How to prove the following inequality? (related to no-arbitrage conditions)

Question

I'm working through a practice book on mathematical finance, but struggling to prove part of a question on no-arbitrage conditions.

In the problem, I'm first given $K_1 < K_2 < K_3 $. Then, the book claims that the following relationship: $x_1 (S - K_1) - x_2 (S - K_2) + x_3 (S- K_3) \geq 0, \forall S \geq 0$,

holds if and only if the two conditions are satisfied: $x_1 - x_2 + x_3 \geq 0$ and $x_1 (K_3 - K_1) - x_2 (K_3 - K_2) \geq 0$.

For the life of me I can't seem to prove why this is the case! I know it should be basic algebra but I'm really struggling for some reason. Anyone have any tips?

Answer 1

This will make sense for a portfolio of three call options with common expiry and with strike prices $0 < K_1 < K_2 < K_3$, where we are long $x_1 > 0$ options struck at $K_1$, short $x_2 > 0$ options struck at $K_2$ and long $x_3> 0$ options struck at $K_3$.

The payoff of this option portfolio at expiration for underlying price $S$ is

$$V(S) = x_1(S-K_1)^+ - x_2(S - K_2)^+ + x_3(S - K_3)^+$$

where $(S-K_j)^+ = \max(S-K_j,0)$ the payoff of a standard European call option.

For $0 \leqslant S \leqslant K_1$ we have $V(s) = 0$.

For $K_1 \leqslant S \leqslant K_2$ we have $V(S) = x_1(S - K_1) \geqslant 0$.

For $K_2 \leqslant S \leqslant K_3$ we have $V(S) = x_1(S - K_1) - x_2(S- K_2)$. This is a linear function joining the points $(\,K_2,\,x_1(K_2 - K_1)\,)$ and $(\,K_3,\,x_1(K_3-K_1)- x_2(K_3-K_2)\,)$.

Consequently we have $V(S) \geqslant 0$ for $K_2 \leqslant S \leqslant K_3$ if and only if

$$\tag{*}x_1(K_3-K_1)- x_2(K_3-K_2) \geqslant 0$$

For $S \geqslant K_3$ we have

$$V(S) = x_1(S- K_1) - x_2(X- K_2) + x_3(S- K_3) \\ = x_1(K_3-K_1)- x_2(K_3-K_2) +(x_1 - x_2 + x_3)(S - K_3),$$

and assuming that inequality (*) holds we have $V(S) \geqslant 0$ for all $S \geqslant K_3$ if and only if

$$\tag{**} x_1 - x_2 + x_3 \geqslant 0$$

Answer 2

Note what you state seems to require that $K_3 \le 0$. I later found that this is due to the question statement not quite being correct. Please read RRL's answer for a solution to the question as intended. Nonetheless, the following does prove part of the conditions for options, but not arbitrage.

Using this, note that

$$x_1\left(S - K_1\right) - x_2\left(S - K_2\right) + x_3\left(S - K_3\right) \ge 0 \tag{1}\label{eq1}$$

becomes after expanding the terms, collecting all those using $S$ and factoring it out to get that

$$S\left(x_1 - x_2 + x_3\right) - x_1 K_1 + x_2 K_2 - x_3 K_3 \ge 0 \tag{2}\label{eq2}$$

To show the "if" part, if

$$x_1 - x_2 + x_3 \ge 0 \tag{3}\label{eq3}$$

then

$$S\left(x_1 - x_2 + x_3\right) \ge 0 \tag{4}\label{eq4}$$

based on $S \ge 0$. With the second condition,

$$x_1\left(K_3 - K_1\right) - x_2\left(K_3 - K_2\right) \ge 0 \tag{5}\label{eq5}$$

expanding it, collecting the terms using $K_3$ and then factoring out $K_3$, gives

$$-x_1 K_1 + x_2 K_2 + K_3\left(x_1 - x_2\right) \ge 0 \tag{6}\label{eq6}$$

From \eqref{eq3}, and assuming that $K_3 \le 0$, we get

$$x_1 - x_2 \ge -x_3 \Rightarrow K_3\left(x_1 - x_2\right) \le -K_3 x_3 \tag{7}\label{eq7}$$

Reversing the inequality shows this can be substituted into \eqref{eq6} to give that

$$-x_1 K_1 + x_2 K_2 - x_3 K_3 \ge 0 \tag{8}\label{eq8}$$

Putting \eqref{eq4} and \eqref{eq8} together gives \eqref{eq2}. Note I didn't use anything regarding the relationship of $K_1$ or $K_2$ in terms of $K_3$ here. As for the requirement for $K_3 \le 0$, note that if $x_1 = x_2 = 0$, then \eqref{eq3} requires that $x_3 \ge 0$. Setting $S = 0$ causes \eqref{eq2} to become

$$0 - 0 + 0 - x_3 K_3 \ge 0 \tag{9}\label{eq9}$$

but if $x_3 \gt 0$, then $K_3 \leq 0$. As the OP has stated in a comment to this answer that

the $K$'s represent different "strike prices" which would in fact generally be greater than $0$

there must be some other conditions or restrictions I'm not aware of, or possibly one of the statements is not presented correctly. For now, this is the best I can do here.

For the "only if" part, I believe the easiest way to show it would possibly be to provide an example where one of \eqref{eq3} or \eqref{eq5} don't hold, then \eqref{eq2} doesn't hold either.