Show local martingale

Question

I have $\exp(\lambda X_t-\frac{\lambda ^2}{2}t)$ is a local martingale, now i have to know if $X_t$ is also a local martingale.

Can anybody help me how i can show this correctly?

Answer 1

Solution 1: Recall the following two statements.

Lemma 1: Let $(Y_t)_{t \geq 0}$ be a supermartingale and $f$ an increasing concave function, then $(f(Y_t))_{t \geq 0}$ is a supermartingale.

Lemma 2: Let $(Y_t)_{t \geq 0}$ be a locale martingale such that $Y_t \geq 0$ for all $t \geq 0$. Then $(Y_t)_{t \geq 0}$ is a supermartingale.

Lemma 1 is a direct consequence of Jensen's inequality and Lemma 2 follows from Fatou's lemma. Since

$$M_t^{\lambda} := \exp \left( \lambda X_t - \frac{\lambda^2}{2} t \right)$$

is, by assumption, a non-negative local martingale, we find from Lemma 2 that $(M_t^{\lambda})_{t \geq 0}$ is a supermartingale. Applying Lemma 1 with $f(x) := \log x$, $x>0$, we obtain that

$$f(M_t^{\lambda}) = \lambda X_t - \frac{\lambda^2}{2} t \tag{1}$$

is a supermartingale. If $\lambda>0$, this implies that $(X_t-|\lambda| t/2)_{t \geq 0}$ is a supermartingale. Hence,

$$\begin{align*} \mathbb{E}(X_t \mid \mathcal{F}_s) &= \lim_{\lambda \downarrow 0} \mathbb{E}(X_t- |\lambda|t/2 \mid \mathcal{F}_s) \\ &\stackrel{\ast}{\leq} \lim_{\lambda \downarrow 0} (X_s-|\lambda| s/2) = X_s \tag{2} \end{align*}$$

for any $s \leq t$ where we have used in $(\ast)$ the supermartingale property of $(X_t-|\lambda|t/2)_{t \geq 0}$. This shows that $(X_t)_{t \geq 0}$ is a supermartingale. On the other hand, for $\lambda<0$, it follows from $(1)$ that $(X_t-|\lambda| t/2)_{t \geq 0}$ is a submartingale. Using the same reasoning as in $(2)$ [now we let $\lambda \uparrow 0$ and use the submartingale property instead of the supermartingale property], we find that $(X_t)_{t \geq 0}$ is a submartingale. Consequently, $(X_t)_{t \geq 0}$ is a martingale.

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Solution 2: This solution requires that the localizing sequences of stopping times can be chosen independently of $\lambda$, i.e. there exists a sequence of stopping times $(\sigma_n)_{n \in \mathbb{N}}$ such that

$$M_{t \wedge \sigma_n}^{\lambda} = \exp \left(\lambda X_{t \wedge \sigma_n}- \frac{\lambda^2}{2} (t \wedge \sigma_n) \right) \tag{3}$$

is a martingale for all $n \in \mathbb{N}$ and $\lambda \in (-1,1)$. Moreover, we assume that $(X_t)_{t \geq 0}$ has continuous sample paths.

Without loss of generality, we may assume that $|X_{t \wedge \sigma_n}| \leq n$; otherwise we replace $\sigma_n$ by $\sigma_n \wedge (\inf\{ t>0; |X_t|>n\})$ (which is, by the optional stopping theorem, still a localizing sequence). The martingale property of $(M_{t \wedge \sigma_n}^{\lambda})_{t \geq 0}$ is equivalent to

$$\int_F M_{s \wedge \sigma_n}^{\lambda} \, d\mathbb{P} = \int_F M_{t \wedge \sigma_n}^{\lambda} \, d\mathbb{P}$$

for all $F \in \mathcal{F}_s$ and $s \leq t$. Differentiating with respect to $\lambda$, we find that

$$\int_F (X_{t \wedge \sigma_n}-\lambda t \wedge \sigma_n) M_{t \wedge \sigma_n} \, d\mathbb{P} = \int_F (X_{s \wedge \sigma_n}-\lambda s \wedge \sigma_n) M_{s \wedge \sigma_n} \, d\mathbb{P}.$$

(Here, we used the definition of $M_t^{\lambda}$, see $(3)$, and that we may interchange differentiation and integration because $|X_{t \wedge \sigma_n})| \leq n$.) Evaluating at $\lambda = 0$ shows that

$$\int _F X_{t \wedge \sigma_n} \, d\mathbb{P} = \int_F X_{s \wedge \sigma_n} \, d\mathbb{P}.$$

Since $s \leq t$ and $F \in \mathcal{F}_s$ are arbitrary, this is equivalent to saying that $(X_{t \wedge \sigma_n})_{n \in \mathbb{N}}$ is a martingale. Hence, $(X_t)_{t \geq 0}$ is a local martingale.