Prove that $S^0=S^1$ so there is one bank account in the market without arbitrage

Question

In a two-period market with four possible scenarios, the risk-free interest rate is $10$%. There is one stock in this market whose prices are described by process $S$: $$S_0 = 100, S_1(\omega_1) = S_1(\omega_2) = 120, S_1(\omega_3) = S_1(\omega_4) = 80$$ $$S_2(\omega_1) = 140, S_2(\omega_2) = 100, S_2(\omega_3) = 100, S_2(\omega_4) = 60$$ Find the martingale measure $P^*$ on = $\Omega =\omega_1, \omega_2, \omega_3, \omega_4$.

My try: $$\begin{cases} 120p_1 +120p_2 +80p_3 + 80p_4=100*1,1 \\ 140p_1+100p_2+100p_3+60p_4=100*1,1^2 \\ p_1+p_2+p_3+p_4=1 \\ p_i \ge 0\end{cases}$$

Unfortunately I get infinitely many solutions but the answer is: $P^*(\omega_1) = 0,6, P^*(\omega_2) = 0,15, P^*(\omega_3) = 0,175, P^*(\omega_4) = 0,075$.

Can someone point out where I'm making a mistake?

Answer 1

Your mistake, ultimately, is that your second equation is meaningless. It only makes sense if nobody can sell/buy after the first period. Instead of that second equation, you should have two equations: one for $\omega_1, \omega_2$ and another for $\omega_3, \omega_4$.

• For $\omega_1, \omega_2$: This means at time $1$, the stock price is $120$. We may either keep the stock, which after one period will be either $140$ with probability $p_1 / (p_1+p_2)$ or $100$ with probability $p_2 / (p_1+p_2)$; or we may put it in the risk-free account to end with $120 \times 1,\!1$. That is, the equation is $$140p_1/(p_1+p_2) + 100p_2/(p_1+p_2) = 120 \times 1,\!1$$
• Considering $\omega_3, \omega_4$ is similar. The stock price starts at $80$ (which is how we know we're in case $\omega_3, \omega_4$), and we either keep holding the stock or we save at the risk-free rate, giving $$100p_3/(p_3+p_4) + 60p_4/(p_3+p_4) = 80\times 1,\!1$$

When everything is finished, we have the equations $$\begin{cases}120p_1+120p_2 + 80p_3 + 80p_4 = 100\times 1,\!1 \\ 140p_1/(p_1+p_2) + 100p_2/(p_1+p_2) = 120 \times 1,\!1 \\ 100p_3/(p_3+p_4) + 60p_4/(p_3+p_4) = 80\times 1,\!1 \\ p_1 + p_2 + p_3 + p_4 = 1\end{cases}$$