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,

  • $\begingroup$ Hi, don't you mean (r +/-0.5 * s^2) for the stock drift? That is a classic convention in most stockmarket processes; and is easily explicable on account of the arithmetic-vs-geometric mean differential. The +/- depending on whether your "r" is log or linear. The lack of the same vol dimension only happens in bonds, because you specified "deterministic rates", that don't really exist for bonds; and nobody has ever pretended exist for stocks. $\endgroup$ – demully Jan 2 '20 at 2:01
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    $\begingroup$ Hello, nope, I don't mean (r -0.5 * s^2). The term (-0.5 * s^2) appears in the solution of the SDE under the risk neutral or real world measure. The term comes from Ito calculus and as you point out, is just "the way that Ito mathematics works" (and there is also intuition in it when you think of Normal vs. Lognormal). I mean that when you start using Stock as the Numeraire (instead of the Money market numeraire, for example), the Radon-Nikodym derivative that you need to use to create a martingale pricing measure induces the drift (r+σ^2). I would like to understand the intuition behind. $\endgroup$ – Jan Stuller Jan 2 '20 at 10:06

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 $r$.

  • $\begingroup$ Beautiful answer, thank you! Ps: I strongly believe that it is important to link mathematics and intuition. The mathematical tools we use to price derivatives must satisfy common sense principles such as no arbitrage. It is a bit of a chicken and egg argument: what comes first, the model or the price? Usually, we calibrate model to a set of existing prices (and then reuse it to price other non-quoted derivatives). Therefore the model is a consequence of market dynamics and we must be able to interlink the dynamics and the model via intuition (otherwise your hedging might fail, etc). $\endgroup$ – Jan Stuller Jan 2 '20 at 14:36
  • $\begingroup$ @JanStuller You are welcome. $\endgroup$ – Daneel Olivaw Jan 2 '20 at 14:58
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    $\begingroup$ @DaneelOlivaw Nice answer. $\endgroup$ – ilovevolatility Jan 4 '20 at 14:13

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.

  • $\begingroup$ Hi @Gordon, can it also be shown that the drift of the money market account in the stock measure is $r+\sigma^2?$ $\endgroup$ – dm63 Jan 4 '20 at 2:37
  • $\begingroup$ @dm63: The money market account does not have the diffusion part, and then the drift will not change. However, $B_t/S_t$ is a martingale. $\endgroup$ – Gordon Jan 4 '20 at 13:46
  • $\begingroup$ thanks, I had intuitively thought that if A/B is a martingale then A and B must have same drift, but it is not true. $\endgroup$ – dm63 Jan 4 '20 at 14:58
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    $\begingroup$ @dm63: that is right: there is an extra quadratic term from the denominator $B$. $\endgroup$ – Gordon Jan 4 '20 at 15:49

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.

  • $\begingroup$ I am not sure the reasoning is fully correct, letting $P_{t,T}$ be a zero coupon bond with maturity $T$, under the stock measure we must have $E_S(1/S_T)=P_{t,T}/S_t$. Otherwise you are saying there exists an asset with constant value $1$ which would entail arbitrage opportunities using the money market account if $r\not=0$ (i.e. if $r>0$ sell the constant asset and put the money in the MMA). $\endgroup$ – Daneel Olivaw Jan 4 '20 at 15:59
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    $\begingroup$ I assumed $r=0$ in my answer. However it should hold for constant $ r>0$. Just replace $S_t$ by $S_t e^{(R(T-t))}$ everywhere. Then $P(t,T)= e^{-r(T-t)} $ satisfying your condition. $\endgroup$ – dm63 Jan 4 '20 at 16:14
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    $\begingroup$ Ok, good point. $\endgroup$ – Daneel Olivaw Jan 4 '20 at 17:03

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