# A case where a stochastic exponential is a true martingale

Let $$(W^{(1)},W^{(2)})$$ be a two-dimensional standard Brownian motion and let $$dV_t = \kappa(\theta - V_t)dt+ \sigma \sqrt{V_t}dW^{(1)}_t$$ where $$\kappa, \theta$$ and $$\sigma$$ are constants such that $$2 \kappa \theta > \sigma^2.$$ Let $$\lambda$$ and $$a$$ be constants and define \begin{align*} L^{(1)}_t &:= \exp \bigg\{-\int_0^t \lambda \sqrt{V_u}dW_u^{(1)}-\frac{1}{2}\int_0^t(\lambda \sqrt{V_u})^2du \bigg\}; \\ L^{(2)}_t &:=\exp \Bigg\{-\int_0^t \frac{1}{\sqrt{1-\rho^2}}\bigg(\frac{\mu -r}{\sqrt{V_u}}-\lambda \rho \sqrt{V_u}\bigg)dW_u^{(2)} \\ &-\frac{1}{2}\int_0^t \bigg[\frac{1}{\sqrt{1-\rho^2}}\bigg(\frac{\mu -r}{\sqrt{V_u}}-\lambda \rho \sqrt{V_u}\bigg) \bigg]^2du \Bigg\}; \\ L_T &:= L^{(1)}_TL^{(2)}_T. \end{align*}

I have to prove that $$L$$ is a martingale. The process $$L$$ is clearly a positive local martingale with $$L_0=1,$$ so it's a supermartingale. Thus, we can prove that $$L$$ is a true martingale by showing that $$E[L_T]=1.$$ As a part of the proof, in this paper: https://www.hindawi.com/journals/ijsa/2006/018130/
(in page 5) they say that:

"as $$W^{(2)}$$ and $$V$$ are independent, and $$0 for every $$t \leq T$$ with probability 1, by conditional expectations we have $$E[L_T]=E[L_T^{(1)}] \tag*{(\star)}."$$ I don't know how to use conditional expectation and independence between $$V_t$$ and $$W_t^{(2)}$$ to prove $$(\star).$$ There are $$V_t$$ terms in $$L^{(1)}_T$$ and $$L^{(2)}_T$$ so they are not independent, I don't understand why $$L^{(2)}_T$$ goes away. Any ideas?

• Hi only some guesses I don't know if it's useful at all. They might be conditioning on the whole path $V_s$ for s in (0,t), so they could split the product but I'm not if this is licit. Then (still guessing) as $L_2$ looks to me like Radon-Nikodym term in the measure change of Girsanov, this might give $E[L_2(t)|(V_s)_{s \in(0,t)}]=1$ integrating over V would give (*). Commented Jan 29, 2020 at 20:56
• @TheBridge: Thank you for the suggestion. I'm not sure if I understand. Do you a know a book or paper where that idea is applied (even if it is in a different context)?
– UBM
Commented Jan 31, 2020 at 17:47

Based on @user159517 proof and @TheBridge suggestion.

First, an observation.

Observation 1. It's well known that if $$\eta_t$$ is a deterministic function, the stochastic exponential $$\{M_t; 0 \leq t \leq T\},$$ where $$M_t:= \exp \bigg\{-\int_0^t \eta_u dW_u-\frac{1}{2} \int_0^t \eta^2_u du \bigg\}$$ is a martingale and since $$E[M_0]=1$$ we must have $$E[M_T]=1.$$

Let $$\mathcal F_T^{W^{(1)}}$$ be the $$\sigma$$-algebra generated by $$W^{(1)}.$$ In order to prove ($$\star$$), the key point is to realize that $$E[L^{(2)}_T | \mathcal F_T^{W^{(1)}}]=1 \quad \text{a.s.} \tag{1}$$
Then \begin{align*} E[L_T]&=E[E[L_T|\mathcal F_T^{W^{(1)}}]] \\ &=E[E[L^{(1)}_TL^{(2)}_T|\mathcal F_T^{W^{(1)}}]] \\ &= E[L^{(1)}_T E[L^{(2)}_T|\mathcal F_T^{W^{(1)}}]] \\ &= E[L^{(1)}_T] \end{align*} and condition ($$\star$$) is proved.

Proof of (1): $$\quad$$ (from @user159517)

Let $$(\Omega_i,\mathcal F_i, \mathbb P_i),i=1,2$$ and take as a probability space $$\Omega := \Omega_1 \times \Omega_2.$$ Also, let $$\mathbb P = \mathbb P_1 \otimes \mathbb P_2$$ be the product measure on $$\Omega.$$ Note that because $$\mathcal F_T^{W^{(1)}}$$ is generated by $$W^{(1)}$$ which is not a function of $$\omega_2$$, the same holds for any $$\mathcal F_T^{W^{(1)}}$$-measurable random variable. As the coefficients in the SDE for $$V$$ are locally Lipschitz, by the Ito theorem we may take $$V$$ to be a strong solution, hence $$V$$ is $$\mathcal F_T^{W^{(1)}}$$-measurable and therefore a function only of $$\omega_1$$.

By the definition of conditional expectation, (1) is proved if we show that for all $$A \in \mathcal F_T^{W^{(1)}}$$ we have $$E[1_A E[L^{(2)}_T | \mathcal F_T^{W^{(1)}}]=E[1_A].$$ So let $$A \in \mathcal F_T^{W^{(1)}}.$$ Then

\begin{align}\int_{\Omega} E[L^{(2)}_T | \mathcal F_T^{W^{(1)}}]1_{A} ~d\mathbb{P} &= \int_{\Omega} L^{(2)}_T 1_{A} ~d\mathbb{P} = \int_{\Omega_1} 1_{A}(\omega_1)\left(\int_{\Omega_2}^{} L_{T}^{(2)}(\omega_1,\omega_2)d\mathbb{P}_2(\omega_2)\right)d\mathbb{P}_1(\omega_1) \\ &= \int_{\Omega_1} 1_{A}(\omega_1)~d\mathbb{P}_1(\omega_1) = \int_{\Omega} 1_{A}~d\mathbb{P}, \end{align}

where in the third equality above, we have used Observation 1 and the fact that $$V = V(\omega_1)$$ to conclude that $$\int_{\Omega_2}^{} L_{T}^{(2)}(\omega_1,\omega_2)d\mathbb{P}_2(\omega_2) = 1$$ for any $$\omega_1 \in \Omega_1$$.

• @user159517: Thank you, I edited the answer.
– UBM
Commented Feb 14, 2020 at 19:57
• very good, deleted my old answer. Commented Dec 20, 2020 at 11:34
• user159517 : Thank you!
– UBM
Commented Dec 20, 2020 at 12:18