# Two probability measures $P,Q$ which do not admit densities wrt Lebesgue measure but have computable $\frac{dP}{dQ}$?

My question is simple as in the title: Is there any example of two probability measures $P$ and $Q$ where $P \ll Q$ which do not have densities w.r.t Lebesgue measure but the Radon-Nikodym derivative $\frac{dP}{dQ}$ is actually analytically available?

In other words, I wonder if there exists an actually computable (in terms of an analytical formula) Radon-Nikodym derivative, without referring each measure's density wrt Lebesgue measure.

I would love to see an answer with nondegenerate $P$ and $Q$, by which I mean $P$ and $Q$ are defined on $\mathbb{R}$ and they are not empirical measures (Otherwise, it must be easy to construct examples with discrete measures).

PS: Also I must add that I am wondering about different probability measures: $P \neq Q$.

More notes: The reason I am asking for two probability measures that do not have densities wrt Lebesgue measure because the answer is trivial otherwise. If $P$ and $Q$ are allowed to be absolutely continuous wrt Lebesgue, then take any two continuous probability densities $p(x),q(x)$ and RN derivative is trivial: $p(x)/q(x)$. I am asking two probability measures which do not have densities (wrt Lebesgue measure) but still has a Radon-Nikodym derivative $dP/dQ$ which has an analytical expression.

• A somewhat silly way to construct examples that meet your current criteria is to take some example measures $P$ and $Q$ that are absolutely continuous with respect to the Lebesgue measure for which the R-N derivative is known and perturb them by a point mass $\delta_{x}$. It's easy to see how this perturbation changes the R-N derivative. – Rhys Steele Jun 29 '18 at 0:00

Let $P$ be the probability measure on $[0,1]$ for which the corresponding distribution function is the Cantor-Lebesgue function. No one ever seems to say it, but P can be understood as the distribution of a $[0,1]$-valued random variable that is equally likely to have 0 or 2 in each entry of its trinary expansion, but never has a 1.

Define $Q$ by $dQ = \chi_{[2,3]} dP$. Then $Q$ is the distribution of a $[0,1]$-valued random variable that certainly has a 2 as the first digit of its trinary expansion and afterward is equally likely to have a 0 or a 2.

I think martingale betting strategies could offer more interesting choices of $Q$ here, but I haven't spent much time thinking about those so I'll avoid going there for now.

These are examples not over $\mathbb{R}$, but might still be of interest: They are cases when $P$ and $Q$ are defined over spaces on which no Lebesgue measure exists.

For example on $\{0,1\}^{n}$, the space of length$-n$ binary strings. Take the two measures $P$=Bernoulli$(p)^{n}$ and $Q$=Bernoulli$(q)^{n}$, i.e. the measures under which the elements of the sequences are iid from Bernoulli with parameter $p$ and $q$ respectively. Then a density can be given by $$\frac{dP}{dQ}(s)=\Big(\frac{p}{q}\Big)^{n_1(s)}\Big(\frac{1-p}{1-q}\Big)^{n-n_1(s)},$$ where $n_1(s)$ is the number of 1's in a string $s$.

Another example is when $P$ and $Q$ are defined over some abstract probability space (also with no Lebesgue measure available) and we know that a certain process $(W_t)_{t\in[0,T]}$ is a Wiener process under $Q$, while under $P$, it behaves as a Wiener process with some drift $h_t$. The density is given by $$\frac{dP}{dQ}\Big|_{\mathcal{F_T}}=\exp\Big[\int_0^Th_tdW_t-\frac{1}{2}\int_0^Th_t^2dt\Big],$$ which for certain forms of the drift, can be further simplified.

• Thanks, this is useful. I was thinking in terms of importance sampling, where $dP/dQ$ is a weight function -- looking for examples where you can sample from $Q$ directly (which does not have a density wrt Lebesgue) and perform importance sampling. – odea Jul 2 '18 at 23:17