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Given the asset price $S_t$ which is defined as follows $$\frac{dS_t}{S_t}= r_tdt+\sigma_tdW_t$$ where $r_t$ is not necessarily deterministic.

What is the strategy of replication of the portfolio with the payoff $\int_0^T \frac{dS_t}{S_t}$ ?

My attempt:

In fact, I can solve this problem only for the special case where $r_t$ is deterministic. For simplicity's sake, I provide the solution for an easier case where $r_t =r$ constant.

Let's $V_t$ the replicating porfolio of $\int_0^T \frac{dS_t}{S_t}$, we have \begin{align} V_t &=e^{-r(T-t)}E^{\Bbb Q}[\int_0^T \frac{dS_u}{S_u}|\mathcal{F}_t] \\ &=e^{-r(T-t)}\int_0^t \frac{dS_u}{S_u}+e^{-r(T-t)}E^{\Bbb Q}[\int_t^T (rdu+\sigma_udW_u)|\mathcal{F}_t] \tag{1}\\ &=e^{-r(T-t)}(\int_0^t \frac{dS_u}{S_u}+r(T-t)) \tag{2}\\ \end{align}

From (2), by applying the Ito's lemma, we obtain easily that \begin{align} dV_t &= re^{-r(T-t)}(\int_0^t \frac{dS_u}{S_u}+r(T-t))dt +e^{-r(T-t)}(\frac{dS_t}{S_t}-rdt) \\ &= r(V_t-e^{-r(T-t)})dt+e^{-r(T-t)} \frac{dS_t}{S_t} \tag{3}\\ \end{align}

From (3), we obseve that we can replicate $V_t$ (which is equal to $\frac{e^{-r(T-t)}}{S_t}S_t+ \frac{V_t-e^{-r(T-t)}}{B_t}B_t$) by investing $e^{-r(T-t)}$ in the asset $S_t$ at time $t$ and the rest of the portfolio $(V_t-e^{-r(T-t)})$ in cash.

Problem:

For the general case where $r_t$ is stochastic, I don't know how to deduce (2) from (1), or (3) from (2).

I guess the strategy in the general case must be investing $P(t,T)$ in the asset $S_t$ ($P(t,T)$ is the zero-coupon bond price between $t$ and $T$) and the rest of the portfolio in cash. But I don't know how to prove that.

The zero-coupon bond $P(t,T)$ is specified by

$$\frac{dP(t,T)}{P(t,T)} = r_tdt + \gamma_t dB_t$$

For simplicity's sake, let's suppose the correlation between $B_t$ and $W_t$ be zero ($\left<dB_t,dW_t\right> = 0$)

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  • $\begingroup$ the formula 1 is false! Vt is real and the integral is stochastic.... $\endgroup$
    – Valometrics.com
    Mar 6, 2021 at 20:47
  • $\begingroup$ @Valometrics.com $V_t$ is indeed real but the integral is not stochastic at time $t$ because all information before $t$ is known, in particular $S_u$ for $u \in (0,t)$ is known. Then $\int_0^t \frac{dS_u}{S_u}$ is deterministic. PS: Perhaps my formula in the question is not clear (I should have written $\int_0^t \frac{dS_u}{S_u}$ instead of $\int_0^t \frac{dS_t}{S_t}$). I modified it. $\endgroup$
    – NN2
    Mar 6, 2021 at 20:52
  • $\begingroup$ I think if you would like to work using the zero coupon bond as numeraire to take into account the stochastic short rate, you will need to specify the correlation structure between the zero and the stock. $\endgroup$
    – user34971
    Mar 8, 2021 at 14:02
  • $\begingroup$ @FridoRolloos I add this information at the end of the question. We can suppose that the correlation between them is 0. I try to avoid specifing the interest rate model if possible. But if it's necessary, you could specify the interest rate model as you want (and also the correlation structure between the interest rate and the stock). $\endgroup$
    – NN2
    Mar 8, 2021 at 14:21
  • $\begingroup$ Yes, I meant you need to specify/assume zeros are traded. The specific dynamics of the zeros is not important for this. $\endgroup$
    – user34971
    Mar 8, 2021 at 15:03

2 Answers 2

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It is actually fairly simple: just hold $\frac{1}{S_t}$ units of the stock at all time! Then, no matter if rates or volatilities are stochastic, the change in value of your portfolio is $\frac{dS_t}{S_t}$ at all time and the terminal value of your portfolio is therefore $$ \int_0^T{\frac{\mathrm{d}S_t}{S_t}} $$

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  • $\begingroup$ I don't agree because according to my proof, we need to hold $\frac{e^{-r(T-t)}}{S_t}$ (and not $\frac{1}{S_t}$) units of the stock $S_t$ at all time, for the case rates are deterministic. Of course, we can have the intuition that we should hold $\frac{P(t,T)}{S_t}$ units of the stock $S_t$ but I would like to have a rigorous proof (based on mathematics) or at least an explication why holding $\frac{P(t,T)}{S_t}$ units of the stock works. $\endgroup$
    – NN2
    Mar 6, 2021 at 22:32
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So it is sufficient to assume the existence of zero coupon bonds. We do not have to specify anything else.

Define first

$$ X_t = \int_0^t \frac{dS_u}{S_u} $$

Then $$ dX_t = \frac{dS_t}{S_t} $$

Note that $$ X_T = \frac{X_T}{P_T} = \int_0^T d \left( \frac{X_t}{P_t} \right) $$ since $P_T = 1$ and $X_0 = 0$.

Under the $T$-forward measure $X_t/P_t$ is a martingale, which again shows the current price of the claim is $0$.

We are mainly interested in the expression in the integrand: $$ d \left( \frac{X_t}{P_t} \right) = \frac{1}{P_t} dX_t - \frac{X_t}{P_t^2} dP_t + O(dt) $$ We are not interested in the Ito terms as they add up to zero (pricing PDE).

So the replication should be $$ \frac{1}{P_t S_t} dS_t - \frac{1}{P_t^2} \left( \int_0^t \frac{dS_u}{S_u} \right) dP_t $$

I think this is the way to do this.

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  • $\begingroup$ My comment is too long, I separate it into 3 parts. Part 1: I still don't get it. From your last two equations, if I understand correctly, you want to prove $$X_T = \int_0^T \frac{dS_u}{P(u,T)S_u} - \int_0^T \left(\frac{X_u}{P^2(u,T)}dP_u\right)$$ But I don't know how to deduce the strategy of repplication from that. $\endgroup$
    – NN2
    Mar 8, 2021 at 21:24
  • $\begingroup$ Part 2: I think there is a confusion here between the process $X_t = \int_0^{\color{red}t} \frac{dS_u}{S_u}$ and the product with payoff $\int_0^{\color{blue}T} \frac{dS_u}{S_u}$. In fact, the value of this product at time $t$ calculated in forward-measure $\mathbb{Q}^T$ or in risk neutral measure $\mathbb{Q}$, is $$V_t = P(t,T) E^{\mathbb{Q}^T} \left( \int_0^T \frac{dS_u}{S_u} | \mathcal{F}_t \right) = E^{\mathbb{Q}} \left( \frac{B_t}{B_T}\int_0^T \frac{dS_u}{S_u} | \mathcal{F}_t \right)$$ $\endgroup$
    – NN2
    Mar 8, 2021 at 21:26
  • $\begingroup$ Part 3: while you are trying to replicate the product with the value at time $t$ $$V_t = X_t = \int_0^{t} \frac{dS_u}{S_u}$$ $\endgroup$
    – NN2
    Mar 8, 2021 at 21:27
  • $\begingroup$ @NN2 Thanks for being critical! Let me take another look at my answer, as you say it's probably on the right track but not yet correct. $\endgroup$
    – user34971
    Mar 9, 2021 at 5:55
  • $\begingroup$ Finally, I found that it's impossible to replicate the product with that payoff at time $t$ for $t<T$ in the case $r_t$ is stochastic. For the case where $r_t$ is stochastic, we can only replicate this product only at time $T$. The answer can be found in the document "Just what you need to know about variance swaps", page 19. $\endgroup$
    – NN2
    Mar 9, 2021 at 10:33

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