Hexa Scholar

What makes a trading strategy self-financing?

A strategy buys and sells shares as prices move. When does it need no money beyond what it cost at the start?

When every purchase is paid from the strategy's own cash account and every sale is paid into it. Then nothing enters or leaves after the start, the value changes only through price moves and interest, and at every time the value equals the initial cost plus the gains so far.

Before you start · 3 quick checks on binomial prices, interest and trading gains, about 3 min
Setup: a strategy, its cash account and outside money
  • Market. The n-period binomial tree of step 6: the stock starts at \(S_0 \gt 0\) and each period its price is multiplied by \(u\) or by \(d\), with \(0 \lt d \lt u\). A cash account grows by the factor \(1+r\) each period, with \(r \gt -1\) and \(d \lt 1+r \lt u\). Times are \(k = 0, 1, \dots, n\).
  • Strategy. At each time \(k \lt n\), after seeing the prices so far, the strategy holds \(\Delta_k\) shares and \(B_k\) dollars in the cash account until time \(k+1\). Both may be negative: a short sale, a loan. At time \(n\) nothing is traded: \(\Delta_n = \Delta_{n-1}\).
  • Value. \(V_k = \Delta_kS_k + B_k\) at every time \(k\). The initial cost is \(V_0\), the money put in at time 0.
  • Outside money. At time \(k\) the cash account earns its interest, pays for shares bought at the price \(S_k\), receives the proceeds of shares sold, and changes by any money \(C_k\) put in from outside (\(C_k \gt 0\)) or taken out (\(C_k \lt 0\)): for \(k = 1, \dots, n\), \[ \begin{aligned} B_k = (1+r)B_{k-1} &- (\Delta_k - \Delta_{k-1})S_k \\ &+ C_k, \end{aligned} \] with \(C_n = 0\), since nothing is traded or taken out at the end.
  • Self-financing. The strategy is self-financing if \(C_k = 0\) at every time \(k = 1, \dots, n-1\) on every path: after the initial cost, no money goes in or out.
  • Gains. The gains up to time \(k\) are what the holdings earned: the shares from price changes, the cash from interest, \[ \begin{aligned} G_k = \sum_{j=1}^{k}\big(&\Delta_{j-1}(S_j - S_{j-1}) \\ &+ r(V_{j-1} - \Delta_{j-1}S_{j-1})\big), \end{aligned} \] since \(V_{j-1} - \Delta_{j-1}S_{j-1} = B_{j-1}\) is the cash held from \(j-1\) to \(j\). Each \(\Delta_{j-1}\) is chosen before the change \(S_j - S_{j-1}\) is known: the share gain is a left-endpoint sum, as on the Itô integral page (step 12).
Fine print: trading assumptions
  • The market is frictionless: no transaction costs, no taxes, no bid-ask spread. Any real number of shares can be bought or sold short at the current price, and any amount can be lent or borrowed at the rate \(r\) per period.
  • The stock pays no dividends, so a share earns only its price change. With dividends, the cash account would also receive them, and a short position would pay them.
  • A strategy may depend on the path so far but not on later prices: the holdings are fixed when a node is reached, and two paths that reach the same price may hold different amounts.
  • "Self-financing" is a statement about money flows only. It does not say the strategy is safe or that its value stays positive.
1 See

Keep the books along one path

At each time, set the shares with the slider and type the new cash balance. Then choose the next price move. The page marks every time at which the cash you typed does not match what the account holds after interest and the trade: that difference is outside money. At time 3 it lays out the books. Cash is typed to the cent at time 0; each period of interest can add one more decimal, so the box takes one more decimal place at each later time, enough for the exact balance a self-financed trade leaves.

"Start a new walk" clears the books and keeps your progress through the goals. "Restart the goals" also goes back to goal 1.

2 Try

A fixed dollar amount, time-0 dollars and continuous time

The questions use the market of stage 1.

3 Prove

The value is the cost plus the gains, at every time

Theorem (self-financing strategies on a binomial tree)

Assumptions. The setup and its fine print: an n-period binomial tree with \(n \ge 2\), \(S_0 \gt 0\), \(0 \lt d \lt u\), \(r \gt -1\) and \(d \lt 1+r \lt u\), a frictionless market, no dividends, and any strategy \((\Delta_k, B_k)\) with outside money \(C_k\) defined by the cash equation, \(\Delta_n = \Delta_{n-1}\) and \(C_n = 0\).

Conclusion. (a) For every strategy, every path and every \(k\) from 1 to \(n\),

\[ \begin{aligned} V_k - V_{k-1} &= \Delta_{k-1}(S_k - S_{k-1}) \\ &\quad + r(V_{k-1} - \Delta_{k-1}S_{k-1}) + C_k, \end{aligned} \] \[ V_k = V_0 + G_k + \sum_{j=1}^{k} C_j . \]

(b) A strategy is self-financing if and only if \(V_k = V_0 + G_k\) for every path and every \(k\) from 1 to \(n\). In particular a self-financing strategy ends with \(V_n = V_0 + G_n\).

Each line of the proof is a claim about every strategy, every path and every \(k\) in the range it names. A candidate line that fails for one strategy is false, even if it holds for others.

Scratch work, not part of the proof.

The converse in line 5 needs the identity at every time. At the end alone it is not enough when there are two or more trading times: in goal 3 of stage 1, money put in at time 1 and taken out at time 2 cancel at time 3.

Key point One period of bookkeeping gives \(V_k - V_{k-1} = \Delta_{k-1}(S_k - S_{k-1}) + rB_{k-1} + C_k\): a trade paid from the cash account changes nothing, and outside money changes the value one for one. Adding the periods gives \(V_k = V_0 + G_k + \sum_{j=1}^{k} C_j\), so the strategy is self-financing exactly when \(V_k = V_0 + G_k\) at every time.
4 Apply

Problems

More problems
Key point To keep a strategy self-financing, set each new cash balance to the old one with interest, minus the cost of the shares bought or plus the proceeds of the shares sold. The value then follows \(V_k - V_{k-1} = \Delta_{k-1}(S_k - S_{k-1}) + r(V_{k-1} - \Delta_{k-1}S_{k-1})\), the tree's form of \(dV = \Delta\,dS + r(V - \Delta S)\,dt\).
Level Advanced undergrad Prerequisites binomial trees (step 6), the Itô integral (step 12) Time ~40 min

Further reading