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?

  • $\begingroup$ Could you maybe clarify whether in "[...] price of $D$ at time $t$ is given by $p_1(t)/p_2(t)$" you mean the price or the payoff? $\endgroup$ – Daneel Olivaw May 27 '17 at 14:42
  • 1
    $\begingroup$ Thank you for pointing it out Daneel. I have price (and not profit or payoff) of D at time t in mind. I have edited the question, to make it more crisp. $\endgroup$ – swanar May 27 '17 at 14:46

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:

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


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

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 $V(t)$:


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.