II. Module 5 Homework [final]: Arbitrage Bounds

Due No due date
Points 12
Questions 4
Time Limit None

Instructions

I always found arbitrage bounds annoying because they seemed to require you to be clever -- oh, payoff combination so and so is always greater than payoff combination such an such, so the price must be greater too. In any sort of real situation, with maybe 5-10 assets, I can't count on being clever enough to see all the possibilities. (Suppose this question were, given put and call options at 10 different strikes, find the arbitrage bounds on a given call. Fun, eh?) Nor, can I be sure I haven't left something out. Wouldn't it be nice if there were a constructive way to find arbitrage bounds?

There is. Arbitrage bounds are a linear program. And linear programs have constructive solutions.

To see the point, state the problem as a search for the discount factor: We want to look among all strictly positive (hence, arbitrage bound) discount factors which price the stock and bond to find the discount factor which generates the largest/smallest price for the option,

$LaTeX: \max_{m},~\text{or}~ \min_{m} C =E(mx^{c})$

s.t.

$LaTeX: m \geq 0$

$LaTeX: S =E(mS_{T})$

$LaTeX: 1 =E(mR^{f})$

meaning, really, that we are looking for the vector LaTeX: m(s) that maximizes or minimizes

$LaTeX: C =\sum\pi(S_{T})m(S_{T})x^{c}(S_{T})$

subject to constraints expressed similarly. Since $LaTeX: \pi(S_{T})m(S_{T})$ always enter together, risk vs risk aversion is irrelevant, and we can choose instead the risk neutral probabilities as a function $LaTeX: \pi^{\ast}(S_{T})=R^{f}\pi(S_{T})m(S_{T}).$

Writing the problem this way

$LaTeX: \min_{\pi ^{\ast }(S_{T})}, ~\text{or}~ \max_{\pi ^{\ast }(S_{T})} C =\frac{1}{R^{f}}\sum \pi ^{\ast }(S_{T})\max (S_{T}-X,0)$ s.t.

$LaTeX: \pi ^{\ast }(S_{T}) \geq 0$

$LaTeX: S =\frac{1}{R^{f}}\sum \pi ^{\ast }(S_{T})S_{T}$

$LaTeX: 1 =\sum \pi ^{\ast }(S_{T})$

With continuous states, we are looking for the function $LaTeX: \pi^*(S_{T})$ that

$LaTeX: \min_{\pi ^{\ast }(S_{T})}, ~\text{or}~ \max_{\pi ^{\ast }(S_{T})} C =\frac{1}{R^{f}}\int\pi ^{\ast }(S_{T})\max (S_{T}-X,0)dS_T$ s.t.

$LaTeX: \pi ^{\ast }(S_{T}) \geq 0$

$LaTeX: S =\frac{1}{R^{f}}\int \pi ^{\ast }(S_{T})S_{T}(S_{T})dS_{T}$

$LaTeX: 1 =\int \pi ^{\ast }(S_{T})dS_{T}$

Now look hard at what we have. You're choosing the vector/function $LaTeX: \pi^*(S_{T}),$ the objective is a linear objective, and the constraints are linear constraints. This is a linear program!

This approach also gives a useful reexpression of the problem. We can look at the risk neutral probabilities that generate arbitrage bounds. If we have a set of real probabilities (like a lognormal), we can divide the real from the risk neutral and find the discount factor, and we can at least think about whether that is a reasonable discount factor for the problem. Often, it is not -- it rises as LaTeX: S declines, it concentrates all mass on a few points, it's all ziggy zaggy, its variance implies a Sharpe ratio in the hundreds and so on. That may suggest further refinements in pricing, when you don't want to jump all the way to continuous time no-arbitrage arguments.

Ok, this is supposed to be a quiz, not a lecture.

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