0
$\begingroup$

I'm introduced to the Radon-Nikodym derivative in the following way:

Probability space $(\Omega, \mathcal{F}, \mathbb{P})$: $\mathbb{P}$ a probability measure on $\Omega$. A random variable $X$ is a deterministic function $X:\Omega\to\mathbb{R}$. Distribution of $X$: determined by how $\mathbb{P}$ assigns probabilities to subsets $\Omega$ and how $X$ maps those to subsets of $\mathbb{R}$.

An example: $\Omega = [0,1]$, and $\mathbb{P}$ is uniform on $\Omega$, i.e., for $0\leq a\leq b\leq 1$: $\mathbb{P}[a,b] = b-a$. Define $X:[0,1]\to\mathbb{R}$ by $X(w) = -\log w$. Under $\mathbb{P}$ the rv $X$ has an $\operatorname{Exp}(1)$ distribution. Alternative measure $Q$ assigns $Q[a,b] = b^2 - a^2$. Under $Q$ the rv $X$ has an $\operatorname{Exp}(2)$ distribution: $$Q\{X\leq x\} = Q\{w:-\log w\leq x\} = Q[e^{-x}, 1] = 1^2 - (e^{-x})^2 = 1 - e^{-2x}$$

The Radon-Nikodym derivative of $Q$ w.r.t. $\mathbb{P}: \,\,\dfrac{dQ}{d\mathbb{P}}(w) = 2w.$

Now there are a couple of things that I don't understand about this example:

  • Why is the notation of a set used when writing down the cdf of $Q$? We have that $Q\{X\leq x\} = 1 - e^{-2x}$, why don't we just write $Q(X\leq x) = 1 - e^{-2x}$?
  • How should I calculate the Radon-Nikodym derivative? I don't understand how you would arrive at $\dfrac{dQ}{d\mathbb{P}}(w) = 2w$? Both $\mathbb{P}$ and $Q$ take intervals as arguments right? Could someone show me how you would compute this derivative?

Thanks!

$\endgroup$
  • $\begingroup$ Formally you should write $Q(\{X\leq x\})$ which is commonly abbreviated by $Q(X\leq x)$. Is there a distinction between $\mathbb P$ and $P$ (falling out of the sky) in your question? If $Q$ has a Radon-Nikodym derivative wrt $P$ then $Q$ and $P$ must be measures on the same measure space. $Q$ and $\mathbb P$ are not measures on the same measure space. $\endgroup$ – drhab Apr 9 '18 at 8:39
1
$\begingroup$

There is no difference between the notations $Q(X\leq x)$ and $Q\{X\leq x\}$. For the second question the Radon Nikodym derivative $f$ of $Q$ with respect to $P$ is defined by the equation $Q(E)=\int_E fdP$ for every measurable set $E$. In order find what this $f$ is it is enough consider the sets $E=[o,x]$ where $0\leq x \leq 1$. Thus we have to find $f$ such that $1-e^{-2x} =Q(X\leq x)=\int_{[0,x]} f(t)dt$. [ Note that $P$ is just the uniform measure (i.e. the Lebesgue measure on $[0,1]$ so $\int_E fdP=\int_E f(y)dy)$. To find $f$ from the equation $1-e^{-2x} =\int_0^{x} f(y)dy$ simply differentiate both sides with respect to $x$. Hence $f(x)=2e^{-2x}$.

$\endgroup$
  • $\begingroup$ $\mathbb P$ is defined on $\Omega=[0,1]$ so not on $\mathbb R$. That means that $Q$ - which is defined on $\mathbb R$ has no Radon-Nikodym derivative wrt $\mathbb P$. Further how does this correspond with $\frac{dQ}{d\mathbb P}(w)=2w$ as posed in the question? $\endgroup$ – drhab Apr 9 '18 at 9:15
  • $\begingroup$ @drhab there is a lot of confusion in the notations used by OP. It seems to me that he wants to compute $\frac {QX^{-1}} {dP}$ and not dQ/dP. Of course, as you have observed, $QX^{-1}$ is not absolutely continuous w.r..t P. What I have computed is the Radon Nikodym derivative of the absolutely continuous part of $QX^{-1}$ w.r..t P. $\endgroup$ – Kavi Rama Murthy Apr 10 '18 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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