Ratio of real world to risk-neutral density

Question

Suppose I have a risk-neutral pdf and a real-world pdf of an asset. Both functions are related by a scaling factor or the sdf which would transform the risk-neutral into the real-world density, is this correct? Would I be able to obtain this scaling factor by simply computing the ratio of the RW/RN density across X? If not, why?

Answer 1

The ratio of any two PDFs (if it can be defined: see below) is just the Radon-Nikodym derivative. The conditions for the Radon-Nikodym derivative to exist are given by the Radon-Nikodym theorem.

Whilst on a first read basis, the rigorous phrasing of the theorem might sound a bit abstract / difficult, the intuition is easy. The theorem says that given two probability measures $\mathbb{P}_1$ and $\mathbb{P}_2$, the Radon-Nikodym derivative $\frac{\partial\mathbb{P}_2}{\partial\mathbb{P}_1}$ exists if the two measures agree on "what is possible", which means they agree on where probabilities are zero and where they are non-zero (see footnote*).

Example 1:

Consider two log-normal random variables $X$ and $Y$, where:

$$\ln{(X)}\sim N\left(\mu -0.5 \sigma^2, \sigma\right), \quad \ln{(Y)}\sim N\left(r -0.5 \sigma^2, \sigma\right)$$

Let the PDF of $X$ be $f_X$ and the PDF of $Y$ be $f_Y$ (and the associated probability measures induced by $X$ and $Y$ are $\mathbb{P}_X$ and $\mathbb{P}_Y$). Note that the support of $X$ and $Y$ are the same: specifically the two random variables have positive densities on the real-line for $x>0$, whilst their densities are zero for $x\leq0$: therefore, these two random variables "agree on what is possible" (any $x>0$): they just assign slightly different probabilities to these outcomes. So the conditions for the Radon-Nikodym derivative to exist are satisfied and we can write:

$$\frac{\partial\mathbb{P}_Y}{\partial\mathbb{P}_X}=\frac{f_Y}{f_X}$$

Example 2:

Consider now a log-normal random variable $X$ and a normal random variable $Z$, where:

$$\ln{(X)}\sim N\left(\mu -0.5 \sigma^2, \sigma\right), \quad Z\sim N\left(\mu -0.5 \sigma^2, \sigma\right)$$

The problem here is that $Z$ has positive densities across the entire real line $\mathbb{R}$ whilst $X$ only has positive densities for $x>0$. So for the case where we want to go from $\mathbb{P}_X$ to $\mathbb{P}_Z$, the conditions for the Radon-Nikodym theorem are not satisfied and the Radon-Nikodym derivative does not exist. Indeed, if we tried to compute the ratio of the two PDFs, we would run into the following issues:

$$\frac{\partial\mathbb{P}_Z}{\partial\mathbb{P}_X}=\frac{f_Z}{f_X}=??$$

The derivative is clearly undefined $\forall x \leq 0$ due to division by zero.

*Footnote: strictly speaking, for $\frac{d\mathbb{P}_2}{d\mathbb{P}_1}$ to exsist, the theorem states that $\mathbb{P}_2$ must be absolutely continuous w.r.t. $\mathbb{P}_1$, which means that whenever $\mathbb{P}_2(A)=0$ for any $A \in \mathcal{F}$, it implies that $\mathbb{P}_1(A)=0$ (but the converse needs not to be true).