# Arrow-Debreu Model and Risk-Neutral Probabilities

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:

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.

• I don't really understand b. – Gordon Feb 12 '16 at 2:56
• 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? – Wolfy Feb 12 '16 at 2:58
• Your part a still has some problem. I posted an answer for this part below. – Gordon Feb 12 '16 at 14:24

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, $\pi_1+\pi_3=1/2$.