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


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

  • $\begingroup$ 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. $\endgroup$
    – Chaos
    Jul 24, 2020 at 10:47
  • $\begingroup$ @RScrlli Can you give a counterexample? I know that by Ito's formula you have that it is a local martingale. $\endgroup$
    – user658409
    Jul 24, 2020 at 15:06

1 Answer 1


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 $t\in[0,1]$.


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