Self-financing strategy

We will review the definition of the self-financing, and give some related examples. One example shows that if all asset prices being only dependent to the same 1-d Brownian motion must have the same Sharpe ratios.

Let the {(N+1)}-dimensional price process {\{S(t)= (S_{i}(t))_{i=0}^{N};t \ge 0\}} be given to denote the asset prices in the market. In particular, the {i}th component of the vector {S(t)}, denoted by {S_{i}(t)}, is the {i}th asset price at time {t}. Let {\{h(t) = (h_{i}(t))_{i=0}^{N}; t\ge 0\}} be an {(N+1)}-dimensional process, where its {i}th component {h_{i}(t)} stands for the number of shares of the {i}th asset in the investor’s portfolio. This implies that, if the investor’s portfolio consists of {h_{i}(t)} shares of {i}th asset at time {t}, then the value at {t} is the sum

\displaystyle   V^{h}(t) = \sum_{i=0}^{N} h_{i}(t) S_{i}(t). \ \ \ \ \ (1)

By Ito’s formula, the local change of the portfolio value is

\displaystyle d V^{h}(t) = \sum_{i=0}^{N} \Big(h_{i}(t) d S_{i}(t) + S_{i}(t) d h_{i}(t) + d \langle h_{i}, S_{i}\rangle (t)\Big).

In other words, the change of portfolio value consists of two parts: the change due to the stock price change

\displaystyle \sum_{i=0}^{N} h_{i}(t) d S_{i}(t)

and the change due to exogenous capital injection

\displaystyle  \sum_{i=0}^{N} \Big( S_{i}(t) d h_{i}(t) + d \langle h_{i}, S_{i}\rangle (t)\Big).

In this below, we will give mathematical definition of strategy and self-financing strategy. Loosely speaking, self-financing strategy is a strategy with zero exogenous injection for all {t>0}.

Definition 1 Let the {(N+1)}-dimensional price process {\{S(t);t \ge 0\}} be given stock price process.

  1. A portfolio strategy (most often simply called a portfolio) is any {\mathbb F^{S}}-adapted {(N+1)}-dimensional process {\{h(t); t \ge 0\}}, and the value process {V^{h}} corresponding to the portfolio {h} is given by (1).
  2. A portfolio strategy {h} is called self-financing if the value process {V^{h}} also satisfies the condition

    \displaystyle   d V^{h}(t) = \sum_{i=0}^{N} h_{i}(t) d S_{i}(t), \hbox{ or equivalently } V^{h}(t) = V^{h}(0) + \sum_{i=0}^{N} \int_{0}^{t} h_{i}(u) d S_{i}(u), \ \ \ \ \ (2)

Remark 1 Note that, if {h} is self-financing, then {h_{0}} can be always computed from {\{h_{i}: i= 1\ldots N\}} and {V^{h}(0)}, by equating (1) and (2).

Usually we assume there is one risk-free asset {S_{0}} with short rate {r_{t}}, i.e.

\displaystyle d S_{0} (t) = S_{0}(t) r({t}) dt, \quad S_{0}(0) =1

and {N} risky assets with drift {\alpha_{i}} and volatility {\sigma_{i}>0}, i.e.

\displaystyle d S_{i} (t) = S_{i}(t) (\alpha_{i}(t) dt + \sigma_{i}(t) d W_{i}(t))

where {W_{i}} is a Brownian motion under a given filtered probability space {(\Omega, \mathcal F, \mathbb P, \{\mathcal F^{S}\})}. In this context, Sharpe ratio is defined as

\displaystyle \gamma_{i}(t) = \frac{\alpha_{i}(t) -r_{i}(t)}{\sigma_{i}(t)}.

Example 1 Suppose that there exists a self-financing portfolio with its value {V} satisfying

\displaystyle d V(t) = V(t) k(t) dt,

where {k} is an adapted process. Then it must hold that {k(t) = r(t)} for all {t}, otherwise there exists an arbitrage opportunity. In particular, if {V} of the above form satisfies {V(0) = 0}, then {V(t) = 0} for all {t>0}.

Example 2 Suppose two stocks with positive volatility both follow Ito process dependent on the same BM, then their Sharpe ratio are equal.

Proof: Suppose two asset prices {S_{i}} follow, for {i = 1,2}

\displaystyle d S_{i} (t) = S_{i}(t) (\alpha_{i}(t) dt + \sigma_{i}(t) d W(t)).

Consider self-financing {h = (h_{0}, h_{1}, h_{2})} with initial {V^{h}(0) = 0}, where {h_{i}(t) = \frac{1}{S_{i}(t)\sigma_{i}(t)}} for {i=1,2}. Then, we have

\displaystyle d V^{h}(t) = (\alpha_{1}(t)/\sigma_{1}(t) - \alpha_{2}(t)/\sigma_{2}(t) + h_{0}(t)S_{0}(t) r(t) )dt.

By Example~1, we conclude {V^{h}(t) = 0} for all {t>0}. So, {h_{0}(t)} can be determined from

\displaystyle \alpha_{1}(t)/\sigma_{1}(t) - \alpha_{2}(t)/\sigma_{2}(t) + h_{0}(t)S_{0}(t) r(t) = 0.

On the other hand, we have

\displaystyle V^{h}(t) = 1/\sigma_{1}(t) - 1/\sigma_{2}(t) + h_{0}S_{0}(t) = 0.

The above two equations implies {\gamma_{1} (t) = \gamma_{2}(t)}.

\Box

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