Necessary and Sufficient Conditions for the Uniform Integrability of the Stochastic Exponential

Abstract

We establish necessary and sufficient conditions for uniform integrability of the stochastic exponential ${{{\mathcal {E}}}}(M)$, where M is a continuous local martingale.

Appendix

Here we introduce four important lemmas with proofs. These lemmas are essential in the proof of Theorems 1 and 2.

Lemma 1

Let $f:[0;+\infty ) \rightarrow (0; +\infty )$ be a non-decreasing function with $\lim _{x\rightarrow \infty }f(x) = \infty$. Then, a non-decreasing, absolutely continuous function $g:[0;+\infty ) \rightarrow [0; +\infty )$ exists which satisfies the following conditions:

• (a) $g(x)\le f(x)$;

• (b) $\lim _{x\rightarrow \infty }g(x) = \infty$;

• (c) $g(x+y)\le g(x)+y+2$.

Proof

Define the function F: $F(x) = \sum ^{\infty }_{k=1}f(k-1)1_{[k-1; k[}(x)$. It is obvious that $F(x)\le f(x)$. Let us denote by $\bigtriangleup F(k)=F(k)-F(k-1)$ the jumps of F. Because f is non-decreasing, the jumps of F will be nonnegative. Now define a non-decreasing sequence $(g_k)_{k\ge 1}$ through the recurrence relationship:

$$\begin{aligned} g_0=0; \, g_1=1\wedge f(0); \, g_2=g_1+(1\wedge \bigtriangleup F(1)); \, g_k = g_{k-1}+(1\wedge \bigtriangleup F(k-1)) \,\, k\ge 2. \end{aligned}$$

As a result, we have points $(k; g_k), \,\,\, k\ge 0$. Define function g by connecting points $(k; g_k)$ with straight lines. It follows from the definition that g is absolutely continuous, non-decreasing, $g(x)\le F(x) \le f(x)$ and $\lim _{x\rightarrow \infty }g(x) = \infty$. It remains to show that $g(x+y)\le g(x)+y+2$. Let us take $x\in [k-1;k[$ and $y\in [n-1;n[$. It is clear that $x+y \le k+n$. Using the definition of the function g, we obtain:

$$\begin{aligned} g(x+y)-g(x)\le g(k+n)-g(k-1)&= \sum _{i=1}^{k+n}1\wedge \bigtriangleup F(i-1) - \sum _{i=1}^{k-1}1\wedge \bigtriangleup F(i-1) = \\&\quad = \sum _{i=k}^{k+n}1\wedge \bigtriangleup F(i-1) \le n+1 \le y+2. \end{aligned}$$

So, by arbitrariness of k and n, $g(x+y)\le g(x)+y+2$. $\square$

Lemma 2

For any random variable $\xi$ such that $P(0\le \xi < \infty )=1$ there exists a positive, non-decreasing and continuous function g with $\lim _{x\rightarrow \infty }g(x) = \infty$, such that $Eg(\xi ) < \infty$.

Proof

If $\xi$ is a bounded random variable, then the proof of Lemma 2 is trivial, so we can consider only the case when $\xi$ is unbounded. In this case, $F_{\xi }(x)<1$ for any $x\in [0;+\infty )$ where $F_{\xi }(x) = P(\xi \le x)$ is the probability distribution function of $\xi$. Let us take $f(x)=\frac{1}{\sqrt{1 - F_{\xi }(x-)}}$. Then we will have:

$$\begin{aligned} Ef(\xi )&= \int ^{\infty }_0 \frac{1}{\sqrt{1 - F_{\xi }(x-)}} {\text {d}}F_{\xi }(x) = \bigg [ -2\sqrt{1-F_{\xi }(x)} \bigg ]^{\infty }_0 \\&\quad - \sum _{0<x<\infty }\Big [ -2\sqrt{1-F_{\xi }(x)} + 2\sqrt{1-F_{\xi }(x-)}-\frac{\Delta F_{\xi }(x)}{\sqrt{1-F_{\xi }(x-)}} \Big ] \\&\quad \le \bigg [ -2\sqrt{1-F_{\xi }(x)} \bigg ]^{\infty }_0 = 2\sqrt{1-F_{\xi }(0)} < \infty . \end{aligned}$$

Here we have used the inequality $-2\sqrt{1-F_{\xi }(x)} + 2\sqrt{1-F_{\xi }(x-)}-\frac{\Delta F_{\xi }(x)}{\sqrt{1-F_{\xi }(x-)}} \ge 0$, which follows from the convexity of the function $y \rightarrow -2\sqrt{1-y}$. Now if we use Lemma 1, we can find an absolutely continuous, positive and non-decreasing function g with $\lim _{x\rightarrow \infty }g(x) = \infty$, such that $g(x)\le f(x)$. The inequalities $g(x)\le f(x)$ and $Ef(\xi ) < \infty$ imply that $Eg(\xi ) < \infty$. $\square$

Lemma 3

Let $\varphi$ be a lower function. Then, the function $\psi (x)=\varepsilon \varphi (\frac{x}{\varepsilon ^2})$ will be a lower function, too.

Proof

It is well known that if $B_t$ is a Brownian motion, then $W_t = \varepsilon B_{t/\varepsilon ^2}$ will be Brownian motion, too. With this if we take $s=t/\varepsilon ^2$, then we will have:

$$\begin{aligned}&P \Big \{ \omega \,\, : \,\, \exists \, t(\omega ), \, W_t< \psi (t) \,\,\, \forall \, t>t(\omega ) \Big \} \\&\quad = P \Big \{ \omega \,\, : \,\, \exists \, s(\omega ), \, B_s < \varphi (s) \,\,\, \forall \, s>s(\omega ) \Big \} = 0. \end{aligned}$$

$\square$

Lemma 4

For any $\varepsilon >0$, there exists $\delta > 0$ sufficiently close to 0 such that the inequality $\delta x^2 + \delta \varepsilon \le \frac{1}{2}(x-1)^2$ holds true for any $x \notin (1-\varepsilon ; 1+\varepsilon )$.

Proof

It is clear that we can take $\delta >0$ sufficiently close to 0 such that

$$\begin{aligned} 1-\varepsilon< \frac{1-\sqrt{2\delta (1+\varepsilon -2\delta \varepsilon )}}{1-2\delta }< \frac{1+\sqrt{2\delta (1+\varepsilon -2\delta \varepsilon )}}{1-2\delta } < 1+\varepsilon . \end{aligned}$$

It is easy to check that for such $\delta$ the condition $|x-1|\ge \varepsilon$ implies the inequality

$$\begin{aligned} (1-2\delta )x^2 - 2x + 1-2\delta \varepsilon \ge 0 \end{aligned}$$

which is equivalent to $\delta x^2 + \delta \varepsilon \le \frac{1}{2}(x-1)^2$. $\square$