# Is it true that if $P(\int_0^T f^2(s) ds<\infty)=1$ then the exponential defines a density?

Let $$f(t)$$ be a progressively measurable process wrt Brownian motion $$B(t)$$ so that $$P\left(\int_0^Tf^2(s)ds<\infty\right)=1$$ Is it true then that the exponential

$$\exp\left(\int_0^T f(s)dB(s)-\frac12\int_0^Tf^2(s)ds\right)$$

defines a density on Wiener space? I know the Novikov condition

$$E\left[\exp\left(\int_0^Tf^2(s)ds\right)\right]<\infty$$

implies that the exponential defines a density. But what is you just have a.s. $$L^2$$?

• A density function must be integrable, but I don't think the a.s. $L^2$ condition implies integrability (of it's exponential), and hence I don't believe ir would suffice. Jul 24, 2020 at 10:47
• @RScrlli Can you give a counterexample? I know that by Ito's formula you have that it is a local martingale.
– user658409
Jul 24, 2020 at 15:06

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 $$(X_t)_{t\in[0,1]}$$ is a $$Q$$-BM. In particular, $$Q(\exists t\in[0,1]:\,X_t=0)>0,$$ but, by standard properties of two-dimensional BM, $$P(\exists t\in[0,1]:\,X_t=0)=0.$$ It is interesting to observe that, however, $$\mathbb{E}\left[\exp\left(\int_0^t f(s)\,ds-\frac12\int_0^t f(s)^2\,ds\right)\right]<\infty$$ for all $$t\in[0,1]$$.