# trading strategy problem - initial capital x buys S over time [0,T] at the constant rate of x/T euros per unit of time

I am looking for clarification to the trading strategy problem where the number of stocks is depending on time.

In the Market with zero safe rate and stock dynamics defined as $$\frac{dS_t}{S_t}=\mu_t dt + \sigma_t dW_t \quad \quad (1)$$ investor with initial capital x buys stock for an interval of time [0,T] at the constant rate of x/T euros per unit of time.

I am calculating the number of shares at time t, first by defining the change rate

$$d \theta_t=\frac{x}{T} \frac{1}{S_t} dt \quad \quad (2)$$

and then getting the function for $\theta$ by integrating (2)

$$\theta_t=\frac{x}{T} \int_0^t \frac{1}{S_t} ds \quad \quad (3)$$

is this approach correct?

Further I want to show that the payoff $V_T$ at T equals $\frac{x}{T} \int_0^T R_{t,T} dt$ where $R_{t,T}$ is the simple return rate between t and T.

Solution manual says that this should be calculated as $$V_T= \int_0^T \theta_t dS_t \quad \quad (4)$$ and integration by parts yields $$V_T= \theta_T S_T - \int_0^T S_t d\theta_t = \int_0^T (S_T-S_t) d\theta_t = \int_0^T (\frac{S_T}{S_t}-1)S_t d\theta_t = \frac{x}{T} \int_0^T R_{t,T} dt \quad (5)$$

The way the payoffs are derived here is unclear to me. My understanding is that the number of shares is different at each time point but only two prices were used here $S_T$ and $S_t$ while integration is with respect to $\theta$. However the price changes over the time interval as well.

Can anybody explain the reasoning used for the payoffs?

Equations (1) to (3) are correct. Your investment strategy is then, $\forall t > 0$ $$X_t = \theta _ t S_t$$ Provided you use this strategy as part of self-financing portfolio you can write the P&L over an infinitesimal time interval as $$dV_t = \theta_ t dS_t$$ assuming zero safe rate, i.e. that any cash required to finance your long stock position (resp. cash received when you are short) does not cost you (resp. earn you) anything.
So that the total P&L over $[0,T]$ reads $V_T = \int_0^T dV_t$ or equivalently: $$V_T = V_0 + \int_0^T \theta_ t dS_t$$ which is equation (4). As you indicate going from (4) to (5) is simply integration by parts. Apply Ito to $\theta_ t S_t$ to obtain $$d (\theta_ t S_t) = \theta_ t dS_t + d\theta_ t S_t$$ because $\theta_ t$ has finite variation. Now integrate on both sides and rearrange terms to end up with $$\int_0^T \theta_ t dS_t = (\theta_T S_T - \theta_0 S_0) + \int_0^T S_t d\theta_ t$$ Hence the final wealth $$V_T = V_0 + \int_0^T \theta_ t dS_t$$ can be re-written (noting that $V_0 = \theta_0 S_0$) $$V_T = \theta_T S_T + \int_0^T S_t d\theta_ t$$ thanks to integration by parts.
• some things are still unclear to me. $$.$$ 1) where the $d (\theta_ t S_t) = \theta_ t dS_t + d\theta_ t S_t$ was applied? $$\theta_T S_T - \int_0^T S_t d\theta_t = \int_0^T S_T d\theta_t -\int_0^T S_t d\theta_t = \int_0^T (S_T-S_t) d\theta_t$$ $$.$$ 2) in the above $\int_0^T S_T d\theta_t$ is the payoff at time T, where we have just one price at T. how would be the $\int_0^T S_t d\theta_t$ interpreted? isn't S_t in this integral used as a constant? different amount of stocks are being bought at different $S_t$ prices over [0,T] though. I am confused here – Michal Apr 30 '16 at 22:18
• I've explicited the use of integration by parts in my answer. Usually the word payoff is reserved form contingent claims, the equation above gives you the terminal wealth if use you the investment strategy buy $\theta_t$ shares $\forall t>0$ as part of a self-financing PF. IMHO you should only try to interpret the final result. – Quantuple May 1 '16 at 9:13