Claim the collection $\left\{\mathcal{E}(aL)_{T}\right\}$ ($T$ stopping time) is uniformly integrable in the proof of Novikov's Condition

Question

Here I have a question in the proof of Novikov's condition from Le Gall's Book Brownian Motion and Stochastic Calculus Page 137:

The proof is very long, and I only have problem in the following part:

Now we have the following :

For continuous local matingale $L_{0}=0$ , $0<a<1$, $\mathcal{E}(aL)=\exp\left(aL_{t}-a^{2}\frac{1}{2}\left\langle L,L\right\rangle_{t}\right)$

We have known that it is a continuous local martingale, By Ito's formula. My goal is to prove $\mathcal{E}(aL)_{T}$, where $T$ is stopping time, is uniformly integrable. One important inequality in the proof is:

$$E\left[1_{\Gamma}\mathcal{E}(aL)_{T}\right]\leqslant E\left[1_{\Gamma}\exp\left(\frac{1}{2}L_{T}\right)\right]^{2a(1-a)}$$ For $\Gamma:=\left\{\mathcal{E}(aL)_{T}>x,x>0\right\}$

I have known that $\exp\left(\frac{1}{2}L_{T}\right)$ is uniformly integrable, then how can I derive that $\mathcal{E}(aL)_{T}$ is also uniformly integrable? (Le Gall's book has omitted this part.)

Answer 1

Let $E_R:=\left\{ \exp \left(L_T/2\right) \gt R\right\}$ where $R$ is fixed. Then $$ \mathbb E\left[1_{\Gamma}\exp\left(\frac{1}{2}L_{T}\right)\right]= \mathbb E\left[1_{\Gamma}1_{E_R}\exp\left(\frac{1}{2}L_{T}\right)\right] +\mathbb E\left[1_{\Gamma}1_{E_R^c}\exp\left(\frac{1}{2}L_{T}\right)\right]\\ \leqslant \mathbb E\left[1_{E_R}\exp\left(\frac{1}{2}L_{T}\right)\right] +R\mathbb P\left(\Gamma\right). $$ Let us fix $\varepsilon\gt 0$. Since the family $\left\{\exp\left(\frac{1}{2}L_{T}\right), T\mbox{ is a stopping time}\right\}$ is uniformly integrable, we can choose $R_\varepsilon$ such that $\mathbb E\left[1_{E_R}\exp\left(\frac{1}{2}L_{T}\right)\right]\lt \varepsilon$. Moreover, we have $$ \mathbb E\left[ \mathcal{E}(aL)_{T}\right]\leqslant \mathbb E\left[\exp\left(\frac{1}{2}L_{T}\right)\right]^{2a(1-a)} $$ by letting $x$ going to $0$ hence $S:=\sup_T\mathbb E\left[ \mathcal{E}(aL)_{T}\right]$ is finite. In particular, by Markov's inequality, $P\left(\Gamma\right)\leqslant S/x$.

We thus have $$ \sup_T\mathbb E\left[1_{\Gamma}\mathcal{E}(aL)_{T}\right]\leqslant \varepsilon^{2a(1-a)}+\left(\frac{R_\varepsilon S}x\right)^{2a(1-a)} $$ hence $$ \lim_{x\to +\infty}\sup_T\mathbb E\left[1_{\Gamma}\mathcal{E}(aL)_{T}\right]\leqslant \varepsilon^{2a(1-a)}. $$

Answer 2

The question here is just to verify the definition of uniform integrability:

By the uniform integrability of $\exp\left(\frac{1}{2}L_{T}\right)$, for each $\epsilon>0$, there exists a $\delta>0$, such that for $\mathbb{P}(\Gamma)<\delta$, $E\left[1_{\Gamma}\exp\left(\frac{1}{2}L_{T}\right)\right]<\epsilon$. Considering the the inequality given in the question, we can also derive that $E\left[1_{\Gamma}\mathcal{E}(aL)_{T}\right]\leqslant\epsilon^{2a(1-a)}$ for such $\Gamma$. Then notice that $\epsilon$ is arbitrarily chosen, then the uniformly integrability of $E\left[1_{\Gamma}\mathcal{E}(aL)_{T}\right]$ has been verified.

Also, notice that $\Gamma$ can be chosen arbitrarily in $\mathcal{F}$, and the uniformly integrability only verifies certain classes of $\Gamma$.