Related: Dumb question: is risk-neutral pricing taking conditional expectation?
- Maybe there's not quite an interpretation given Lewis' triviality result if $E^Q[X]$ is a real world conditional expectation like $E^Q[X] = E^P[X|\text{something}]$ . Not sure.
As I recall the one-step binomial model goes like this:
The time periods are now $t=0$ and later $t=1$.
We have
2.1. a stock that pays off $u$ for going up or $d$ for going down for an initial investment of $S_0$, the price of 1 unit of the stock, where $u > d > 0$ and $S_0 > 0$.
2.2. a claim $X$ on the stock (where $X$ is, say, a European call option or something). The payoffs are $X_u$ for up and $X_d$ for down for an initial investment of 1 unit of the claim.
2.3. and a bond that pays $1+R$ for an investment of $1$ ($R$ is rate of return right?), for $R>0$
We assume both going up and going down have positive probability. We have real world probability $\mathbb P$ and risk-neutral probability $\mathbb Q$. The probabilities are $P(up)=p_u$, $P(down)=p_d$, $Q(up)=q_u$, $Q(down)=q_d$.
For no arbitrage we must have $d < 1+R < u$ (or $d \le 1+R \le u$ or something).
From no arbitrage we can compute $q_u$ and $q_d$ in terms of $u,d,R$ and then we don't need $p_u$ and $p_d$ except for the assumption that both real world probabilities are positive or something. (and then $q_u$ and $q_d$ are positive too or something.)
- Question 1: What exactly is/how exactly do we interpret the Radon-Nikodym derivative $\frac{d \mathbb Q}{d \mathbb P}$ ?
- 1.1. Its formula/equation/whatever appears to be $\frac{d \mathbb Q}{d \mathbb P} = \frac{q_u}{p_u}1_u + \frac{q_d}{p_d}1_d$, so it's some asset with payoffs $\frac{q_u}{p_u}$ and $\frac{q_d}{p_d}$, expected value of 1 and replicating portfolio $(x,y)$ of
$$x=\frac{1}{1+R}\frac{u(\frac{q_d}{p_d})-d(\frac{q_u}{p_u})}{u-d}$$ $$y=\frac{1}{S_0}\frac{\frac{q_u}{p_u}-\frac{q_d}{p_d}}{u-d}$$
This seems to be some replicating portfolio that is expected to payoff 1 at $t=1$.
1.2. Well, $(x,y)=(\frac{1}{1+R},0)$ seems to give the same payoff but with -100% lower risk
1.3. Guess: It's a hypothetical stock that pays off $\frac{q_u}{p_u}$ for up and $\frac{q_d}{p_d}$ for down. In particular, what's so hypothetical about this is that we don't necessarily know $p_u$ and $p_d$.
1.4. Guess: In re (2) below, I think $\frac{d \mathbb Q}{d \mathbb P}$ is like a specific case of $X\frac{d \mathbb Q}{d \mathbb P}$ with $X_u=1=X_d$ so like...we choose $X$ as like...a bond with rate of return $0$? idk
- For $X\frac{d \mathbb Q}{d \mathbb P}$ in the one-step binomial model...
we could say that the price of $X$ uses not
$$E[X] = X_up_u+X_dp_d$$
but rather
$$E^{\mathbb Q}[X] = E[X\frac{d \mathbb Q}{d \mathbb P}] = X_u\frac{q_u}{p_u}p_u + X_d\frac{q_d}{p_d}p_d$$
Question 2: So what exactly is/how exactly do we interpret '$X\frac{d \mathbb Q}{d \mathbb P}$', i.e. $X$ multiplied by the Radon-Nikodym derivative $\frac{d \mathbb Q}{d \mathbb P}$ ?
2.1. Its formula/equation/whatever appears to be $\frac{d \mathbb Q}{d \mathbb P} = X_u\frac{q_u}{p_u}1_u + X_d\frac{q_d}{p_d}1_d$, so it's some asset with payoffs $X_u\frac{q_u}{p_u}$ and $X_d\frac{q_d}{p_d}$
2.2. Not sure what's its replicating portfolio. Not sure we need one since we're using real world probabilities.
2.3. Its real world expected payoff is equal to $X$'s risk neutral expected payoff.
2.4. Guess: Once again, it's a hypothetical stock but this time the payoffs are $X_u\frac{q_u}{p_u}$ for up and $X_d\frac{q_d}{p_d}$ for down. Again, in particular, what's so hypothetical about this is that we don't necessarily know $p_u$ and $p_d$. So if the option/claim price is $\frac{1}{1+R}E^{\mathbb Q}[X]$, then I could tell you to, instead of buying the claim or investing in its replicating portfolio, buy a stock that will payoff $X_u\frac{q_u}{p_u}$ for up and $X_d\frac{q_d}{p_d}$ for down and has an initial price of $\frac{1}{1+R}E^{\mathbb Q}[X]$ or something. The thing is we don't necessarily know $p_u$ and $p_d$, so we can't quite identify a similar stock.
