Let $S\left(t\right)$ be a tradable financial security that doesn't generate cash flow (eg no dividend). $S\left(t\right)$ follows an unknown stochastic process.

We now have a financial derivative that pays $\frac{S\left(T_2\right)} {S\left(T_1\right)}$ at $t=T_2$, where $0<T_1<T_2$

Assume interest rate $r_t$ is not constant.

What's the present value of this financial derivative at $t=0$ ?

My attempt so far:

$V\left(0\right)=E^{\mathbb{Q}}\left[e^{-\int_{0}^{T_2}r_t\,dt} \frac{S\left(T_2\right)}{S\left(T_1\right)}\right]$

I believe my next step should be to get rid of the discount factor term. Any idea how can I do that?

  • $\begingroup$ Do you truly know nothing about $S(t)$? $\endgroup$ – will Jul 22 '17 at 8:41
  • $\begingroup$ @will Yes, you only need to know that $S(t)$ is a stochastic process. The model of $S(t)$ would not affect the answer. $\endgroup$ – chengcj Jul 22 '17 at 9:57
  • $\begingroup$ Do you have any of its properties? Or is the question asking you to write down the integral representation of the expectation? $\endgroup$ – will Jul 22 '17 at 10:10
  • 1
    $\begingroup$ This question has something to do with converting risk neutral measure to forward measure. Then after some manipulation, the present value of the derivative should be a function of $r_t$. Present value of the derivative should end up not dependant on $S(t)$ $\endgroup$ – chengcj Jul 22 '17 at 10:18
  • $\begingroup$ That is only the case if we can make a bunch of assumptions on S. You need to give us all the assumptions you're allowed to make. $\endgroup$ – will Jul 22 '17 at 10:19

We assume a Black-Scholes world except the dynamics of the stock price, namely:

  • No arbitrage opportunities.
  • No dividend payments from the stock.
  • Existence of a riskless asset yielding the risk free rate $-$ which here we assume non-constant, $(r_t)_{t \geq 0}$.
  • Possibility to borrow and lend infinitely at the risk-free rate.
  • Possibility to buy and sell infinitely the stock $-$ even fractional amounts.
  • No transaction costs.

We also assume that the stock is tradable and that the derivative is attainable $-$ we basically assume we are in the standard pricing setting except for the stock price dynamics.

Then the price at time $t=0$, $V(0)$, of the derivative is given by:

$$ V(0) = P(0,T_1)$$

where $P(0,T_1)$ is the price of a riskless zero-coupon contracted at time $t=0$ and maturing at time $t=T_1$ $-$ which is effectively a function of the rate $r_t$ and is independent of $S(t)$.

Financial proof: the financial derivative you describe delivers a quantity $w$ of the stock at time $T_2$, where:

$$ w = \frac{1}{S(T_1)}$$

Thus $w$ will only be known at time $T_1$, when you will buy $w$ shares of the stock. But at that time, the value of such a position is trivially equal to $\$1$. Thus you only need to have $\$1$ at time $T_1$ to settle the trade at maturity $T_2$; no further transactions are needed. The value today of $\$1$ at $T_1$ is simply equal to the value of a zero-coupon bond contracted at $t=0$ and maturing at $T_1$. Hence:

$$ V(0) = P(0,T_1)$$

Mathematical proof: under the assumptions listed at the beginning, by the law of iterated expectations, adaptedness of the stock price with respect to a suitable filtration $(\mathcal{F})_{t \geq 0}$ and the martingality property of discounted stock prices under the risk-neutral measure $\mathbb{Q}$, we obtain:

$$ \begin{align} V(0) & = E^{\mathbb{Q}}\left[e^{-\int_0^{T_2}r_t\,dt} \frac{S(T_2)}{S(T_1)}\right] \\[6pt] & = E^{\mathbb{Q}}\left[E^{\mathbb{Q}}\left[e^{-\int_0^{T_2}r_t\,dt} \frac{S(T_2)}{S(T_1)}|\mathcal{F}_{T_1}\right]\right] \\[6pt] & = E^{\mathbb{Q}}\left[e^{-\int_0^{T_1}r_t\,dt}\frac{1}{S(T_1)}E^{\mathbb{Q}}\left[e^{-\int_{T_1}^{T_2}r_t\,dt} S(T_2)|\mathcal{F}_{T_1}\right]\right] \\[6pt] & = E^{\mathbb{Q}}\left[e^{-\int_0^{T_1}r_t\,dt}\frac{1}{S(T_1)}S(T_1)\right] \\[9pt] & = P(0,T_1) \end{align} $$

  • $\begingroup$ This is only if you assume that $\mathrm{d}S_t = S_t r_t \mathrm{d}t + S_t \sigma \mathrm{d}W_t$. If you assume that, it should be obvious, since the discounting at the end will be $e^{\int_0^{T_2} r_t \mathrm{d}t}$, while while the expected value of $S(t)$ is $S_0 e^{\int_0^{t} r_\tau \mathrm{d}\tau}$ - i.e. the extra discounting from $T_1$ to $T_2$ will exactly match the expected performance, so you're left with the discounting up to the start of the period. This requires many assumptions about $S(t)$ though. $\endgroup$ – will Jul 22 '17 at 11:26
  • $\begingroup$ @will umm but the martingality property of the discounted stock price is independent of the posited stock price dynamics. $\endgroup$ – Daneel Olivaw Jul 22 '17 at 11:31
  • $\begingroup$ You're assuming $S(t)$ is matingale. I asked if we knew nothing about the process, and chengcj responded with "Yes, you only need to know that $S(t)$ is a stochastic process. The model of $S(t)$ would not affect the answer". $\endgroup$ – will Jul 22 '17 at 11:33
  • 1
    $\begingroup$ @will well I am assuming we are using risk-neutral pricing theory, which is independent of the stock price dynamics. $\endgroup$ – Daneel Olivaw Jul 22 '17 at 11:35
  • 1
    $\begingroup$ @will I am assuming no dividens as the OP did not mention them. $\endgroup$ – Daneel Olivaw Jul 22 '17 at 11:36

An alternative proof: The contract may be replicated by waiting until $T_1$ and then investing one dollar in the stock. Hence its value must be the same as a zero coupon bond priced at t maturing at $T_1$.

The above holds for any stock dynamics and rate dynamics.


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.