Risk-Neutral Probabilities in a Two-Asset Arrow-Debreu Model
Summary
The document considers a one-period, four-state Arrow-Debreu model with two assets and zero interest. It derives constraints on risk-neutral probabilities by equating each asset’s current price to its expected payoff under those probabilities. The resulting equations require half the probability mass to fall in each pair of states associated with an asset’s payoff, with a third equation linking the states in which the first asset pays its higher value.
For the independence question, the original attempt incorrectly treats independence as linear independence of payoff vectors. The document does not provide a valid probability choice or a completed argument for independence. Its worked derivation for the probability constraints also relies on comparing coefficients across asset payoff values, which should be distinct for that step to identify the sums. Thus, the material is useful for illustrating state-price constraints, but leaves the key independence task unresolved.
Key ideas
- Risk-neutral probabilities make discounted expected payoffs equal current asset prices.
- The two asset pricing equations impose constraints on sums of state probabilities.
- The probability constraints alone do not establish statistical independence of asset payoffs.
- The proposed independence argument confuses independence of random variables with linear independence of vectors.
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# Arrow-Debreu Model and Risk-Neutral Probabilities
# 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.
## Answer by Gordon (score 1, accepted)
https://quant.stackexchange.com/a/23257
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, $\pi_1+\pi_3=1/2$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.