suppose \frac{d\nu}{d\mu} exists. Then, for any measurable injection h : E \to F,
\frac{d\nu \circ h^{-1}}{d\mu \circ h^{-1}} (x) = \frac{d\nu}{d\mu} (h^{-1} (x)) \quad \mu \circ h^{-1} \text{-a.e.}
8.3 Kernel densities
8.3.1 Marginalised densities
Let (E, \mathcal E) and (F, \mathcal F) be standard measurable spaces, \mu, \nu be measures on E and L, K be kernels (E, \mathcal F) \to \mathbb R. Suppose \frac{d\nu}{d\mu} exists, and \frac{dL(x, dy)}{dK(x, dy)} exists and is measurable wrt the product sigma algebra \mathcal{E} \otimes \mathcal{F}.
Denote the marginal measures by \mu K(A) = \int \mathbb 1_A(y) \mu(dx)K(x, dy) and \nu L(A) = \int \mathbb 1_A(y) \nu(dx)L(x, dy).
Let M: (F, \mathcal E) \to \mathbb R be the disintegration kernel of \mu K such that \mu (dx) K(x, dy) = \eta(dy) M(y, dx) (\eta is a measure on (F, \mathcal F)) - shorthand for “\forall g\in \mathcal E \otimes F, \int g(x,y) \mu (dx) K(x, dy) = \int g(x, y) \eta(dy) M(y, dx)”.
Then the density of \nu L wrt \mu K is given by:
\frac{d\nu L}{d\mu K}(y) = \int M(y, dx) \frac{d\nu}{d\mu}(x) \frac{dL}{dK}(x, y)