Consider a 2-period binomial tree where the derivative price is $f$ and the stock price is $S$. Also, let the bond be deterministic with continuous growth rate $r$ and initial value $B_0$. binomial tree

Recall the replicating strategy is at each time $t_i$ hold $\phi_i = \frac{f_{i+1}^{up} - f_{i+1}^{down}}{S_{i+1}^{up} - S_{i+1}^{down}}$ units of the stock and $\psi_i = B_0^{-1} e^{-r(i+1)\Delta t}(f_{i+1}^{up} - \phi_i S_{i+1}^{up})$ units of the bond. In particular, the value of the portfolio at time $0$ is $V_0 = \phi_0 S_0 + \psi_0 B_0$. When we arrive at time tick 1, lets say our stock price went up to $S_3$. Before rebalancing, our portfolio is worth $V_0|_{end} = \phi_0 S_3 + \psi_0 B_0e^{r \Delta t}$, and after rebalancing it is $V_1 = \phi_1 S_3 + \psi_1 B_0e^{r \Delta t}$. In order for this to be self-financing, we must have $V_1 - V_0|_{end} = 0$. However, \begin{align*} V_1 - V_0|_{end} & = (\phi_1 - \phi_0)S_3 + (\psi_1 - \psi_0)B_0e^{r \Delta t} \\ & = (\phi_1 - \phi_0)S_3 + \left(B_0^{-1} e^{-2r\Delta t}(f_7 - \phi_1 S_{7}) - B_0^{-1} e^{-r\Delta t}(f_{3} - \phi_0 S_{3})\right)B_0e^{r \Delta t} \\ & = (\phi_1 - \phi_0)S_3 + e^{-r\Delta t}(f_7 - \phi_1 S_{7}) - (f_{3} - \phi_0 S_{3}) \\ & = \phi_1 S_3 + e^{-r\Delta t}(f_7 - \phi_1 S_{7}) - f_{3} \\ & = \frac{f_{7} - f_{6}}{S_7 - S_6} S_3 + e^{-r\Delta t}(f_7 - \frac{f_{7} - f_{6}}{S_7 - S_6} S_{7}) - f_{3} \\ & = \frac{1}{S_7 - S_6} \left((f_{7} - f_{6})S_3 + (S_7 - S_6)e^{-r\Delta t}f_7 - e^{-r\Delta t}(f_{7} - f_{6}) S_{7} - (S_7 - S_6)f_{3} \right)\\ & = \frac{1}{S_7 - S_6} \left((f_{7} - f_{6})S_3 - S_6e^{-r\Delta t}f_7 + e^{-r\Delta t}f_{6} S_{7} - (S_7 - S_6)f_{3} \right)\\ & \neq 0. \end{align*}

It seems a lot of effort is put into self-financing strategies, and in fact the binomial representation theorem is used to prove the existence of them in the binomial model. Am I missing something?


2 Answers 2


I'd add a comment if I could but don't have enough reputation. How do you know your final equation is not equal to zero. The $f_i$ have not yet been calculated in terms of $S$. Certainly the $f_i$ that are in the final set of nodes are known since they are defined in terms of the payoff and the terminal price. The ones in intermediate steps have to be calculated.

  • $\begingroup$ For those, who are still wondering, why Muaddib was completely right. The easiest way for me to see this is the following: The risk free measure q is calculated from the known process S and gives S3 = exp(-r delta t) (q S6 - (1-q) S7). Now the price process f as well has to be a martingale with respect to q and therefor: f3 = exp(-r delta t) (q f6 - (1-q) f7). Plug these equivalences into the term in the brackets and the whole term vanishes. $\endgroup$ Commented Oct 29, 2021 at 7:29

The binomial model certainly is self-financing. First, get the value at every node by working backwards using risk-neutral evaluation.

Then at each step and node, you get the value in the up node and the down node from where you are. You can fit a straight line as a function of stock through the two. You hold stocks and bonds to fit this straight line with a replicating portfolio. The cost of set-up is precisely the price at the node you are currently at and it is self-financing.

I think you need to read an account that's a little more discursive and less equationy (see eg my book Concepts) or actually do an example numerically.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.