Deriving P&L for a Constant-Rate Stock Purchase Strategy
Summary
The document considers an investor who spends an initial amount evenly over a fixed interval to buy shares of a stock with stochastic price dynamics. The number of shares accumulated changes over time according to the spending rate divided by the current stock price. Under the stated model, this share-count process is treated as having finite variation.
For a self-financing position with zero interest on cash, incremental portfolio P&L is the current share holding multiplied by the stock price change. Integrating this relation gives total P&L; integration by parts connects it to the terminal stock value and the costs of purchases made along the way. The source’s accepted explanation contains a sign inconsistency in its rearranged integration-by-parts expression, while the question’s expression using terminal-minus-purchase price leads to the stated return integral. The derivation depends on the financing and zero-rate assumptions and does not address transaction costs or other trading frictions.
Key ideas
- A constant spending rate buys a time-varying number of shares based on the stock’s current price.
- For a self-financing strategy with zero safe rate, incremental P&L is the share holding times the price change.
- Integration by parts relates accumulated trading gains to terminal holdings and purchase costs.
- The return-integral form relies on the stated assumptions and excludes transaction costs.
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Full text
# trading strategy problem - initial capital x buys S over time [0,T] at the constant rate of x/T euros per unit of time
# 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?
## Answer by Quantuple (score 1, accepted)
https://quant.stackexchange.com/a/25734
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.