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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.

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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.

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  • $\begingroup$ I don't really understand b. $\endgroup$ – Gordon Feb 12 '16 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 '16 at 2:58
  • $\begingroup$ Your part a still has some problem. I posted an answer for this part below. $\endgroup$ – Gordon Feb 12 '16 at 14:24
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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, $\pi_1+\pi_3=1/2$.

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  • $\begingroup$ Thank you by the way I attempted b but not sure if that is right $\endgroup$ – Wolfy Feb 12 '16 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 '16 at 20:58

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