Mathematics Asked by user658409 on November 29, 2021
Let $f(t)$ be a progressively measurable process wrt Brownian motion $B(t)$ so that $$Pleft(int_0^Tf^2(s)ds<inftyright)=1$$
Is it true then that the exponential
$$expleft(int_0^T f(s)dB(s)-frac12int_0^Tf^2(s)dsright)$$
defines a density on Wiener space? I know the Novikov condition
$$Eleft[expleft(int_0^Tf^2(s)dsright)right]<infty$$
implies that the exponential defines a density. But what is you just have a.s. $L^2$?
An easy counterexample is provided by the Bessel process (of dimension $2$). Let $X_t=|W_t|$, where $W$ is a two-dimensional Brownian motion started at some unit vector. Then, by Ito, $$ dX_t=frac{1}{2X_t},dt+dB_t,quad X_0=1, $$ for a one-dimensional BM $B$. Consider $f(t)=-frac{1}{2X_t}$ and note that this satisfies your assumption $P(int_0^1 f(u)^2,du<infty)=1$. If the stochastic exponential would define a density, then, by Girsanov, there were an equivalent probability measure $Q$ such that $(X_t)_{tin[0,1]}$ is a $Q$-BM. In particular, $$ Q(exists tin[0,1]:,X_t=0)>0, $$ but, by standard properties of two-dimensional BM, $$ P(exists tin[0,1]:,X_t=0)=0. $$ It is interesting to observe that, however, $$ mathbb{E}left[expleft(int_0^t f(s),ds-frac12int_0^t f(s)^2,dsright)right]<infty $$ for all $tin[0,1]$.
Answered by julian on November 29, 2021
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