Let $(M_t)$ be a nonnegative martingale in a probability space $(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, \mathbb{P} )$ given by \begin{equation} dM_t = M_t \sigma_t dW_t \end{equation} for some bounded, predictable process $\sigma$. For $0 \leq t \leq T$, let $\Sigma (t,T)$ be the unique $\mathcal{F}_t$-measurable non-negative random variable such that \begin{equation} \mathbb{E} [ \sqrt{M_T} | \mathcal{F}_t ] = \sqrt{M_t} \exp \Bigg\{\frac{- (\Sigma (t,T))^2 (T-t)}{8} \Bigg\} \quad\text{ a.s.}. \end{equation} Suppose that $a$ and $b$ are non-negative constants such that $a \leq \sigma_t \leq b$ a.s., for all $t \geq 0$. We want to show that $a \leq \Sigma(t,T) \leq b$ a.s., for all $0 \leq t < T$.

(The question asks us to first consider the case that $\sigma$ and $W$ are independent, and then extend to the general case by a change of measure, but I have no idea how to use this at all.)

What I have done so far:

I solved the SDE and obtain $M_t = M_0 \exp \big\{ \int_{0}^{t} \sigma_s \,dW_s - \int_{0}^{t} \sigma^2_s \,ds \big\}$, but this has no help in relating to the conditional expectation.

  • $\begingroup$ May I wonder, where the question is coming from? $\endgroup$
    – Ilya
    Nov 30 '14 at 19:05
  • $\begingroup$ @Ilya From some exercises in a probability book.... $\endgroup$
    – Richard
    Nov 30 '14 at 19:07
  • 1
    $\begingroup$ Well, that was clear - I was asking for the book's name as an exercise sounds interesting to me (unless you want to keep it in secret) $\endgroup$
    – Ilya
    Nov 30 '14 at 19:17
  • $\begingroup$ It is from one of those books written by Professor Chris Rogers (my lecturer). He didn't disclose which of his books is this question from. $\endgroup$
    – Richard
    Nov 30 '14 at 19:37
  • $\begingroup$ Are you sure about the $\frac{3}{8}$? If I'm not mistaken, it should read $\frac{1}{8}$. (Just consider the case where $\sigma_s$ is constant.) $\endgroup$
    – saz
    Dec 6 '14 at 13:17

It follows from Itô's formula that the solution to the SDE

$$dM_t = M_t \sigma_t dW_t$$


$$M_t = M_0 \exp \left( \int_0^t \sigma_s \, dW_s - \frac{1}{2} \int_0^t \sigma_s^2 \, ds \right).$$

By assumption, $M_0 \geq 0$. Hence,

$$\begin{align*} \sqrt{M_T} &= \sqrt{M_0} \exp \left( \frac{1}{2} \int_0^T \sigma_s \, dW_s - \frac{1}{4} \int_0^T \sigma_s^2 \, ds \right) \\ &= \underbrace{\sqrt{M_0} \exp \left( \int_0^T \frac{\sigma_s}{2} - \frac{1}{2} \int_0^t \left( \frac{\sigma_s}{2} \right)^2 \right)}_{=:N_T} \exp \left( - \frac{1}{8} \int_0^T \sigma_s^2 \, ds \right). \tag{1} \end{align*}$$

Note that $(N_t)_{t \geq 0}$ is a martingale and that $\mathbb{E}N_T=1$. Define a probability measure $\mathbb{Q}$ by $d\mathbb{Q} := N_T \, d\mathbb{P}$ (i.e. $\mathbb{Q}$ has density $N_T$ with respect to $\mathbb{P}$). Now we use the following

Lemma: Let $\beta \geq 0$ be a random variable, $\int \beta \, d\mathbb{P}=1$ and $d\mathbb{Q} := \beta \, d\mathbb{P}$. Then it holds for any $\sigma$-algebra $\mathcal{F}$: $$\mathbb{E}_{\mathbb{Q}}(X \mid \mathcal{F}) = \frac{\mathbb{E}(X \cdot \beta \mid \mathcal{F})}{\mathbb{E}(\beta \mid \mathcal{F})}. \tag{2}$$ (Here $\mathbb{E}_{\mathbb{Q}}( \cdot \mid \mathcal{F})$ denotes the conditional expectation with respect to the probability measure $\mathbb{Q}$.)

In our setting, $\beta = N_T$ and $X = \exp \left(- \frac{1}{8} \int_0^T \sigma_s^2 \, ds \right)$. Consequently, $$\begin{align*} \mathbb{E}(\sqrt{M_T} \mid \mathcal{F}_t) &\stackrel{(1)}{=} \mathbb{E}(X \cdot \beta \mid \mathcal{F}_t) \stackrel{(2)}{=} \mathbb{E}(\beta \mid \mathcal{F}_t) \cdot \mathbb{E}_{\mathbb{Q}}(X \mid \mathcal{F}). \tag{3} \end{align*}$$

Since $(N_t)_{t \geq 0}$ is a martingale (with respect to $\mathbb{P}$), we have

$$\begin{align*} \mathbb{E}(\beta \mid \mathcal{F}_t) &= \mathbb{E}(N_T \mid \mathcal{F}_t) = N_t = \sqrt{M_t} \cdot \exp \left( \frac{1}{8} \int_0^t \sigma_s^2 \, ds \right). \end{align*}$$

Plugging this into $(3)$ yields

$$ \mathbb{E}(\sqrt{M_T} \mid \mathcal{F}_t) = M_t \cdot \mathbb{E}_{\mathbb{Q}}\left[ \exp \left( - \frac{1}{8} \int_t^T \sigma_s^2 \, ds \right) \right].$$

Finally, using that $a \leq \sigma_s \leq b$, we get

$$\exp \left( - \frac{1}{8} b^2 (T-t) \right) \leq \mathbb{E}_{\mathbb{Q}}\left[ \exp \left( - \frac{1}{8} \int_t^T \sigma_s^2 \, ds \right) \right] \leq \exp \left(- \frac{1}{8} a^2 (T-t) \right)$$

and this proves the claim if we set

$$\Sigma(t,T) := \sqrt{- \frac{8}{T-t} \ln \left\{\mathbb{E}_{\mathbb{Q}}\left[ \exp \left( - \frac{1}{8} \int_t^T \sigma_s^2 \, ds \right) \right] \right\}}.$$

  • $\begingroup$ How would you check by Novikov's condition to ensure that $(N_t)$ is indeed a martingale. I mean, it is not mentioned as a condition that $\mathbb{E} ( \exp \{ \frac{1}{2} \int_{0}^{T} (\frac{\sigma_s}{2})^2 \,ds) < + \infty$. $\endgroup$
    – Richard
    Dec 6 '14 at 15:22
  • $\begingroup$ @Richard Note that $\sigma$ is bounded. $\endgroup$
    – saz
    Dec 6 '14 at 15:58
  • $\begingroup$ Oh, Yeah... Too many conditions in the question. It is often easy to neglect some. $\endgroup$
    – Richard
    Dec 6 '14 at 16:02
  • $\begingroup$ Can you please help me with some of my other questions?math.stackexchange.com/questions/1050319/… $\endgroup$
    – Richard
    Dec 6 '14 at 20:24
  • $\begingroup$ @Richard Perhaps tomorrow; too tired right now. $\endgroup$
    – saz
    Dec 6 '14 at 21:08

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