A set of interesting options arbitrage topics

A set of interesting options arbitrage topics – Financial Economics Notes by Pan Deng

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When I was watching the video a few days ago, I found such a question

The solution given by the author

Because the option cash flow is a convex function,

$$\begin{gathered} K=40，50，70 \\ V(K)=10,20,30 \\ 50=\frac{2}{3}40+\frac{1}{3}70 \\ 20=V(50)\ge \frac{2}{3}V(40)+\frac{1}{3}V(70)=16.67 \\ Thereisarbitrage\; space,\; sell350,buy240,170 \end{gathered}$$

But the author of the question arbitrarily takes $\frac{2}{3}$and $\frac{1}{3}$The approach is not necessarily convincing, but it is advisable to essentially want the sum of the cash flows of the linear part to be equal (that is, the cash flows of the two asset portfolios bought and sold are equal, and then offset each other)

Draw the cash flow of the three options and the cash flow of the combination on the graph.

It can be seen that when the price is less than or equal to 40, the combined cash flow is stable, which is the idea of ​​linear partial cash flow offset;

my solution

As a rigorous financial scholar, we must not blindly set $\frac{2}{3}$and $\frac{1}{3}$proportion;

According to the no-arbitrage pricing idea, if there is an arbitrage opportunity to construct an asset portfolio, there must be
$$\begin{array}{rl}Pi& =a{f}_a+b{f}_b+c{f}_c\\ & =a(\max (40-x,0)-10)+b(\max (50-x,0)-20)+c(\max (70-x,0)-30)\ge 0\end{array}$$
Among them, the equal sign is not always true; write the above formula into a piecewise function
$$\begin{gathered} \{\begin{array}{ll}a(30-x)+b(30-x)+c(40-x)\ge 0,& x\le 40\\ -10a+b(30-x)+c(40-x)\ge 0,& 40<x\le 50\\ -10a+-20b+c(40-x)\ge 0,& 50<x\le 70\\ -10a+-20b-30c\ge 0,& 70<x\end{array} \\ \Rightarrow \{\begin{array}{ll}(a+b+c)x\le 30a+30b+40c,& x\le 40\\ (a+b)x\le -10a+30b+40c,& 40<x\le 50\\ cx\le -10a-20b+40c,& 50<x\le 70\\ a+2b+c\le 0,& 70<x\end{array} \end{gathered}$$
Observing the above formula, we can see that the right side of the inequality is all parameters, and the left side of the inequality is a linear function (except the last inequality); suppose given $a,b,c$ , then the value on the right side of the inequality is determined, and the monotonicity on the left side of the inequality is also determined; because the left side of the inequality is either monotonically increasing or monotonically decreasing, so the boundary is brought in, if the boundary value is satisfied, then all values ​​are satisfied, and then {
$$\{\begin{array}{l}0\le 30a+30b+40c\\ 0\le -10a-10b\\ 0\le -50a-10b+40c\\ 0\le -60a-20b+40c\\ 0\le -10a-20b-10c\\ 0\le -10a-20b-30c\end{array}$$
Obviously this is a three-dimensional linear programming, which is difficult to solve, but because $a,b,c$ is just a ratio, not a specific number, let$a=1$ , transform the problem into a two-dimensional linear programming
$$\{\begin{array}{l}0\le 3+3b+4c\\ 0\le -5-b+4c\\ 0\le -3-b+2c\\ 0\le -1-2b-c\\ 0\le -1-2b-3c\\ b\le 0\end{array}$$
Change the inequality sign into an equal sign and draw it in the picture

According to the relationship between each line and 0, the solution space can be found in the figure, then this solution space is the proportional choice for arbitrage, and it should be noted that there is not only one kind of arbitrage proportional choice ;

Written as a mathematical expression, it is
$$\{\begin{array}{l}0\le -1-2b-3c\\ 0\le -5-b+4c\\ 0\le 3+3b+4c\end{array}$$

In this solution set, randomly find a ratio $a=1,b=-3,c=1.6$ , draw the combined cash flow

So, is it because my solution is more rigorous, arbitrage is no longer so magical, it is just a veil outside the essence of mathematics! ! !