What's the point of discounting in risk-neutral pricing?

Question

Let $\phi$ be a self-financing strategy that replicates a time $T$ option payoff $X$ on stock $S$. By definition of a trading strategy, $\phi$ is previsible. Finally, let $V_t$ be the time $t$ value of the portfolio implementing $\phi$.

Usual theorem: If $S_te^{-rt}$ is a $\mathbb{Q}$-martingale then $V_t = \mathbb{E}_\mathbb{Q}[e^{-r(T-t)}X|\mathcal{F}_t]$.

Note if the price of the option were anything other than $V_t$, we would have arbitrage.

My Claim: If $S_t$ is a $\mathbb{Q}$-martingale then $V_t = \mathbb{E}_\mathbb{Q}[X|\mathcal{F}_t]$.

No discounting necessary, and again this $\phi$ is a replicating self-financing strategy, so the value of the option must again be $V_t$ for all $t$.

Proof of My Claim. Consider discrete time and let $S_t$ be a $\mathbb{Q}$-martingale. By definition of a self-financing strategy, $\Delta V_{t+1} = \phi_{t+1}\Delta S_{t+1}$. where $\Delta V_{t+1} = V_{t+1} - V_t$. Hence \begin{align*} \mathbb{E}_\mathbb{Q}[\Delta V_{t+1}|\mathcal{F}_t] =\mathbb{E}_\mathbb{Q}[\phi_{t+1}\Delta S_{t+1}|\mathcal{F}_t] =\phi_{t+1}\mathbb{E}_\mathbb{Q}[\Delta S_{t+1}|\mathcal{F}_t] = 0, \end{align*} where the second equality is because $\phi$ is previsible. So $V_t$ is a $\mathbb{Q}$-martingale, and \begin{align*} V_t = \mathbb{E}_\mathbb{Q}[V_T | \mathcal{F}_t] = \mathbb{E}_\mathbb{Q}[X | \mathcal{F}_t], \end{align*} where the second equality is because $\phi$ replicates $X$.

What am I missing? Why do we only consider discounted stock prices, and hence what's the point of discounting the expectation? The only good reason I can think of for considering only discounted stock prices is that martingale representation theorem guarantees the existence of self-financing strategies in this case. But still, the usual theorem is valid for any self-financing strategy, so the discounting still seems unnecessary.

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Update: Okay, my previous example indeed assumed I was trading in only one stock, in which it's generally not possible to have a replicating self-financing strategy (right?). Let me prove a new, more general claim, again asserting discounting is not necessary by trading in two stocks $S^1$ and $S^2$ and requiring them both to be $\mathbb{Q}$-martingales. This will be slightly different than the "usual" theorem above in that we'll require both stocks to be martingales, but this seems valid. Furthermore, if we had an option payoff that was a function of two stocks, we could use this same claim to trade in only those two stocks; i.e., no "extra" stocks needed to trade in.

Again let $\phi$ be a self-financing strategy that replicates a time $T$ option payoff $X$ on stock $S^1$, but now $\phi$ trades in both $S^1$ and $S^2$. By definition of a trading strategy, $\phi$ is previsible. Finally, let $V_t$ be the time $t$ value of the portfolio implementing $\phi$.

Updated claim: If the vector $S = (S_1, S_2)$ is a $\mathbb{Q}$-martingale; i.e., each component of $S$ is a $\mathbb{Q}$-martingale, then $V_t = \mathbb{E}_\mathbb{Q}[X|\mathcal{F}_t]$.

Proof. Let $S = (S_1, S_2)$ be a $\mathbb{Q}$-martingale. By definition of a self-financing strategy, $\Delta V_{t+1} = \phi_{t+1}\cdot\Delta S_{t+1}$, where $\Delta V_{t+1} = V_{t+1} - V_t$ and "$\cdot$" is the dot product. Hence \begin{align*} \mathbb{E}_\mathbb{Q}[\Delta V_{t+1}|\mathcal{F}_t] =\mathbb{E}_\mathbb{Q}[\phi_{t+1}\cdot\Delta S_{t+1}|\mathcal{F}_t] =\phi_{t+1}\cdot\mathbb{E}_\mathbb{Q}[\Delta S_{t+1}|\mathcal{F}_t] = 0, \end{align*} where the second equality is because $\phi$ is previsible. So $V_t$ is a $\mathbb{Q}$-martingale, and \begin{align*} V_t = \mathbb{E}_\mathbb{Q}[V_T | \mathcal{F}_t] = \mathbb{E}_\mathbb{Q}[X | \mathcal{F}_t], \end{align*}

Does this seem correct? If so, it indeed seems discounting (in general, by some numeraire) would be necessary for the "usual" theorem, but this seems just as valid.

Answer 1

You only have one asset in your portfolio which means that you can only statically hedge. By the definition of self financing, $V_0=\phi_0 S_0$, $V_1=V_0+\phi_1 (S_1-S_0)$, and $V_1= \phi_1 S_1$. Putting these last two together, $V_0=\phi_1 S_0 $. Hence $\phi_1=\phi_0$ and you have a static position. Intuitively, this is because you cannot trade in another asset. If you could dynamically trade another asset, then your logic breaks down since no asset has a zero rate of return (in general).

The previous result can be generalized. By no arbitrage, all assets denominated in terms of other assets are martingales in the measure induced by the denominating asset.

In the "Risk Neutral" case, the denominating asset is a bond or money market account, and hence the "discounted" $S_t$ is a martingale (or, perhaps more clearly, $\frac{S_t}{M_t}$ is a martingale where $M_t$ is the money market account).

Regardless of the chosen denominating asset, the option pricing formula will remain the same. You cam prove this to yourself by solving the Black Scholes option price using the stock rather than the money market as the denominating asset.

So no, you can't simply take the expected value of the terminal payoff under the measure under which $S_t$ is a martingale as in general there is no asset which has zero return.

Edit in response to OP's update:

I have two stocks to trade and only two stocks to trade. I am trying to replicate the payoff of a function of one of the stocks; say $g(S_1 ^ T)$. My self financing portfolio is $X=\Delta_1 S_1 + \Delta_2 S_2$ and $dX=\Delta_1 dS_1 + \Delta_2 dS_2$. I will assume for simplicity that both stocks have dynamics of the form $dS=\alpha S dt+\sigma S dW_t $ and that they are independent. Then $$dX=\Delta_1 \alpha_1 S_1 dt+\Delta_1 \sigma_1 S_1 dW_1+\Delta_2 \alpha_2 S_2 dt+\Delta_2 \sigma_2 S_2 dW_2 $$ Substituting $X=\Delta_1 S_1 +\Delta_2 S_2$, $$dX=\Delta_1 \alpha_1 S_1 dt+\Delta_1 \sigma_1 S_1 dW_1+ \alpha_2 (X-\Delta_1 S_1)dt+\sigma_2 (X-\Delta_1 S_1) dW_2 $$ Letting $\hat{X}=\frac{X}{S_2}$, $$d\hat{X}=\frac{dX}{S_2}-\frac{X}{S_2 ^2} dS_2+\frac{X}{S_2 ^3}\sigma^2 S_2^2 dt-\frac{dXdS_2}{S_2 ^2}$$ $$=\Delta_1 \alpha_1 \frac{S_1}{S_2} dt+\Delta_1 \sigma_1 \frac{S_1}{S_2} dW_1+ \alpha_2(\hat{X}-\Delta_1 \frac{S_1}{S_2})dt+(\hat{X}-\Delta_1 \frac{S_1}{S_2})\sigma_2 dW_2-\hat{X} \alpha_2 dt-\hat{X} \sigma_2 dW_2+\hat{X} \sigma_2 ^2 dt -\Delta_2 \sigma_2 ^2 dt$$ $$=\Delta_1 d\hat{S}$$ Where $\hat{S}=\frac{S_1}{S_2}$.

Now the problem reduced to finding a measure under which $d\hat{S}$ is a martingale since then the value of the option is $$S_2 ^ 0 \hat{X}_0= S_2 ^ 0 \mathbb{\hat{E}}[\hat{X}_T]$$

The differential of $d\hat{S}$ is $$d\hat{S}=\hat{S}\left(\alpha_1 -\alpha_2\right)dt + \hat{S}( \sigma_1 dW_1-\sigma_2 dW_2)+\hat{S} \sigma^2 dt $$

We try to find a pair $\theta_1, \theta_2 $ that solves

$$ \hat{S}( \sigma_1 (dW_1+\theta_1 dt)-\sigma_2 (dW_2+\theta_2 dt))=\hat{S}\left(\alpha_1 -\alpha_2\right)dt + \hat{S}( \sigma_1 dW_1-\sigma_2 dW_2)+\hat{S} \sigma^2 dt $$

This can be simplified to $$\sigma_1 \theta_1-\sigma_2 \theta_2=\alpha_1-\alpha_2+\sigma_2 ^2 $$ There is no unique solution, so the market is incomplete. However, note that your proposed solution $\theta_1=\frac{\alpha_1}{\sigma_1}$, $\theta_2=\frac{\alpha_2}{\sigma_2}$ is not one of the possible solutions.

Answer 2

It seems to me that if it were a martingale, one could make a lot of money (in expectation) by shorting it and investing the proceeds in the risk neutral asset/bank account.