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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 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 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. $\endgroup$ – Frido Rolloos Mar 8 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 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. $\endgroup$ – Frido Rolloos Mar 8 at 15:03
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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 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 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 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 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. $\endgroup$ – Frido Rolloos Mar 9 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 at 10:33

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