Consider one period Arrow-Debreu model with $N = 2$ and $M = 4$ shown in Figure 3.5 and take $R = 0$.

a.) Show that any risk neutral probability $\hat{\pi} = (\hat{\pi}_1, \hat{\pi}_2, \hat{\pi}_3, \hat{\pi}_4)$ satisfies $$\begin{cases} \hat{\pi}_1 + \hat{\pi}_2 = \frac{1}{2}\\ \hat{\pi}_3 + \hat{\pi}_4 = \frac{1}{2}\\ \hat{\pi}_1 + \hat{\pi}_3 = \frac{1}{2}\\ \end{cases}$$ b.) Recall the notion of independent random variables. Find a risk neutral probability that makes the random variables of the price of two assets independent.

Figure: enter image description here

Attempted solution for a.) $$p_1 = \frac{D_1 + D_2}{2} = D_1\pi_1 + D_2\pi_2 + D_1\pi_3 + D_2\pi_4$$ So, $$p_1 = \frac{D_1 + D_2}{2} = D_1(\pi_1 + \pi_3) + D_2(\pi_2 + \pi_4)$$ Now for $p_2$ we have $$p_2 = \frac{D_1 + D_2}{2} = D_1\pi_1 + D_1\pi_2 + D_2\pi_3 + D_2\pi_4$$ and so $$p_2 = \frac{D_1 + D_2}{2} = D_1(\pi_1 + \pi_2) + D_2(\pi_3 + \pi_4)$$ by comparing the coefficients for $D_1$ and $D_2$ and given that $R = 0$ implies that $\hat{\pi} = \pi$ we can deduce that $$\begin{cases} \hat{\pi}_1 + \hat{\pi}_2 = \frac{1}{2}\\ \hat{\pi}_3 + \hat{\pi}_4 = \frac{1}{2}\\ \hat{\pi}_1 + \hat{\pi}_3 = \frac{1}{2}\\ \end{cases}$$

Not sure if this is rigorous enough, any suggestions is greatly appreciated.

Attempt for b.) Assume $p_1$ and $p_2$ are dependent. Then the vector set $D_1$ and $D_2$ contains only the zero vector by definition of dependence. But this is a contradiction to the fact that $$\begin{cases} \hat{\pi}_1 + \hat{\pi}_2 = \frac{1}{2}\\ \hat{\pi}_3 + \hat{\pi}_4 = \frac{1}{2}\\ \hat{\pi}_1 + \hat{\pi}_3 = \frac{1}{2}\\ \end{cases}$$ Hence $p_1$ and $p_2$ must be independent.

  • $\begingroup$ I don't really understand b. $\endgroup$
    – Gordon
    Feb 12, 2016 at 2:56
  • $\begingroup$ That makes both of us I am trying to work on the Markovian option I posted by back solving right now, know anything about those? $\endgroup$
    – Wolfy
    Feb 12, 2016 at 2:58
  • $\begingroup$ Your part a still has some problem. I posted an answer for this part below. $\endgroup$
    – Gordon
    Feb 12, 2016 at 14:24

1 Answer 1


For part a). As you posted, \begin{align*} (\pi_1+\pi_2)D_1 + (\pi_3+\pi_4)D_2 = \frac{D_1+D_2}{2}.\tag{1} \end{align*} Moreover, \begin{align*} \pi_3+\pi_4 = 1 - (\pi_1+\pi_2).\tag{2} \end{align*} Then \begin{align*} (\pi_1+\pi_2)(D_1-D_2)=\frac{D_1-D_2}{2}. \end{align*} That is, $\pi_1+\pi_2=1/2$. Similarly, $\pi_1+\pi_3=1/2$.

  • $\begingroup$ Thank you by the way I attempted b but not sure if that is right $\endgroup$
    – Wolfy
    Feb 12, 2016 at 20:48
  • $\begingroup$ I do not really know b) and not in a position to answer. To prove independence, you need to show the probability of the joint event is the product of the individual probability for all cases. You may deselect my answer to wait for some others. $\endgroup$
    – Gordon
    Feb 12, 2016 at 20:58

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