Consider a general probability space $(\Omega, \mathcal{F}, \mathbb{S})$, on which two or more other probability measures, $\mathbb{P}_1$, $\mathbb{P}_2$,…,$\mathbb{P}_j$,…,$\mathbb{P}_n$ are defined that are absolutely continuous with respect to one another.
Consider a normally distributed random variable $X\sim N(0, \sigma)$ with its density function denoted $f_{X}$. Consider a probability measure $\mathbb{P_1}$ defined by the cumulative density function (CDF) of $X$:
\begin{equation}
\label{eq1}
\mathbb{P}_1(A):=\int_{-\infty}^{h=a}\frac{1}{\sigma \sqrt{ 2\pi}}e^{\frac{-(h)^2}{2\sigma^2}}dh
\tag{1}
\end{equation}
(to be clear, $X$ is defined on Borel-measuruable sets on $\mathbb{R}$, above $a \in \mathbb{R}$ and $A$ is defined as the event: $\{A \in \mathcal{F} : X \leq a\}$).
Next consider the following function of $X$:
\begin{equation}
\label{eq2}
g(X):=\frac{1}{k}e^{\frac{(k^2-1)X^2}{2k^2\sigma^2}}
\tag{2}
\end{equation}
Using the "law of the unconscious statistician", we have that:
\begin{equation}
\label{eq3}
\mathbb{E}^{\mathbb{P_1}}\left[g(X)\right]=\int_{-\infty}^{\infty}g(h)f_{X}(h)dh=\int_{-\infty}^{\infty}\frac{1}{k\sigma\sqrt{ 2\pi}}e^{\frac{-(h)^2}{2\sigma^2 k^2}}dh
\tag{3}
\end{equation}
Since in the last expression in \ref{eq3} we recognize a PDF of a normally distributed random variable with mean zero and variance $\sigma^2k^2$, we have that $\mathbb{E}^{\mathbb{P_1}}\left[g(X)\right]=1$. Furthermore, restricting $k>0$ results in $g(X)>0$, and under this restriction, $g(X)$ is a valid Radon-Nikodym derivative. We can define a new measure $\mathbb{P}_2$ as follows (below $\mathbb{I}_{\{.\}}$ is an indicator function):
\begin{equation}
\label{eq4}
\mathbb{P}_2(A):=\mathbb{E}^{\mathbb{P}_1}\left[g(X)\mathbb{I}_{\{A\}}\right]
\tag{4}
\end{equation}
We can say that under the measure $\mathbb{P}_2$: $X^{\mathbb{P}_2}\sim N(0, k^2\sigma^2)$.
Consider now a different probability space $(\Omega, \mathcal{F_t}, \mathbb{S})$ and assume that $\mathbb{S}$ is a probability measure induced by the standard Wiener process $W_t$. Define $\mathbb{P}_1$ as:
\begin{equation}
\label{eq5}
\mathbb{P_1}(A):=\mathbb{P_1}(W_t\leq a: a \in\mathbb{R})=\int_{-\infty}^{h=a}\frac{1}{\sqrt{ 2\pi t}}e^{\frac{-(h)^2}{2t}}dh=\int_{-\infty}^{h=a}f_{W_t}(h)dh
\tag{5}
\end{equation}
Consider the following function of $W_t$:
\begin{equation}
\label{eq6}
g(W_t):=\frac{1}{k}\exp{\left(\frac{W_t^2\left(k^2-1\right)}{2tk^2}\right)}
\tag{6}
\end{equation}
Above, $k>0$ is some constant. Repeating the same argument, we can conclude that $g(W_t)$ is a valid Radon-Nikodym derivative (i.e. integrates to 1 and is positive).
We can write $\mathbb{P}_2(A):=\mathbb{E}^{\mathbb{P}_1}\left[g(W_t)\mathbb{I}_{\{A\}}\right]$ and conclude that:
\begin{equation}
\label{eq7}
W(t)^{\mathbb{P}_2}\sim N(0,k^2t)
\tag{7}
\end{equation}
Question: in the text-book "Stochastic Calculus and Financial Application" by J.M. Steele, in the first paragraph on page 221, the author states (without proof) that given two processes $\sigma_1 W_t$ and $\sigma_2 W_t$, the measures $\mathbb{P}_{\sigma_1}$ and $\mathbb{P}_{\sigma_2}$ are singular whenever $\sigma_1 \neq \sigma_2$. Setting $\sigma_1=1$ and $\sigma_2=k$, I am unable to prove / explain why the construction above wouldn't produce two equivalent probability measures.
Best Answer
One event where they differ is $$A:=\{\langle W_{t}\rangle=\sigma_{1}t\}$$
since $\mathbb{P}_{\sigma_1}(A)=1$ and $\mathbb{P}_{\sigma_2}(A)=0$.
see here for more Laws of $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$ : singular?
This only for each fixed time-slice whereas in Brownian motion we consider it over intervals.