When Stock Prices and Their Powers Can Serve as Numeraires

# Why is a unit of stock worth $S_T$ a valid numeraire? Is $S_T^2$ a valid numeraire?


# Why is a unit of stock worth $S_T$ a valid numeraire? Is $S_T^2$ a valid numeraire?












When we switch numeraires from $M$ to $S$, we change the measure from $\mathbb{Q}^M$ to $\mathbb{Q}^S$. I derived a closed-form formula for the payoff $S_T(S_T - K)\mathbf{1}_{\{S_T > K\}}$ here. How do I quickly test if $f(S_T)$ qualifies as a numeraire? Intuitively, the discounted values $f(S_T)/M(T)$ must be martingales under $\mathbb{Q}^M$. Are my explanations on point?

- How can I argue that a unit of stock (non-dividends) $S_T$ is indeed a valid numeraire?

The random variable

$$ \begin{align*} Z_T = \frac{S_T/S_0}{M_T/M_0}=\exp\left(-\frac{\sigma^2}{2}T + \sigma W^{\mathbb{Q}^M}_T\right) \end{align*} $$

which is a $\mathbb{Q}^M$ martingale, with $\mathbb{E}^{\mathbb{Q}^M}[Z] = 1$. So, I can construct $\mathbb{Q}^S(A) = \mathbb{E}^{\mathbb{Q}^M}[Z1_A]$, it is a valid measure.

- How can I argue that a dividend-paying stock fails to qualify as a numeraire?

The random variable

$$ \begin{align*} Z_T = \frac{S_T^{div}/S_0^{div}}{M_T/M_0}=\exp\left(\left(-q -\frac{\sigma^2}{2}\right)T + \sigma W^{\mathbb{Q}^M}_T\right) \end{align*} $$

which not a $\mathbb{Q}^M$ martingale, with $\mathbb{E}^{\mathbb{Q}^M}[Z] = e^{-qT}$. So, we can't construct $\mathbb{Q}^{S^{div}}$.

- Are powers of the stock price $S_T^{\alpha}$, $\alpha > 1$ valid numeraires?

## Answer by Andrea (score 8, accepted)

https://quant.stackexchange.com/a/81292

You say it correctly: if $N_t$ is the old numeraire, any asset $M$ such that $\frac{M_t}{N_t}$ is an N-martingale is a valid numeraire (as long as it is positive).

This means that a numeraire must be a tradable-asset and $S_t^2$ is not (it has convexity and you can price an option on it, but $S_t^2$ is never the price process of a tradable asset).

You can see why the restriction: take a simple model with a bank account $B_t$ (deterministic rates = 0) and a non dividend paying stock.

When you change measure, the drift of $S_t$ changes, an only that. Both assets ($B_t$ and $S_t$) must stay martingales when discounted by any numeraire.

- If $S_t$ is a numeraire, this means that $\frac{B_t}{S_t}$ and $\frac{S_t}{S_t}=1$ must be martingales. And it is possible since this is really only 1 constraint (the one for $B_t$).

- If you allowed $S_t^2$ to be a numeraire, then you could not make both $\frac{B_t}{S_t^2}$ and $\frac{S_t}{S^2_t}=\frac{1}{S_t}$ martingales at the same time by changing only one drift.

- If the stock pays divided, the reinvested stock is a valid numeraire.

## Answer by Frido (score 5)

https://quant.stackexchange.com/a/81297

Too long for a comment:

Just to add to Andrea's answer, if $X_T$ is any state variable (not necessarily tradable), and $F(X_T) > 0$, then you can use $E_t [ F(X_T) ]$ as a numéraire and clearly $F(X_T) = E_T^\mathbb Q [ F(X_T) ]$. In your example $S_T^2 = E_T^\mathbb Q [ S_T^2]$.

One of my favourite and not so well-known application of this is that in a stochastic volatility model setting with $\rho = 0$ between the spot price and the instantaneous volatility, $dS_t = \sigma_t S_t dW_t$, the volswap price is given by $$ E_t^\mathbb Q [\sigma_{t,T} ] = E_t^\mathbb H [ (\log S_T)_+ ] = E_t^\mathbb Q \left[ \frac{ \sqrt{S_T} }{E_t[ \sqrt{S_T} ]} (\log S_T)_+ \right] $$ where $\sigma^2_{t,T} := \frac{1}{T-}\int_t^T \sigma^2_u du$ and the "half-measure" $\mathbb H$ is the measure under which $S_t/E_t [ \sqrt{S_T} ]$ is a martingale.

To see this you can first assume $\sigma$ is deterministic to get a feel for the calculations, and then in the stoch vol case for $\rho = 0$ apply conditioning.

(I've taken $r=0$ but principle remains the same for $r\neq 0$.)