# Illustrating the change of measure in Black-Scholes-Merton

Say that we have the following environment: \begin{align} dS_t &= \mu S_t dt + \sigma S_t dZ_t \\ dB_t &= r B_t dt \end{align} where $$S_t$$ is the price of a stock, $$B_t$$ is the price of a risk-free bond and $$\left( Z_t \right)_{t \geq 0}$$ is a standard Brownian motion. It's easy enough to get the densities for either $$S_t/S_0$$ or $$ln S_t - ln S_0$$ under both the physical and risk-neutral measures.

What I would like to know, though, is how I could tie those densities to the stochastic discount factor. I know that the SDF is going to take the form $$M_t = M_0 \exp \left( -r\int_0^t ds - \frac{\eta^2}{2}\int_0^t ds - \eta \int_0^t dZ_s \right)$$ where $$\eta = \frac{\mu - r}{\sigma}$$ is the Sharp ratio. Specifically, this makes $$M_tS_t$$ and $$M_tB_t$$ martingales under the physical measure.

I know that, intuitively, $$M_t$$ is going to increase the density for lower values of $$S_t/S_0$$ (and likewise for $$ln S_t - lnS_0$$). However, is there a way I can illustrate this?

I know that the ratio of risk-neutral to physical densities for $$S_t/S_0$$ is going to have an exponential appearance, but I'm not sure how to formally tie it back to some aspect of the model. Anyone knows how to visualize this?

• Hi Stéphane; You could draw the quotient of the two densities. If I remember correctly, the shift (slope?) would be driven by the risk aversion (MPOR aka Sharpe ratio) – Kermittfrog Dec 6 '20 at 5:48

## The model

In a Black-Scholes world, we have under $$\mathbb{P}$$ \begin{align*} \text{d}S_t &= \mu S_t\text{d}t+\sigma S_t\text{d}W_t^\mathbb P \hspace{1.7cm} \implies\mathbb{E}^\mathbb{P}[S_t]=S_0e^{\mu t}, \\ \text{d}M_t &= -r M_t\text{d}t+\varphi M_t\text{d}W_t^\mathbb P \hspace{1cm} \implies\mathbb{E}^\mathbb{P}[M_t]=e^{-r t}, \\ \text{d}M_tS_t &= (\varphi+\sigma) M_tS_t\text{d}W_t^\mathbb P \hspace{1.6cm} \implies\mathbb{E}^\mathbb{P}[M_tS_t]=S_0, \end{align*} where $$M_0=1$$, $$\text{d}B_t=rB_t\text{d}t$$ and $$\varphi=-\frac{\mu-r}{\sigma}$$ is the Girsanov kernel.

Under $$\mathbb{Q}$$, we have \begin{align*} \text{d}S_t &= \mu S_t\text{d}t+\sigma S_t\text{d}W_t^\mathbb Q \hspace{1.7cm} \implies\mathbb{E}^\mathbb{Q}[S_t]=S_0e^{r t}, \\ \text{d}\frac{S_t}{B_t} &= \sigma \frac{S_t}{B_t}\text{d}W_t^\mathbb Q \hspace{3.2cm} \implies\mathbb{E}^\mathbb{Q}\left[\frac{S_t}{B_t}\right]=S_0. \end{align*}

## The densities

In general, $$\text{d}X_t=mX_t\text{d}t+sX_t\text{d}Z_t$$ with $$X_0=x_0$$ gives rise to a process where $$X_t$$ is log-normally distributed for every $$t$$ with probability density function \begin{align*} f_{X_t}(x)=\frac{1}{\sqrt{2\pi}x\sqrt{s^2t}}\exp\left(-\frac{1}{2}\left(\frac{\ln(x/x_0)-\left(m-\frac{1}{2}s^2\right)t}{\sqrt{s^2t}}\right)^2\right). \end{align*}

## Plots

I plot below the density of the stock price $$S_t$$ (under $$\mathbb{P}$$ and $$\mathbb{Q}$$) and the density of the SDF $$M_t$$ (under $$\mathbb{P}$$ of course). I use $$T=1$$ (one year), $$\mu=0.12$$, $$r=0.01$$, $$\sigma=0.3$$ and $$S_0=1$$. Plotting the densities from $$S_T=0$$ up to $$S_T=4$$ is enough for the densities to numerically integrate to one.

As you see, the SDF puts the most weight on low values of $$S_T$$. That makes sense. The SDF is driven by marginal utility and marginal utility and risk aversion are high in bad states of nature. Similarly, the risk-neutral density of the stock price puts more weight on economically bad states of nature and reduces the likelihood of good events. Thereby, it reduces the expected future stock price from $$S_0e^{\mu t}$$ to $$S_0e^{rt}$$.

The full picture would be this

Computing the expectation corresponding these densities, $$\int_0^4 xf(x)\text{d}x$$, we indeed get $$S_0e^{\mu T}\approx1.1275$$ for the $$\mathbb{P}$$-density, $$e^{-rt}\approx0.9899$$ for the density of $$M_t$$, $$S_0e^{rT}\approx1.01$$ for the density of $$S_T$$ under $$\mathbb{Q}$$ and, of course, $$S_0=1$$ for the densities of $$M_TS_T$$ and $$\frac{S_T}{B_T}$$.