In the one-step binomial model...

For $\frac{d \mathbb Q}{d \mathbb P}$,

I think it's $\frac{d \mathbb Q}{d \mathbb P} = \frac{q_u}{p_u}1_u + \frac{q_d}{p_d}1_d$, 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 $t=1$.

Question: Any intuition?

Well, $(x,y)=(\frac{1}{1+R},0)$ seems to give the same payoff but with -100% lower risk

For $X\frac{d \mathbb Q}{d \mathbb P}$,

we could say that the price of X uses not

$$E[X] = X_up_u+X_dp_d$$

but rather

$$E^{\mathbb Q}[X] = E[X\frac{d \mathbb Q}{d \mathbb P}] = X_u\frac{q_u}{p_u}p_u + \frac{q_d}{p_d}p_d$$

Question: So what's a '$X\frac{d \mathbb Q}{d \mathbb P}$'?

It pays $X_u\frac{q_u}{p_u}$ or $X_d\frac{q_d}{p_d}$ then replicating portfolio is...then idk. Not sure we need one since we're using real world probabilities anyhoo


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.


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