# What exactly is/How exactly do we interpret the binomial model's Radon-Nikodym derivative?

Related: Lewis' triviality result?

As I recall the one-step binomial model goes like this:

1. The time periods are now $$t=0$$ and later $$t=1$$.

2. 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 $$R>0$$

1. 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 $$P(up)=p_u$$, $$P(down)=p_d$$, $$Q(up)=q_u$$, $$Q(down)=q_d$$.

2. For no arbitrage we must have $$d < 1+R < u$$ (or $$d \le 1+R \le u$$ or something).

3. 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.)

1. 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

1. 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.

In economics, the Radon-Nikodym density $$\frac{d \mathbb Q}{d \mathbb P}$$ of the risk-neutral measure $$\mathbb Q$$ with respect to the physical measure $$\mathbb P$$ is the price of Arrow-Debreu securities. It is a price, not a claim.

In the binomial setting, there are two AD securities, $$1_u$$ and $$1_d$$. The former entitles the holder to 1 unit of numeraire if, and only if, state $$u$$ realizes. The no-arbitrage price of $$1_u$$ is $$\frac{q_u}{p_u}1_u$$ (say $$R = 1$$). Similarly for $$1_d$$.

In general, the price of the AD portfolio $$1_{\Omega'}$$---which pays off 1 unit of numeraire if, and only if, state $$\omega \in \Omega'$$ realizes---is $$E^{\mathbb P}[ 1_{\Omega'} \frac{d \mathbb Q}{d \mathbb P}] = \mathbb Q (\Omega').$$ So $$E^{\mathbb P}[\frac{d \mathbb Q}{d \mathbb P}] = 1$$ is then the price of a bond/risk-free security (discount accordingly if $$r \neq 1$$).

Extending from $$1_{\Omega'}$$ to a general $$X$$ gives the usual risk-neutral pricing formula $$E^{\mathbb Q}[X]$$, as you have written out.

• Thanks for answering 1st question. Wait so what exactly is a '$1_{\Omega'} \frac{d \mathbb Q}{d \mathbb P}$' ? That's my 2nd question. I'll edit post to be more clear
– BCLC
Jul 25 '20 at 8:17
• Edited post to hopefully be even clearer
– BCLC
Sep 14 at 4:15
• wait but actually... $1_{\Omega'} \frac{d \mathbb Q}{d \mathbb P}$ is a price not a claim again?
– BCLC
Sep 14 at 4:16