Intuition for Stock Price Numeraire Drift

Question

I would like to ask whether there is an intuition for the drift of price processes under the Stock numeraire.

I find it intuitive that the martingale measure under the Money Market numeraire induces the drift "r" to all price processes (via the appropriate change of measure): with the money market compounding continuously at rate "r", all prices need to drift at this rate "r", otherwise the price processes discounted by the money market numeraire would not be martingales (i.e. any price process that wouldn't drift at "r" would give rise to arbitrage between Spot and Forwards, i.e. there would be miss-pricing of Forwards under the money market numeraire if the price process didn't drift at "r").

Same holds for the Discount bond numeraire under deterministic rates (because the Bond numeraire under deterministic rates turns out to be the money market numeraire scaled by a constant).

However, I haven't managed to build similar reasoning for the Stock price numeraire.

We know that the Stock price process under the Stock numeraire is:

\begin{align*} \frac{dS}{S} &= rdt + \sigma dW_t\\ &=\big(r+\sigma^2\big)dt + \sigma d \widehat{W}_t. \end{align*}

Above, $W_t$ is a standard Brownian motion under the risk-neutral measure associated with the Money market numeraire, whilst $\widehat{W_t}$ is a standard Brownian motion under the pricing measure associated with the Stock numeraire.

Why does the Stock numeraire induce the drift:

\begin{align*} &\big(r+\sigma^2\big) \end{align*}

Why would (intuitively) being able to borrow at the rate of the stock mean that price processes must have this drift?

Thank you so much,

Answer 1

As a general principle, I would be wary of economic or financial interpretations of change of measure techniques. Changing numéraires is merely a mathematical tool to ease pricing, see for example the last part of this answer. Nevertheless, here’s my take on your question.

Think of a numéraire as the basic financial asset of your economy, namely a store of value. In real life, you can put your money in a deposit account, or a money market account. Now, these are considered risk-free (or at least, we assume that), hence they only yield a risk-free rate $r$ with no return volatility.

Consider now an economy where your basic financial asset is a stock $S$: for example, when your employer pays your salary every month, instead of putting it into a deposit account, it buys shares for you. In a Black-Scholes setting, note that: $$\begin{align} V^S\left(\frac{dS_t}{S_t}\right)&=V^S\left(\sigma d\widehat{W}_t\right) \\ &=E^S\left(\sigma^2d[\widehat{W},\widehat{W}]_t\right) \\[3pt] &=\sigma^2dt \end{align}$$ Hence the variance of your return is $\sigma^2$ per infinitesimal unit of time. Thus if the stock is the basic store of value of your economy, it is understandable that economic agents would ask to be compensated for the risk they are taking and expect a higher return than a simple risk-free rate $r$.

Addendum: generally speaking, in an economy endowed with measure $P$, which is associated to a certain asset numéraire $N$, individual risk preferences are not accounted for through $N$ (contrary to the physical measure, whose numéraire directly represents aggregated risk preferences). We might interpret the risk of $N$ as the flat risk aversion of the economy $P$.

Answer 2

The drift is the expectation of the return over an infinitesimal interval. Let $Q$ be the risk-neutral measure and $Q^S$ be measure associated with the stock price numeraire defined by \begin{align*} \frac{dQ^S}{dQ}\big|_t = \frac{S_t}{B_t S_0}, \end{align*} where $B_t=e^{rt}$ is the value at time $t$ of the money-market account. Moreover, let $E$ and $E^S$ be expectation operators corresponding to measures $Q$ and $Q^S$. Then, \begin{align*} E\left(\frac{S_{t+\Delta t}-S_t}{S_t}\mid \mathscr{F}_t \right) &= E\left(e^{(r-\frac{1}{2}\sigma^2)\Delta t + \sigma(W_{t+\Delta t} -W_t)}-1\mid \mathscr{F}_t \right)\\ &=e^{r \Delta t} - 1 \approx r \Delta t. \end{align*} Similarly, \begin{align*} E^S\left(\frac{S_{t+\Delta t}-S_t}{S_t}\mid \mathscr{F}_t \right) &= E\left(\frac{dQ^S}{dQ}\big|_{t+\Delta t}\left( \frac{dQ^S}{dQ}\big|_{t}\right)^{-1}\frac{S_{t+\Delta t}-S_t}{S_t}\mid \mathscr{F}_t \right)\\ &=E\left(\frac{S_{t+\Delta t} B_t}{S_t B_{t+\Delta t}}\frac{S_{t+\Delta t}-S_t}{S_t}\mid \mathscr{F}_t \right)\\ &=E\left(\left(\frac{S_{t+\Delta t}}{S_t}\right)^2 e^{-r\Delta t} - \frac{S_{t+\Delta t}}{S_t} e^{-r\Delta t}\mid \mathscr{F}_t \right)\\ &=e^{(r+\sigma^2)\Delta t} -1 \approx (r+\sigma^2)\Delta t. \end{align*} That is, under the respective probability measure, the drift is the expectation of the return, over an infinitesimal interval.

Answer 3

I have a take on the intuition part of the question. Isn't it a simple consequence of Jensen's inequality? Thus, assuming $r=0$ for simplicity, we have in the money market measure: $E(S_T)=S_t$, but then $E(1/S_T)>1/S_t$ by Jensen since $1/x$ is convex. Now in the stock measure, we must force $E_S (1/S_T)=1/S_t$ to create the correct martingale, but then by "reverse Jensen" we must have $E_S(S_T)>S_t$. The amount by which the inequality exceeds equality is related to the standard deviation, intuitively.

Answer 4

When you try and discount everything by the stock, every price process now has extra gamma PnL naturally just due to the presence of the stochastic discount factor (due to it's quadratic variation). Every payoff now is a non linear function of this stock and thus after delta hedging earns this extra PnL. Simply applying Ito's lemma on the discounted payoff will result in an extra PnL term here.

Thus every portfolio now has an extra drift equal to the quadratic variation of the stock price. This PnL is a natural consequence of stochastic discounting, and can be looked at as compensation for the risk that stochastic discounting brings. This is accounted for in the drift.