# What is the difference between state prices and stochastic discount factor?

I was reading a paper on arbitrage and it was mentioned that a positive SDF implies no arbitrage and later on it said that positive state prices imply no arbitrage. I am new to this topic and i am confused with the concept.

The two are very similar. To understand the difference notice that given a discrete sample space $\Omega=\{\omega_1,\omega_2…\omega_S\}$ the price of any payoff can be computed if we define the state prices $q$ (or prices of Arrow-Debreu securities,i.e., securities that pay 1 in one state and 0 in all other states. For instance $q_i(\omega_i)=1$ and $q_i(\omega_j)=0$ for $i\neq j$), as $$P_t(X_{t+1})=\sum_{s=1}^S q_s X_{t+1}(\omega_s)$$ If we multiply and divide each state price by the physical probabilities $p(\omega)$ we obtain $$P_t(X_{t+1})=\sum_{s=1}^S p(\omega_s)\frac{q_s}{p(\omega_s)} X_{t+1}(\omega_s)\equiv \sum_{s=1}^S p(\omega_s)m_{t+1}(\omega_s) X_{t+1}(\omega_s)=E^p[m_{t+1}X_{t+1}]$$ where $m_{t+1}(\omega)$ is the stochastic discount factor. The relationship between state prices and SDF is, therefore, $q_s=p(\omega_s)m_{t+1}(\omega_s).$
• By the way, one thing that I always found very fascinating is that we are able to recovery $q$ from options’ butterfly spreads, but given the fact that $q=pm$ we are not yet able to have a model-free way to distinguish $p$ from $m$, e.g. if q=0.4 both p=0.2 and m=2 and p=0.4 and m=1, and infinitely many others, work. – fni Dec 5 '16 at 8:40