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

• 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:

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.

• Why not just ask a new question? Sep 1 '21 at 13:44
• @BobJansen editing will bump right? or do you mean asking a new question brings more attention than editing an old question?
– BCLC
Sep 1 '21 at 13:56
• 1. Yes, but that’s a bad reason to edit. 2. A new question, which this seems to be, will also be on top? Sep 1 '21 at 13:58
• @BobJansen i've never heard of this kind of practice or rule. is this dependent on the vote total? i imagine i can delete and re-ask questions if my question has a negative score for instance just to avoid the downvotes
– BCLC
Sep 14 '21 at 3:52
• You have 120 accounts on StackExchange, you should know better. These are the guidelines for editing. Your edit clearly doesn't conform with this guideline. I see that you've added a bounty which I don't want to let go to waste so in this particular instance I wont rollback but rolling back would be the right thing to do here. Sep 14 '21 at 5:57