I have been toying around to get some understanding of what the stochastic discount factor look likes in Black-Scholes-Merton and how it relates to the exponential process in Girsanov's theorem. I find that the stochastic discount factor is the exponential process in Girsanov's Theorem discount at the risk-free rate, i.e. it scales Girsanov's exponential process by $\exp(-rt)$.

Does anyone have an intuition about this? I mean, the math should check out just fine, but I'm not sure if there is deeper meaning at play. Anyhow, I sketched out my work below.

Sketch of the work

Let's call $S_t$ the price of a stock, $B_t$ the price of a risk-free bond and $M_t$ the stochastic discount factor. We have the following dynamics: \begin{align} \frac{dS_t}{S_t} &= \mu dt + \sigma dZ_t \\ \frac{dB_t}{B_t} &= r dt \end{align} with $(Z_t)_{t \geq 0}$ a standard Brownian motion. If we apply Girsanov's theorem, we get a process of the following form for the change of measure: \begin{align} A_t &= \exp \left( -\int_0^t \eta_s dZ_s - \frac{1}{2}\int_0^t \eta_s^2 ds \right) \\ \forall t \; \eta_t = \eta \Rightarrow A_t &= \exp \left( -\eta Z_t - \frac{1}{2}\eta^2 t \right) \\ \Rightarrow dln A_t = ln A_t - ln A_0 &= -\frac{1}{2}\eta^2dt -\eta dZ_t \\ \Rightarrow \frac{dA_t}{A_t} &= -\eta dZ_t. \end{align} However, I know that $M_t B_t$ must be a martingale under the physical measure, hence $M_t$ must be a diffusion of the form $\frac{dM_t}{M_t} = -rdt + \phi(.) dZ_t$. Using the fact that $M_t S_t$ must also be a matringale, we get \begin{align} \frac{dM_tS_t}{M_tS_t} &= \frac{dS_t}{S_t} + \frac{dM_t}{M_t} + \frac{dS_t}{S_t}\frac{dM_t}{M_t}\\ &= \left(\mu dt + \sigma dZ_t \right) + \left(- r dt + \phi(.) dZ_t \right) + \left( \sigma \phi(.) dt \right) \\ \Rightarrow E^\mathbb{P} \left( \frac{dM_tS_t}{M_tS_t} \right) &= \left( \mu - r + \sigma \phi(.) \right)dt = 0 \\ \Leftrightarrow \mu - r + \sigma \phi(.) &= 0 \Leftrightarrow \phi(.) = - \frac{\mu -r}{\sigma}. \end{align} In this model, if we worked a bit, we could show that $\eta = \frac{\mu - r}{\sigma}$, hence \begin{align} \frac{dM_t}{M_t} &= -rdt + \frac{dA_t}{A_t} = -rdt - \eta dZ_t \\ \Rightarrow M_t &= M_0 \exp \left( -\int_0^t \eta dZ_s - \frac{1}{2} \int_0^t \eta^2 ds -rt \right) \\ &= M_0 A_t \exp(-rt). \end{align} Hence, the stochastic discount factor is just a scaled version of $A_t$ here: $M_t/M_0 = \exp(-rt) A_t/A_0$.


1 Answer 1



The stochastic discount factor (SDF) really has two jobs to do: it needs to incorporate the time value of money (discount) and take the riskiness of cash flows into account (stochastic). Thus, it can make sense to split the SDF into its two components, $$M_t=e^{-rt}A_t,$$ where $A_t$ does the risk compensation. Girsanov's theorem is concerned with $A_t$ only.

The change of measure technique (Girsanov theorem) relates the driftless martingale $A_t$ to a Radon-Nikodym derivative, $A_T=\frac{\text{d}\mathbb Q}{\text d\mathbb P}$. But because the SDF is not a martingale, the deterministic discount factor $e^{rt}$ essentially corrects that drift.

The disillusioned financial economist may simply say that $e^{-rt}$ is just a correction term that we need fiddle around with all the time. For example, remember Breeden and Litzenberg's (1978) formula, $$f_{S_t}(x)=e^{rT}\frac{\partial^2 C}{\partial K^2}\bigg|_{K=x}.$$ We simply need somewhere $e^{rT}$ to ensure that the risk-neutral drift of $S_t$ is $S_0e^{rt}$.

Girsanov Theorem and Black Scholes

Following Björk, let's $\varphi$ be the (constant) Girsanov kernel and set $\text{d}A_t=\varphi A_t\text{d}W_t^\mathbb P$ with $A_0=1$. Clearly, $A_t$ is a $\mathbb P$-martingale with $\mathbb{E}^\mathbb P[A_t]=1$. Girsanov tells us that when we set $\frac{\text d\mathbb Q}{\text d\mathbb P}=A_T$, then $W_t^\mathbb Q=W_t^\mathbb P-\varphi t$.

In the Black-Scholes world, $\text{d}S_t=\mu S_t\text{d}t+\sigma S_t\text{d}W_t^\mathbb P$. Using the Girsanov kernel, we get $\text{d}S_t=(\mu+\sigma\varphi) S_t\text{d}t+\sigma S_t\text{d}W_t^\mathbb Q$. This suggests that the (negative) Sharpe ratio (aka market price of risk) $\varphi=-\frac{\mu-r}{\sigma}$ would be a decent idea, that is $\text{d}S_t=rS_t\text{d}t+\sigma S_t\text{d}W_t^\mathbb Q$.

Links to Asset Pricing

As you say, $S_tM_t$ is a $\mathbb P$-martingale, i.e. $$S_t=\mathbb{E}_t^\mathbb P\left[\frac{M_T}{M_t}S_T\right],$$ which looks like our beloved Euler equation. Now, with our decomposition of $M_t=e^{-rt}A_t,$ we get

$$S_t= \mathbb{E}_t^\mathbb P\left[\frac{M_T}{M_t}S_T\right]= e^{-r(T-t)}\mathbb{E}_t^\mathbb P\left[\frac{A_T}{A_t}S_T\right]=e^{-r(T-t)}\mathbb{E}_t^\mathbb Q\left[S_T\right].$$

Because $\mathbb{E}^\mathbb P[A_t]=1$, we get from $M_t=e^{-rt}A_t$ that $$e^{rt}=\frac{1}{\mathbb{E}^\mathbb P[M_t]},$$ which reminds us of $R_f=\frac{1}{\mathbb{E}[m]}$ in discrete time.

Note, in this Black-Scholes world, everything is log-normally distributed: $S_t$, $M_t$, $A_t$, $M_tS_t$, ...

  • $\begingroup$ In your formulation, I think the risk-neutral drift should be $\mu + \sigma \phi$ since $\mu + \sigma \phi = \mu + r - \mu = r$. I'm also expecting $r < \mu$ and $\phi < 0$ would then require the positive sign in the drift. $\endgroup$
    – Stéphane
    Commented Dec 7, 2020 at 3:43
  • $\begingroup$ @Stéphane thanks for spotting the typo. You are, of course, absolutely correct $\endgroup$
    – Kevin
    Commented Dec 7, 2020 at 8:27

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