Background Information:

This question follows from here

It is tempting to write $$V_0(X) = \beta\left[\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}\right)X(u) + \left(\frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}\right)X(d)\right]$$ as

$$V_0(X) = E_Q[\beta X]$$ where the expectation is taken with respect to the new purely formal probability measure $Q$ defined by $$Q(u) = \frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}$$ and $$Q(d) = \frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}$$

Note that $Q(u) + Q(d) = 1$; $Q$ will be a probability measure provided these values are non-negative.


Show why $ \ Q(u),Q(d) \geq 0$

My reasoning:

Suppose $$Q(u) = \frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)} < 0$$ then either $\beta^{-1}S_0 - S_1(d) > 0$ and $S_1(u) - S_1(d) < 0$ or $\beta^{-1}S_0 - S_1(d) < 0$ and $S_1(u) - S_1(d) > 0$. In either case there would be an arbitrage opportunity.

Consider the first case where $$\beta^{-1}S_0 - S_1(d) > 0 \ \ \text{and} \ \ S_1(u) - S_1(d) < 0$$ Then I believe we would short $\beta^{-1}S_0$ and use the proceeds to go long in a bond.

I am not sure if this is sort of the correct reasoning I need to fulfill the question. Any suggestions are greatly appreciated.


1 Answer 1


You've said it, $Q$ will be a probability measure only if $Q(u)$ and $Q(d)$ are non-negative. This only happens in the absence of arbitrage opportunities.

For instance, if $Q(u)<0$, then either $\beta^{-1}S_0 > S_1(d) > S_1(u)$ or $ S_1(u) > S_1(d) > \beta^{-1} S_0$. In other words, either the risk-free asset strictly outperforms the risky one in all states of the world or the very opposite situation. In any case, this constitutes an arbitrage opportunity since both assets share the same initial price $S_0$. Shorting the under-performing one to go long the other is a zero cost strategy with a postive payout in all states of the world, hence the static arbitrage opportunity. You can use a similar rationale to show that $Q(d)$ cannot be negative in the absence of arbitrage.

Now, if you take the argument from the beginning it gives you that: assuming there is no arbitrage opportunity (which is precisely what allowed you to write the first expression you state for $V_0(x)$), then, there exists a probability measure $Q$ such that one can write $V_0(X) = \Bbb{E}^Q[\beta X]$.

  • $\begingroup$ Just curious, but why did you say "hence the static arbitrage". What do you mean by static? $\endgroup$
    – Wolfy
    Oct 13, 2016 at 2:33
  • $\begingroup$ @Wolfy this will help: quant.stackexchange.com/questions/19035/… $\endgroup$
    – Quantuple
    Oct 13, 2016 at 6:27
  • $\begingroup$ For your last statement, are you essentially verifying why we can write $V_0(X) = E^{Q}[\beta X]$? $\endgroup$
    – Wolfy
    Oct 13, 2016 at 21:12

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