equities - Replicating a portfolio with a certain payoff function

Assume there are two stocks $S_1$ with price $p_1(t)$ and $S_2$ with price $p_2(t)$ where $t$ indicates time. Assume, there is a hypothetical derivative $D$, which is such that, price of $D$ at a time $t$ is given by $p_1(t)/p_2(t)$.

Is it possible to find a dynamic hedging strategy using stocks and bonds to construct this derivative?

If yes, then how?

Answer

A general hedging strategy

Let assume that $S_1(t)$ and $S_2(t)$ are the price processes of your 2 stocks and that they follow a Geometric Brownian Motion (GBM):

$$\forall \, i \in \{1,2\}, dS_i(t) =\mu_iS_i(t)dt + \sigma_iS_i(t)dW_i(t)$$

We assume both stocks have an instant correlation of $\rho$:

$$dW_1(t)dW_2(t)=\rho dt$$

Let also $V(t)$ be the value or (fair) price of a derivative that depends on prices $S_1(t)$ and $S_2(t)$ at time $t$. We construct a self-financing portfolio made up of $w_0(t)$ derivative contracts, $w_1(t)$ shares of stock $1$ and $w_2(t)$ shares of stock $2$. Its value at $t$, $\Pi(t)$, is given by:

$$\Pi(t) = w_0(t)V(t)+w_1(t)S_1(t)+w_2(t)S_2(t)$$

The portfolio value, being self-financing, evolves according to:

$$d\Pi(t) = w_0(t)dV(t) + w_1(t)dS_1(t) + w_2(t)dS_2(t)$$

We then have $-$ dropping time:

$$dV = \frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S_1}dS_1 + \frac{\partial V}{\partial S_2}dS_2 + \frac{1}{2}\frac{\partial^2 V}{\partial S_1^2}dS_1^2 + \frac{1}{2}\frac{\partial^2 V}{\partial S_2^2}dS_2^2 + \frac{\partial^2 V}{\partial S_1 \partial S_2}dS_1dS_2$$

In the above differential equation, multiplied back by weight $w_0(t)$, the random element is:

$$\frac{\partial V}{\partial S_1}w_0\sigma_1S_1dW_1 + \frac{\partial V}{\partial S_2}w_0\sigma_2S_2dW_2$$

Now, in $w_1(t)S_1(t)+w_2(t)S_2(t)$, the random element is:

$$w_1\sigma_1S_1dW_1 + w_2\sigma_2S_2dW_2$$

We are hedging the derivative $V(t)$, hence our portfolio must be riskless and earn the risk-free rate:

• From the riskless condition, random fluctuations must be cancelled. Using the portfolio value equation, we derive the hedging strategy:

$$\begin{align} w_0(t) & = \frac{\Pi(t)}{V(t)-S_1(t)\frac{\partial V}{\partial S_1}-S_2(t)\frac{\partial V}{\partial S_2}} \\[12pt] w_1(t) &= -w_0(t)\frac{\partial V}{\partial S_1} \\[12pt] w_2(t) &= -w_0(t)\frac{\partial V}{\partial S_2} \end{align}$$

• We let $B(t) = e^{rt}$ be a riskless bond earning the risk-free rate $r$. Given the portfolio must earn $r$ and assuming $\Pi(0)$ is normalized so as to be equal to $1$, then $B(t)$ is a solution to the risk-free return constraint :

$$d\Pi(t) = r\Pi(t) dt$$

The final expression of the hedging portfolio is:

$$\begin{align} w_0(t) & = \frac{B(t)}{V(t)-S_1(t)\frac{\partial V}{\partial S_1}-S_2(t)\frac{\partial V}{\partial S_2}} \\[12pt] w_1(t) &= -w_0(t)\frac{\partial V}{\partial S_1} \\[12pt] w_2(t) &= -w_0(t)\frac{\partial V}{\partial S_2} \end{align}$$

Your question

Now, in the particular case of your derivative and interpreting strictly your original question:

" [...] price of [the derivative] $D$ at a time $t$ is given by $\frac{S_1(t)}{S_2(t)}$. "

The price is then given by:

$$V(t) = \frac{S_1(t)}{S_2(t)}$$

We have $\partial V/\partial S_1 = 1/S_2 = V/S_1$ and $\partial V/\partial S_2 = -S_1/S_2^2 = -V/S_2$, hence the hedging strategy would be:

$$\begin{align} w_0(t) &= \frac{B(t)}{V(t)} \\[12pt] w_1(t) &= -\frac{B(t)}{V(t)}\frac{\partial V}{\partial S_1} = -w_0(t)\frac{\partial V}{\partial S_1} \\[12pt] w_2(t) &= -\frac{B(t)}{V(t)}\frac{\partial V}{\partial S_2} = -w_0(t)\frac{\partial V}{\partial S_2} \end{align}$$

However, I am not sure that positing the price $V(t)$ is the correct approach: the price of the derivative at $t$ should be derived from its payoff function and the PDE resulting from the risk-free return condition. After a few steps, we would get the following PDE $-$ dropping time:

$$\begin{align} rV & = \frac{\partial V}{dt} + \frac{\partial V}{\partial S_1}rS_1 + \frac{1}{2}\frac{\partial^2V}{\partial S_1^2}\sigma_1^2S_1^2 + \frac{\partial V}{\partial S_2}rS_2 + \frac{1}{2}\frac{\partial^2V}{\partial S_2^2}\sigma_2^2S_2^2 + \frac{\partial^2V}{\partial S_1 \partial S_2}\sigma_1\sigma_2S_1S_2\rho \\[12pt] & = \frac{\partial V}{dt} + rw_1S_1 + \frac{1}{2}\frac{\partial^2V}{\partial S_1^2}\sigma_1^2S_1^2 + rw_2S_2 + \frac{1}{2}\frac{\partial^2V}{\partial S_2^2}\sigma_2^2S_2^2 + \frac{\partial^2V}{\partial S_1 \partial S_2}\sigma_1\sigma_2S_1S_2\rho \end{align}$$

Solving it would yield the value of $V(t)$. Note that weight $w_0$ does not appear in the PDE above because all three weights $w_0$, $w_1$ and $w_2$ can be written as a function of $w_0$, hence the term $w_0$ ends up being cancelled.

In this particular case, by replacing the derivatives by their specific expression $-$ which we can derive from the fact that $V(t) = S_1(t)/S_2(t)$ $-$ we obtain:

$$\begin {align} & \: rV = 0 + rV + 0 - rV + \sigma_2^2V - \sigma_1\sigma_2\rho V \\[12pt] \Leftrightarrow & \: r = \sigma_2^2 - \sigma_1\sigma_2\rho \end{align}$$

Hence we would be imposing a constraint on "market" parameters. The correct way to proceed would be to derive an expression for $V(t)$ given the PDE above.

A final example: $V(T)=\frac{S_1(T)}{S_2(T)}$

Let assume that your derivative has the following payoff function:

$$V_T=V(T)=\frac{S_1(T)}{S_2(T)}$$

Switching to martingale pricing tools, we know that:

$$\forall \, t \in [0,T], V_t = \mathbb{E}^{\mathbb{Q}}\left[e^{-r(T-t)}\frac{S_1(T)}{S_2(T)}|\mathcal{F}_t\right]$$

Where $\mathbb{Q}$ is the risk-neutral measure. Let now define $X_t = X(t) = S_1(t)/S_2(t)$. Applying Ito's lemma $-$ and dropping time:

$$\begin{align} dX_t = \frac{dS_1}{S_2} - \frac{S_1dS_2}{S_2^2} + \frac{S_1dS_2^2}{S_2^3} - \frac{dS_1dS_2}{S_2^2} \\[12pt] \Leftrightarrow \frac{dX_t}{X_t} = \left(\mu_1-\mu_2-\sigma_1\sigma_2\rho+\sigma_2^2\right)dt +\sigma_1dW_1^{\mathbb{Q}} + \sigma_2dW_2^{\mathbb{Q}} \end{align}$$

Under the risk-neutral measure, $\mu_1=\mu_2=r$. Letting $W_i(t) = W_i^{\mathbb{Q}}(t)$, after a few steps we get:

$$\begin{align} X_T = X_te^{(\sigma_2^2-\sigma_1^2-4\sigma_1\sigma_2\rho)\frac{T-t}{2}+\sigma_1W_1(T-t)+\sigma_2W_2(T-t)} \end{align}$$

Applying Ito's lemma to $W_1(t)W_2(t)$ and using the martingale property of the Ito integral, we get:

$$\mathbb{Cov}[W_1(T-t),W_2(T-t)] = \rho (T-t)$$

Now, letting:

$$\begin{align} & Z(t) = (\sigma_2^2-\sigma_1^2-4\sigma_1\sigma_2\rho)\frac{t}{2}+\sigma_1W_1(t)+\sigma_2W_2(t) \\[6pt] & \mathbb{E}[Z(t)] = (\sigma_2^2-\sigma_1^2-4\sigma_1\sigma_2\rho)\frac{t}{2} \\[6pt] & \mathbb{V}[Z(t)] = \sigma_1^2t + \sigma_2^2t+2\sigma_1\sigma_2\rho t \end{align}$$

We obtain:

$$\mathbb{E}[X_T|\mathcal{F}_t] = X_te^{\sigma_2^2(T-t)-\sigma_1\sigma_2\rho (T-t)}$$

Hence

$$\forall \, t \in [0,T], V_t = \frac{S_1(t)}{S_2(t)}e^{(\sigma_2^2-\sigma_1\sigma_2\rho-r)(T-t)}$$

We have came back to the constraint on market parameters derived in section "Your question":

$$r = \sigma_2^2 - \sigma_1\sigma_2\rho \quad \Rightarrow \quad V_t = \frac{S_1(t)}{S_2(t)}$$

The hedging strategy would then be:

$$\begin{align} & w_0(t) = \frac{S_2(t)}{S_1(t)}e^{(\sigma_1\sigma_2\rho-\sigma_2^2)(T-t)+rT} \\[12pt] & w_1(t) = -\frac{e^{rt}}{S_1(t)} \\[12pt] & w_2(t) = \frac{e^{rt}}{S_2(t)} \end{align}$$

You can easily check that $w_0(t)V(t)+w_1(t)S_1(t)+w_2(t)S_2(t) = B(t)$.

We finally can check that:

$$\begin{align} & \frac{\partial V}{dt} = -(\sigma_2^2-\sigma_1\sigma_2\rho-r)V \\[12pt] & \frac{\partial V}{\partial S_1}rS_1 = rV \\[12pt] & \frac{1}{2}\frac{\partial^2V}{\partial S_1^2}\sigma_1^2S_1^2 = 0 \\[12pt] & \frac{\partial V}{\partial S_2}rS_2 = -rV \\[12pt] & \frac{1}{2}\frac{\partial^2V}{\partial S_2^2}\sigma_2^2S_2^2 = \sigma_2^2V \\[12pt] & \frac{\partial^2V}{\partial S_1 \partial S_2}\sigma_1\sigma_2S_1S_2\rho = -\sigma_1\sigma_2\rho V \end{align}$$

Terms cancel and we are left with the following identity equation:

$$rV = rV$$

Hence $V(t)$ as derived above is a solution to the pricing PDE.

Further references

Further relevant references from Wikipedia and user @Gordon on deriving the hedging strategy and the pricing PDE for a derivative $V(t)$: