Let $\lambda$ be the market price of risk: $\frac{a - r}{\sigma}$, and define $Y(t) = e^{-\lambda W(t) - (r + \frac{\lambda^2}{2})t}$. Let $V^h(t)$ be the value process of any self-financing portfolio. The market is a standard Black Scholes.

Why is $Y(t)V^h(t)$ a martingale?

Evidently, $Y(t)$ solves the GMB $dY(t) = -rY(t)dt - \lambda Y(t)dW(t)$. Since $h$ is self-financing, we also obtain a differential for that process. Then we can apply the Ito formula using these two differentials, and obtain the differential of their product, yes? For it to be a martingale, the $dt$-term must disappear, yet, it does not disappear in my calculations. So are my calculations wrong, or is my method off?


Remember that a martingale is always defined with respect to some probability measure. Your confusion comes from the fact that you are not keeping track of the measures that are introduced.

$\lambda = \frac{a-r}{\sigma}$ indeed reflects some market price of risk, in the sense that, for the following diffusion model under the physical measure $\Bbb{P}$ \begin{align} \frac{dS_t}{S_t} &= a dt + \sigma dW_t^\Bbb{P} \\ &= (r + \lambda \sigma) dt + \sigma dW_t^\Bbb{P} \end{align} $\lambda$ can be seen as a risk-premium (excess return per units of volatility) for the risky asset $S$.

Suppose now that we were to define a measure $\Bbb{Q} \sim \Bbb{P}$ such that $S_t \sim GBM(r, \sigma)$ under $\Bbb{Q}$ (hence making the risk-premium disappear). This measure is defined through the following Random-Nikodym derivative: $$ \left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert_{\mathcal{F}_t} = \mathcal{E}[-\lambda W_t^\Bbb{P}] = e^{rt} Y_t $$ with your $$ Y_t = e^{-\lambda W_t^\Bbb{P}-(r+\frac{1}{2}\lambda^2)t} $$

Indeed, Girsanov tells us that $$ W_t^\Bbb{P} - \langle W^\Bbb{P} , -\lambda W^\Bbb{P} \rangle_t = W_t^\Bbb{P} + \lambda t$$ is a $\Bbb{Q}$ Brownian motion

On the other hand, the fundamental theorem of asset pricing tells you that, in the absence of arbitrage opportunity, the discounted value of any self-financing portfolio should be a martingale under the equivalent measure $\Bbb{Q}$, hence: $$ \Bbb{E}^\Bbb{Q}[e^{-rt} V_t^h \mid \mathcal{F}_s] = \Bbb{E}^\Bbb{Q}_s [ e^{-rt} V_t^h ] = e^{-rs} V_s^h $$

From Girsanov theorem for conditional expectations (sometimes called abstract Bayes formula) we have that, for an $(\mathcal{F}_t,\Bbb{Q})-$measurable process $X_t$ $$ \Bbb{E}^\Bbb{Q}_s [ X_t ] = \frac{ \Bbb{E}_s^\Bbb{P} \left[ X_t Z_t \right] }{ \Bbb{E}_s^\Bbb{P} [ Z_t ] } $$ where we have used the shorthand notation $$ Z_t = \left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert_{\mathcal{F}_t} $$ Applying this to $X_t = e^{-rt}V_t^h$ and the equivalent measures $\Bbb{P}$ and $\Bbb{Q}$ defined earlier (hence $Z_t = e^{rt} Y_t = \mathcal{E}[-\lambda W_t^\Bbb{P}]$ a $\Bbb{P}$-martingale) yields: $$ \Bbb{E}^\Bbb{Q}_s [ e^{-rt} V_t^h ] = \frac{ \Bbb{E}_s^\Bbb{P} \left[ V_t^h Y_t \right] }{ \Bbb{E}_s^\Bbb{P} [ e^{rt} Y_t ] } $$ hence isolating the expectation on the RHS and using the fact that $e^{-rt}V_t^h$ is a $\Bbb{Q}$-martingale, $$ \Bbb{E}_s^\Bbb{P} \left[ V_t^h Y_t \right] = \underbrace{\left(e^{rs} Y_s\right)}_{Z_t \text{ is a } \Bbb{P}-\text{martingale }} \underbrace{e^{-rs} V_s^h}_{e^{-rt}V_t^h \text{ is a } \Bbb{Q}-\text{martingale }} = Y_s V_s^h $$ and $V_t^h Y_t$ is indeed a $\Bbb{P}$-martingale (but not a $\Bbb{Q}$-martingale).

[Remark] Of course you should reach the same result with your approach consisting in using the martingale representation theorem (i.e. calculating the differential of the product $Y_t V_t^h$ and making sure that the resulting drift is zero). The thing is that both processes should be expressed under the same measure for it to work, here $\Bbb{P}$. In your post, you actually mix Brownians defined under $\Bbb{P}$ and Brownians defined under $\Bbb{Q}$.


Alternative to @Quantuple's answer, we can also proceed as follows.

We assume that, under the assumed probability measure (e.g., the real world measure), the underlying stock process $\{S_t, t \ge 0\}$ satisfy an SDE of the form \begin{align*} dS_t = S_t (adt + \sigma dW_t), \end{align*} where $\{W_t, t\ge 0\}$ is a standard Brownian motion. Moreover, let $B_t=e^{rt}$ be the money-market account value at time $t$. We assume that the value process $\{V_t^h, t \ge 0\}$ of the self-financing portfolio is of the form \begin{align*} V_t^h = \alpha_t B_t + \beta_t S_t, \end{align*} where $\alpha_t$ and $\beta_t$ are units of the money-market account and the underlying stock.

Note that, \begin{align*} dV_t^h &= \alpha_t dB_t + \beta_t dS_t\\ &=r\alpha_t B_t dt + a\beta_t S_tdt + \sigma \beta_t S_t dW_t. \end{align*} From the assumption, \begin{align*} dY_t = -rY_tdt - \lambda Y_t dW_t. \end{align*} Then \begin{align*} &\ d\left(Y_tV_t^h\right) \\ =&\ Y_t dV_t^h + V_t^h dY_t + d\langle Y, V^h\rangle_t\\ =&\ Y_t\left( r\alpha_t B_t dt + a\beta_t S_tdt + \sigma \beta_t S_t dW_t\right)+V_t^h\left( -rY_tdt - \lambda Y_t dW_t \right) -\lambda \sigma\beta_tY_t S_t dt\\ =&\ Y_t\left[\left(r\alpha_t B_t+a\beta_t S_t-rV_t^h- \lambda \sigma\beta_tS_t\right)dt + \left(\sigma \beta_t S_t -\lambda V_t^h \right)dW_t \right]\\ =&\ Y_t\left[\left(r\alpha_t B_t+r\beta_t S_t-rV_t^h\right)dt + \left(\sigma \beta_t S_t -\lambda V_t^h \right)dW_t \right]\\ =&\ Y_t\left(\sigma \beta_t S_t -\lambda V_t^h \right)dW_t. \end{align*} That is, $\{Y_tV_t^h, t \ge0\}$ is a martingale.

  • $\begingroup$ This is indeed more straight to the point and hopefully clearer for the OP. $\endgroup$ – Quantuple Jan 20 '17 at 8:17

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