Radon-Nikodym Theorem

Assume throughout this section that measures are defined on a common measurable space $\Omega$ .

lemma: if $P,Q$ are measures such that $P \le Q$ then for any non-negative random variable $X$ , $P[X] \le Q[X]$ .

proof utilizes the standard machine. The statement is valid for indicator functions. Therefore it is valid for simple functions, and non-negative measurable functions, and for integrable functions.

definition: [absolute continuity] measures $P, Q$ are such that $P << Q$ iff for any measurable set, if $QA = 0$ then $PA = 0$ .

the existence proof below is from Von Neumann

theorem: [existence] if $P,Q$ are probability measures on a common measurable space and such that $Q << P$ , then there is a random variable $Z$ such that $P[|Z|] < \infty$ and for which

$Q(\mathbb{I}_A) = P(Z \mathbb{I}_A)$ .

proof: consider the average measure $R = \frac{1}{2} (P + Q)$ . By standard properties of measure, the lemma above, and the Schwartz inequality,

$\left| P[X] \right| \le P [|X|] \le 2R|X| = 2 R[1 \cdot |X|] \le \left( 2 R[1] R[|X|^2] \right)^{1/2} = 2 || X ||_{2,R}$ . Since the function $f(X) = \frac{1}{2} Q[X]$ is norm-continuous linear map on the normed linear space $L_2(R)$ there is a $U \in L_2(R)$ such that

$\frac{1}{2} Q[X] = R[UX] = \frac{1}{2} Q[UX] + \frac{1}{2} P[UX]$ .

$Q[X(1-U)] = P[XU]$

Choosing $X = \mathbb{I}_A$ where $A$ is a measurable set,

$0 \le R[\mathbb{I}_A U] = \frac{1}{2} Q[\mathbb{I}_A] \le \frac{1}{2} R[\mathbb{I}_A]$ .

Since this holds for all measurable sets, $0 \le U \le 1$ $R$ -a.s. Since $P,Q << R$ this condition also holds $P$ -a.s and $Q$ -a.s. Taking $X = \mathbb{I}_{U = 1}$ ,

$0 = Q[X(1-U)] = P[XU] = P[X]$ and since $Q << P$ , $0 = Q[X]$ .

Take $Y_n = \sum_{i = 0..n} U^i$ . Then $\mathbb{I}_A Y_n \in L_2(R)$ and

$Q[\mathbb{I}_A (1-U^{n+1})] = Q[\mathbb{I}_A Y_n (1 - U)] = P[\mathbb{I} Y_n U] = P[\mathbb{I} (Y_{n+1} - 1) U]$

Since $U \in [0,1)$ $Q,P$ -a.s, $1-U^{n+1}$ is non-decreasing with limit 1 ( $Q$ -a.s) and $Y_{n+1} - 1$ is non-increasing with limit $X$ ( $P$ -a.s.) and so by the MCT,

$Q[\mathbb{I}_A] = P[X \mathbb{I}_A]$

The importance of the Radon-Nikodym theorem is that it allows translation between probability measures. The variable $Z$ can be viewed as a renormalization of the probability, with renormalization possible when $Q << P$ . A number of useful properties

lemma: the Radon-Nikodym derivative is a.s unique.

Suppose $QA = P[X \mathbb{I}_A] = P[Y \mathbb{I}_A]$ . Then $P[(X-Y) \mathbb{I}_A] = 0$ . Choose $A = [X > Y]$ . Then $X \le Y$ a.s. Dually $Y \le X$ a.s., and so $X = Y$ a.s

The (a.s) uniqueness motivates identification of a standard candidate as $\frac{d Q}{d P}$ .

lemma: If $P,Q,R$ are probability measures then

if $Q,R << P$ and $\lambda \in (0,1)$ then $\partial_P \left(\lambda Q + (1-\lambda) R \right) = \lambda \partial_P Q + (1-\lambda) \partial_P R$
if $R << Q << P$ then $\partial_P R = \partial_Q R \cdot \partial_P Q$ .
if $R \sim Q ( Q << R << Q$ ) then $\partial_Q R \cdot \partial_R Q = 1$ .

these properties are a consequence of the uniqueness of the RN derivative and linearity properties the measure:

Suppose $Q,R << P$ . Then there are $X,Y$ such that

$Q[\mathbb{I}_A] = P[\mathbb{I}_A X]$

$R[\mathbb{I}_A] = P[\mathbb{I}_A Y]$

and when $\lambda \in (0,1)$ , by linearity properties,

$(\lambda Q + (1-\lambda) R)[\mathbb{I}_A] = \lambda Q[\mathbb{I}_A] + (1- \lambda) R[\mathbb{I}_A] = \lambda P[\mathbb{I}_A X] + (1-\lambda) P[\mathbb{I}_A Y] = P[\mathbb{I}_A(\lambda X + (1- \lambda) Y) ]$

Suppose $R << Q << P$ . Then

$Q[\mathbb{I}_A] = P[\mathbb{I}_A X]$

$R[\mathbb{I}_A] = Q[\mathbb{I}_A Y]$

Since $Y$ is non-negative and measurable choose a sequence of non-negative non-decreasing simple functions $\left< s_n \right>$ such that $\lim s_n = Y$ a.s. Then by MCT $Q[\mathbb{I}_A Y] = \lim_n Q[\mathbb{I}_A s_n]$ . Then since for non-negative simple function $Q[\mathbb{I}_A s_n] = P[\mathbb{I}_A s_n X]$ . Since $\lim_n \mathbb{I}_A s_n X = \mathbb{I}_A YX$ , by MCT $R[\mathbb{I}_A] = Q[\mathbb{I}_A Y] = P[\mathbb{I}_A XY]$

For the final result, choose $P=R$ . Since $1 = \partial_R R$ , the conclusion follows.

Radon-Nikodym Theorem

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