<p>In economics, the Radon-Nikodym density <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span> of the risk-neutral measure <span class="math-container">$\mathbb Q$</span> with respect to the physical measure <span class="math-container">$\mathbb P$</span> is the <strong>price of Arrow-Debreu securities</strong>. It is a <em>price</em>, not a claim. </p>

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

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

<p>Extending from <span class="math-container">$1_{\Omega'}$</span> to a general <span class="math-container">$X$</span> gives the usual risk-neutral pricing formula <span class="math-container">$E^{\mathbb Q}[X]$</span>, as you have written out.</p>

CC BY-SA 4.0
2020-07-25T08:17:42.293Z
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
CC BY-SA 4.0
2021-09-14T04:15:04.837Z
Edited post to hopefully be even clearer
BCLC
CC BY-SA 4.0
2021-09-14T04:16:19.857Z
wait but actually... $1_{\Omega'} \frac{d \mathbb Q}{d \mathbb P}$ is a price not a claim again?
BCLC
CC BY-SA 4.0
2019-07-09T23:49:51.060Z
2019-07-09T23:49:51.060Z
Michael
<p>Related: <a href="https://en.wikipedia.org/wiki/Lewis%27s_triviality_result" rel="nofollow noreferrer">Lewis' triviality result</a>?</p>
<hr />
<p>As I recall the one-step binomial model goes like this:</p>
<ol>
<li><p>The time periods are now <span class="math-container">$t=0$</span> and later <span class="math-container">$t=1$</span>.</p>
</li>
<li><p>We have</p>
</li>
</ol>
<ul>
<li><p>2.1. a stock that pays off <span class="math-container">$u$</span> for going up or <span class="math-container">$d$</span> for going down for an initial investment of <span class="math-container">$S_0$</span>, the price of 1 unit of the stock, where <span class="math-container">$u &gt; d &gt; 0$</span> and <span class="math-container">$S_0 &gt; 0$</span>.</p>
</li>
<li><p>2.2. a claim <span class="math-container">$X$</span> on the stock (where <span class="math-container">$X$</span> is, say, a European call option or something). The payoffs are <span class="math-container">$X_u$</span> for up and <span class="math-container">$X_d$</span> for down for an initial investment of 1 unit of the claim.</p>
</li>
<li><p>2.3. and a bond that pays <span class="math-container">$1+R$</span> for an investment of <span class="math-container">$1$</span> (<span class="math-container">$R$</span> is rate of return right?), for <span class="math-container">$R&gt;0$</span></p>
</li>
</ul>
<ol start="3">
<li><p>We assume both going up and going down have positive probability. We have real world probability <span class="math-container">$\mathbb P$</span> and risk-neutral probability <span class="math-container">$\mathbb Q$</span>. The probabilities are <span class="math-container">$P(up)=p_u$</span>, <span class="math-container">$P(down)=p_d$</span>, <span class="math-container">$Q(up)=q_u$</span>, <span class="math-container">$Q(down)=q_d$</span>.</p>
</li>
<li><p>For no arbitrage we must have <span class="math-container">$d &lt; 1+R &lt; u$</span> (or <span class="math-container">$d \le 1+R \le u$</span> or something).</p>
</li>
<li><p>From no arbitrage we can compute <span class="math-container">$q_u$</span> and <span class="math-container">$q_d$</span> in terms of <span class="math-container">$u,d,R$</span> and then we don't need <span class="math-container">$p_u$</span> and <span class="math-container">$p_d$</span> except for the assumption that both real world probabilities are positive or something. (and then <span class="math-container">$q_u$</span> and <span class="math-container">$q_d$</span> are positive too or something.)</p>
</li>
</ol>
<hr />
<ol>
<li><strong>Question 1:</strong> What exactly is/how exactly do we interpret the Radon-Nikodym derivative <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span> ?</li>
</ol>
<ul>
<li>1.1. Its formula/equation/whatever appears to be <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P} = \frac{q_u}{p_u}1_u + \frac{q_d}{p_d}1_d$</span>, so it's some asset with payoffs <span class="math-container">$\frac{q_u}{p_u}$</span> and <span class="math-container">$\frac{q_d}{p_d}$</span>, expected value of 1 and replicating portfolio <span class="math-container">$(x,y)$</span> of</li>
</ul>
<p><span class="math-container">$$x=\frac{1}{1+R}\frac{u(\frac{q_d}{p_d})-d(\frac{q_u}{p_u})}{u-d}$$</span>
<span class="math-container">$$y=\frac{1}{S_0}\frac{\frac{q_u}{p_u}-\frac{q_d}{p_d}}{u-d}$$</span></p>
<p>This seems to be some replicating portfolio that is expected to payoff 1 at <span class="math-container">$t=1$</span>.</p>
<ul>
<li><p>1.2. Well, <span class="math-container">$(x,y)=(\frac{1}{1+R},0)$</span> seems to give the same payoff but with -100% lower risk</p>
</li>
<li><p>1.3. Guess: It's a hypothetical stock that pays off <span class="math-container">$\frac{q_u}{p_u}$</span> for up and <span class="math-container">$\frac{q_d}{p_d}$</span> for down. In particular, what's so hypothetical about this is that we don't necessarily know <span class="math-container">$p_u$</span> and <span class="math-container">$p_d$</span>.</p>
</li>
<li><p>1.4. Guess: In re (2) below, I think <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span> is like a specific case of <span class="math-container">$X\frac{d \mathbb Q}{d \mathbb P}$</span> with <span class="math-container">$X_u=1=X_d$</span> so like...we choose <span class="math-container">$X$</span> as like...a bond with rate of return <span class="math-container">$0$</span>? idk</p>
</li>
</ul>
<hr />
<ol start="2">
<li>For <span class="math-container">$X\frac{d \mathbb Q}{d \mathbb P}$</span> in the one-step binomial model...</li>
</ol>
<p>we could say that the price of <span class="math-container">$X$</span> uses not</p>
<p><span class="math-container">$$E[X] = X_up_u+X_dp_d$$</span></p>
<p>but rather</p>
<p><span class="math-container">$$E^{\mathbb Q}[X] = E[X\frac{d \mathbb Q}{d \mathbb P}] = X_u\frac{q_u}{p_u}p_u + X_d\frac{q_d}{p_d}p_d$$</span></p>
<p><strong>Question 2:</strong> So what exactly is/how exactly do we interpret '<span class="math-container">$X\frac{d \mathbb Q}{d \mathbb P}$</span>', i.e. <span class="math-container">$X$</span> multiplied by the Radon-Nikodym derivative <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span> ?</p>
<ul>
<li><p>2.1. Its formula/equation/whatever appears to be <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P} = X_u\frac{q_u}{p_u}1_u + X_d\frac{q_d}{p_d}1_d$</span>, so it's some asset with payoffs <span class="math-container">$X_u\frac{q_u}{p_u}$</span> and <span class="math-container">$X_d\frac{q_d}{p_d}$</span></p>
</li>
<li><p>2.2. Not sure what's its replicating portfolio. Not sure we need one since we're using real world probabilities.</p>
</li>
<li><p>2.3. Its real world expected payoff is equal to <span class="math-container">$X$</span>'s risk neutral expected payoff.</p>
</li>
<li><p>2.4. Guess: Once again, it's a hypothetical stock but this time the payoffs are <span class="math-container">$X_u\frac{q_u}{p_u}$</span> for up and <span class="math-container">$X_d\frac{q_d}{p_d}$</span> for down. Again, in particular, what's so hypothetical about this is that we don't necessarily know <span class="math-container">$p_u$</span> and <span class="math-container">$p_d$</span>. So if the option/claim price is <span class="math-container">$\frac{1}{1+R}E^{\mathbb Q}[X]$</span>, then I could tell you to, instead of buying the claim or investing in its replicating portfolio, buy a stock that will payoff <span class="math-container">$X_u\frac{q_u}{p_u}$</span> for up and <span class="math-container">$X_d\frac{q_d}{p_d}$</span> for down and has an initial price of <span class="math-container">$\frac{1}{1+R}E^{\mathbb Q}[X]$</span> or something. The thing is we don't necessarily know <span class="math-container">$p_u$</span> and <span class="math-container">$p_d$</span>, so we can't quite identify a similar stock.</p>
</li>
</ul>

<mathematical-economics><finance><probability><asset-pricing><stochastic-calculus>
What exactly is/How exactly do we interpret the binomial model's Radon-Nikodym derivative?
BCLC
# What exactly is/How exactly do we interpret the binomial model's Radon-Nikodym derivative?

(Asked by: username_0 on 2018-03-26)

Related: [Lewis' triviality result](https://en.wikipedia.org/wiki/Lewis%27s_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 $R>0$

3. 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 $P(up)=p\_u$, $P(down)=p\_d$, $Q(up)=q\_u$, $Q(down)=q\_d$.
4. For no arbitrage we must have $d < 1+R < u$ (or $d \le 1+R \le u$ or something).
5. 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

---

2. 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.

**Tags:** mathematical-economics, finance, probability, asset-pricing, stochastic-calculus

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## 1 answers

### Answer by username_1 on 2019-07-09. Score: 2

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.

> **Comments:**
> **username_0**: 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  
> **username_0**: Edited post to hopefully be even clearer  
> **username_0**: wait but actually... $1_{\Omega'} \frac{d \mathbb Q}{d \mathbb P}$ is a price not a claim again?  

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