Short story: the process for Stock price squared is not a martingale when discounted by the money-market numeraire under the risk-neutral measure. How can we then compute derivative prices on $S_t^2$ under the risk neutral measure? Wouldn't this lead to arbitrage?

Long story: I found some great posts on power options, for example Finding price of the power option. Whilst the maths is clear, I am still somewhat confused about the concept: starting with a simple option on Stock price squared, I do not fully comprehend how an optional claim can be priced within the regular B-S framework, when the price process for $S_t^2=S_0^2exp((2r-\sigma^2)t+2\sigma W_t)$ is not a martingale when discounted by $e^{rt}$ under the risk-neutral money-market numeraire.

I consider a single period model with zero rates. As outlined in the post What is the Risk Neutral Measure?, in the one-period model, the risk-neutral measure arises from no arbitrage assumption in the model. We assume that initially, the stock price is $S_0$ and after one period it can be either $S_u=S_0*u$ or $S_d=S_0*d$, with $u$ and $d$ being some multiplicative factors. Pricing a derivative claim with pay-off function $V(.)$ on the underlying stock $S_t$ via replication gives rise to:

$$V_0 = \left(V(S_u) \left( \frac{1 -d}{u-d} \right) + V(S_d) \left(\frac{u-1}{u-d} \right) \right)$$

Imposing $u \leq 1 \leq d$ will ensure that there is no arbitrage in the one-period model. Furthermore, as a consequence of the condition $u \leq 1 \leq d$, we get that $0 \leq \frac{1 -d}{u-d} \leq 1$ and $0 \leq \frac{u-1}{u-d} \leq 1$. Therefore, we can define $p_u:=\frac{1 -d}{u-d}$, $p_d:=\frac{u-1}{u-d}$ and we can call $p_u$ and $p_d$ "probabilities": indeed, in the one-period model, $p_u$ & $p_d$ form the discrete (risk-neutral) probability measure.

Now, the interesting point is that pricing the claim $V(.)$ on $S_t^2$ via replication in the one-period model actually leads to a different probability measure:

(i) Upper state: $S_{t_1}^2=S_0^2u^2$, denoting risk-free bond as $B$ we have $B_{t_1}=B_{t_0}=1$ since rates are zero and the option pay-off is $V_u=V((S_0u)^2)=[S_0^2u^2-K]^+$.

(ii) Lower state: $S_{t_1}^2=S_0^2d^2$, $B_{t_1}=B_{t_0}=1$, $V_d=V((S_0d)^2)=[S_0^2d^2-K]^+$.

Trying to replicate the payoff $V(S_{t_1}^2)$ in both states via the underlying stock and the risk-free bond, we get two equations with two unknowns (x = number of stocks, y = number of bonds I wanna hold to replicate option pay-off):

$$(i) V_u=x*S_0^2u^2+y*1$$

$$(ii) V_d=x*S_0^2d^2+y*1$$

Solving the system of equations yields:

$$ x=\frac{V_u-V_d}{S_0^2(u^2-d^2)}, y=\frac{u^2V_d-d^2V_u}{u^2-d^2}$$

Which then gives the claim price as (after some basic algebraic simplifications):


Setting $p_u^*:=\frac{1-d^2}{u^2-d^2}$ and $p_d^*:=\frac{u^2-1}{u^2-d^2}$, the above can be re-written as:


In other words, the replication argument gives rise to some new probability measure where $p_u^*=\frac{1-d^2}{u^2-d^2}\neq p_u=\frac{1-d}{u-d}$ and $p_d^*=\frac{u^2-1}{u^2-d^2}\neq p_d=\frac{1-d}{u-d}$.

Instead, we actually have that $p_u^*=p_u \frac{1+d}{u+d}$ and $p_d^*=p_d \frac{1+u}{u+d}$.

Question: So going back to the the start and considering the thread Finding price of the power option, how come we can price power options under the B-S classical risk-neutral measure? That would be equivalent to saying that under the one period model (with rates being zero), the price of the claim $V(S^2_t)$ could be computed as $V_0=\mathbb{E}^Q[V_t(S_t^2)]=p_uV_u(S_t^2) + p_dV_d(S_t^2)$, which does not produce the correct result (indeed, above we instead get that $V_0=\mathbb{E}^{Q_2}[V_t(S_t^2)]=p_u^*V_u(S_t^2) + p_d^*V_d(S_t^2)$).

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    $\begingroup$ Let $\mathcal{F}_t$ be the natural filtration of your stock price $S_t$ and $\xi$ be any $\mathcal{F}_T$-measurable and integrable random variable. For simplicity, think of $\xi=f(S_T)$ as a function of the terminal stock price (option payoff). You can simply use the pricing formula $V_t=B_t\mathbb{E}^\mathbb{Q}[\frac{\xi}{B_T}|\mathcal{F}_t]$, where $B_t=\exp\left(\int_0^t r_s\mathrm{d}s\right)$ is a bank account. This price process gives rise to an $\mathbb Q$ martingale and hence thus not introduce arbitrage in the market! So, $\xi$ (or $S_t^2$) does not need to correspond to a martingale. $\endgroup$
    – Kevin
    Commented Jul 26, 2020 at 17:38
  • $\begingroup$ Thank you, @KeSchn. What about the at-the-money forward example I illustrate in the question? It does not appear to be a Q-martingale when discounted by the bank account. $\endgroup$ Commented Jul 26, 2020 at 17:44
  • $\begingroup$ The discounted value process for the squared stock price, $\frac{V_t}{B_t}=\mathbb E^\mathbb Q\left[\frac{S_T^2}{B_T}\bigg| F_t\right]$, is a $\mathbb Q$-martingale by construction because, in general, $M_t=\mathbb{E}[X\mid F_t]$ is a martingale by the tower law: $\mathbb E[M_{t+1}\mid F_t]=\mathbb E\big[\mathbb E[X\mid F_{t+1}]\mid F_t\big]=\mathbb E[X\mid F_t]=M_t$. $\endgroup$
    – Kevin
    Commented Jul 26, 2020 at 17:52
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    $\begingroup$ The process $\frac{S_t^2}{B_t}$ is not a $\mathbb Q$-martingale. The discounted value process $\frac{V_t}{B_t}=\frac{B_t\mathbb{E}^\mathbb{Q}[\frac{S^T}{B_T}|F_t]}{B_t}=\mathbb{E}^\mathbb{Q}[\frac{S^T}{B_T}|F_t]$ is though. The price process, $V_t=B_t\mathbb{E}^\mathbb{Q}[\frac{S_T}{B_T}|F_t]$, includes a conditional expectation itself. Because $\frac{V_t}{B_t}$ is a $\mathbb Q$ martingale, we can use standard risk-neutral pricing. The tower law merely states that $E[E[X|F]|G]=E[X|F]$ is $F$ is a sub-sigma-algebra of $G$ (the smallest $\sigma$-algebra (representing less information) wins). $\endgroup$
    – Kevin
    Commented Jul 26, 2020 at 18:24
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    $\begingroup$ A forward that pays $S_T^2$? Well, consider $V_t=B_t\mathbb{E}^\mathbb{Q}[\frac{S_T^2}{B_T}\mid\mathcal{F}_t]$. Suppose $B_t=e^{rt}$ and $S_t$ is a geometric Brownian motion. Then, $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[S_T^2|\mathcal{F}_t]=e^{-r(T-t)}S_0^2e^{2\left(r-\frac{1}{2}\sigma^2\right)T}\mathbb{E}^\mathbb{Q}[e^{2\sigma W_T}|\mathcal{F}_t]=e^{-r(T-t)}S_0^2e^{2\left(r-\frac{1}{2}\sigma^2\right)T}e^{2\sigma^2(T-t)+2\sigma W_t}=e^{-r(T-t)}S_t^2e^{2\left(r-\frac{1}{2}\sigma^2\right)(T-t)}e^{2\sigma^2(T-t)}=S_t^2e^{\left(r+\sigma^2\right)(T-t)}$. More elegant approach: numéraire change ;) $\endgroup$
    – Kevin
    Commented Jul 26, 2020 at 18:50

1 Answer 1


Consider a financial market with a filtered probability space $\left(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb P\right)$ satisfying usual conditions equipped with a stock price process $S_t$. Suppose there exists a risk-free asset who is governed by $\mathrm{d}B_t=r_tB_t\mathrm{d}t$.

Suppose the market is free of arbitrage, i.e. there exists a probability measure $\mathbb Q\sim\mathbb P$ such that $$ \mathbb{E}^\mathbb{Q}\left[\frac{S_{t}}{B_{t}}\Bigg|\mathcal{F}_s\right]=\frac{S_s}{B_s}$$ for $s\leq t$.

Let $\xi$ be an integrable and $\mathcal{F}_T$-measurable random variable representing the time-$T$ payoff of some claim (contract). It's typically a function of the terminal stock price $S_T$. What's the fair (i.e. arbitrage-free) time-$t$ price for such a claim? Let's study the following value (or price) process $$V_t=B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi}{B_T}\bigg|\mathcal{F}_t\right].$$ So intuitively, the fair price of a contract $\xi$ equals the expected discounted payoff expressed in terms of the numéraire ($B_t$). Of course, $V_T=\xi$, thus $V_t$ replicates the payoff $\xi$. If interest rates are deterministic, we can pull $B_T$ out of the expectation.

The discounted value process, $\frac{V_t}{B_t}$, is a $\mathbb{Q}$-martingale by construction. It follows immediately from the tower law. We simply note that $$\mathbb{E}^\mathbb{Q}\left[\frac{V_t}{B_t}\bigg|\mathcal{F}_s\right]=\mathbb{E}^\mathbb{Q}\left[\mathbb{E}^\mathbb{Q}\left[\frac{\xi}{B_T}\bigg|\mathcal{F}_t\right]\bigg|\mathcal{F}_s\right]=\mathbb{E}^\mathbb{Q}\left[\frac{\xi}{B_T}\bigg|\mathcal{F}_s\right]=\frac{V_s}{B_s}.$$ In general, if $X$ is an integrable random variable, then $M_t=\mathbb{E}[X|\mathcal{F}_t]$ is a martingale.

Now the catch: the augmented market (with stock, bond and value process) consists of assets whose discounted values are martingales. Thus, by using the first FTAP, we know the market remains free of arbitrage and $V_t$ is a way of replicating $\xi$ without creating an arbitrage opportunity.

If there exists a perfect hedge for $\xi$ (which is self-financing), then $V_t$ has the same price as this perfect hedge for every time point $t\leq T$ (law of one price). Thus, the value process is in fact independent of the equivalent martingale measure (if several exist) for replicable payoffs (this, by the way, points to the second FTAP). In fact, for every equivalent martingale measure $\mathbb Q$, the map $\xi\mapsto B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi}{B_T}\bigg|\mathcal{F}_t\right]$ defines a linear pricing functional.

Example: pricing an asset paying $\xi=S_T^2$ under constant interest rates and geometric Brownian motion dynamics (without dividends). The answer is simply $$V_t=B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi}{B_T}\bigg|\mathcal{F}_t\right]=S_t^2e^{(r+\sigma^2)(T-t)}.$$ Details to this calculation are in the comments. This formula makes intuitive sense: after a numeraire change, $V_t$ is just the expected value of the stock price under the stock measure under which the stock price grows at the drift rate $r+\sigma^2$. Importantly, this price is model-dependent. A standard forward paying $S_T$ is model-independent (and follows from the definition of the equivalent martingale measure). Also, the price of a power claim paying $S_T^2$ is not just delta one but has volatility exposure.

In a one period, binomial setting, the stock moves from $S_0$ to either $S_0u$ or $S_0d$. A hedging portfolio invests $\Delta$ in the stock and $M$ in the bond, i.e. $\Pi_0=\Delta S_0+MB$ and $\Pi_T=\Delta S_T+M$ (in your notation, $x=\Delta$ and $y=M$). You try to replicate a general payoff $V$. Then, solve \begin{align*} \begin{cases} V_u = S_u\Delta+M, \\ V_d = S_d\Delta+M. \end{cases} \end{align*} Solving this system leads to a discrete analogue of a delta hedge: \begin{align*} \Delta &= \frac{V_u-V_d}{S_u-S_d}, \\ M &= \frac{uV_u+dV_d}{u-d}. \end{align*} Thus, $$ \Pi_0=\Delta S_0+MB=\frac{V_u-V_d}{S_u-S_d}S_0+\frac{uV_u+dV_d}{u-d}B=\frac{1-Bd}{u-d}V_u+\frac{Bu-1}{u-d}V_d.$$ You see, the probabilities are independent of the payoffs $V_u$ and $V_d$. Now, simply set $V_u=S_0^2u^2$ and $V_d=S_0^2d^2$ and you're done.

  • $\begingroup$ Thank you so much for this post. Apologies for keeping asking additional questions. I wanted to ask about the one period model: on the binomial tree with terminal pay-off $S_T^2$, one doesn't need to assign probabilities to calculate the option price: one can solve for the option price by replication (i.e. two equations with two unknowns: solving for the number of stocks and risk-free bonds that will replicate the option pay-off in both states). When I use this replication strategy, i keep getting a different pricing measure than the risk- neutral Measure as a result - how do we explain that? $\endgroup$ Commented Jul 27, 2020 at 7:48
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    $\begingroup$ @JanStuller Please look at my edit: you seem to have built a tree for $S^2$ instead of $S$. $\endgroup$
    – Kevin
    Commented Jul 27, 2020 at 9:09
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    $\begingroup$ I agree with this. The deflated price process $V_t/B_t$, rather than the process $S_t^2/B_t$, is a martingale under the risk-neutral measure. $\endgroup$
    – Gordon
    Commented Jul 27, 2020 at 14:07
  • $\begingroup$ With your approval @Gordon, there can't be any remaining doubt! :) In fact, the price process is defined to be a martingale when discounted and thus is compatible with risk-neutral pricing and the absence of arbitrage. $\endgroup$
    – Kevin
    Commented Jul 27, 2020 at 14:51

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