# Understanding this example/explanation of the Radon-Nikodym derivative.

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. $\mathbb{P}: \,\,\dfrac{dQ}{d\mathbb{P}}(w) = 2w.$

• 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!

• 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. – drhab Apr 9 '18 at 8:39

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 $f(x)=2e^{-2x}$.

• $\mathbb P$ 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? – drhab Apr 9 '18 at 9:15
• @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. – Kavi Rama Murthy Apr 10 '18 at 4:44
• @drhab give a look to my answer – Buddy_ Dec 14 '19 at 17:32

It seems to me that 2$$\omega$$ is given by: $$\mathbb P(X \le x) = 1-e^{-x}$$ $$\mathbb Q(X \le x) = 1-e^{-2x}$$ $$d\mathbb P=e^{-x}, d\mathbb Q=2e^{-2x} \to \frac{d\mathbb Q}{d\mathbb P}=2e^{-x}$$

• $\mathbb P$ and $\mathbb Q$ are both measures defined on probability space $(\Omega=[0,1],\mathcal B)$ where $\mathcal B$ stands for the collection of Borel subsets of $[0,1]$. For every $[a,b]\subseteq[0,1]$ we have $\mathcal Q([a,b])=b^2-a^2=\int_a^b\int2\omega\mathcal P(d\omega)$ showing that $\frac{d\mathcal Q}{d\mathbb P}=2\omega$ (as stated in the example). To find this derivative random variable $X$ can be missed and is IMV only confusing. Btw we do not have $d\mathbb P=xe^x$ but $dP_X=e^{-x}dx$ where $P_X(B):=\mathbb P(X\in B)$. Likewise $dQ_X=2e^{-2x}dx$. So $\frac{dQ_X}{dP_X}=2$ – drhab Dec 15 '19 at 12:35
• @drhab there have to be some errors because $dQ/dP=2e^{-x}$ – Buddy_ Dec 15 '19 at 12:45
• Yes, there you are right. We have $\frac{dQ_{X}}{dP_{X}}=\frac{2e^{-2x}dx}{e^{-x}dx}=2e^{-x}$. But that is not the same derivative as $\frac{d\mathbb Q}{d\mathbb P}=2\omega$. – drhab Dec 15 '19 at 13:01
• I disagree. $\mathbb Q$ has Radon-Nikodym derivative wrt $\mathbb P$ and $Q_X$ has a Radon-Nikodym derivative wrt $P_X$. Two distinct observations. In general a measure $\mu$ can only have such a derivative wrt measure $\nu$ if both measures are defined on the same probability space. – drhab Dec 15 '19 at 13:25