# 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 ${\gamma_{1} (t) = \gamma_{2}(t)}$. $\Box$