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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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:

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

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @will I am assuming no dividens as the OP did not mention them. $\endgroup$ – Daneel Olivaw Jul 22 '17 at 11:36
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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.

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