Dumb question: is risk-neutral pricing taking conditional expectation? $\tag{1}$

In trying to recall intuition for risk-neutral pricing, I think I read that we should price derivatives risk-neutrally because the risk is already incorporated in the stock or something. I also think I remember NNT/Nero (Nassim Nicholas Taleb) saying something about how certain information is irrelevant in the expected price of oil if the information is public.

This made me think of risk-neutral pricing in terms of conditional expectation.

Simple I guess but it wasn't discussed in classes since conditional expectation and Radon-Nikodym was taught after one period model.

From what I recall of the one period model:

$$(\Omega, \mathscr F, \mathbb P) = ((u,d),2^\Omega, \text{real world}))$$ Bonds: $$\{B_t\}$$ $$B_0=1, B_1 = 1+R$$ Stocks: $$\{S_t\}$$ $$S_0 \in (0,\infty)$$ $$\mathbb P(S_1(u) = S_0u) = p_u > 0$$ $$\mathbb P(S_1(d) = S_0d) = p_d = 1 - p_u$$ European call option: $X$ $$X(u) = S_1(u) - K$$ $$X(d) = 0$$ $$\text{Price process:} \ \{\Pi(X,t)\}$$

where $t=0,1, u > 1+R > d > 0$.

It can be shown that

$$\Pi(X,0) = \frac{1}{1+R}E^{\mathbb Q}[X] = \frac{1}{1+R}(q_uX(u) + q_dX(d))$$

where $q_u, q_d$ are the risk-neutral probabilities under $\mathbb Q$, equivalent to $\mathbb P$

Also, I think $\sigma(S_1) = \sigma(X) = \{\emptyset, \Omega, \{u\}, \{d\}\}$

Dumb question rephrased:

$$\exists Z \in \mathscr L^1(\Omega, \mathscr F, \mathbb P) \ \text{s.t.} \ E^{\mathbb Q}[X] = E^{\mathbb P}[X|Z]? \tag{2}$$

Well, the left hand side is a constant while the right hand side a random variable so I'm not sure that that would make sense

How about

$$\exists Z \in \mathscr L^1(\Omega, \mathscr F, \mathbb P) \ \text{s.t.} \ E^{\mathbb Q}[X] = E^{\mathbb Q}[E^{\mathbb P}[X|Z]] \tag{3}?$$

  • For $(2)$,

  • One thing I did:

  1. $E^{\mathbb Q}[X]$ is constant and thus $Z-$measurable $\forall \ Z \in \mathscr L^1(\Omega, \mathscr F, \mathbb P)$

  2. $$\int_z E^{\mathbb Q}[X] d \mathbb P = \int_z X d \mathbb P \ \forall \ z \ \in \ \sigma(Z)$$

$$\iff E[E^{\mathbb Q}[X]1_z] = E[X1_z] \ \forall \ z \ \in \ \sigma(Z)$$ $$\iff E^{\mathbb Q}[X]E[1_z] = E[X1_z] \ \forall \ z \ \in \ \sigma(Z)$$ $$\iff E^{\mathbb Q}[X]\mathbb P(z) = E[X1_z] \ \forall \ z \ \in \ \sigma(Z)$$ $$\iff \mathbb P(z) = \frac{E[X1_z]}{E^{\mathbb Q}[X]} \ \forall \ z \ \in \ \sigma(Z)$$

There doesn't seem to be such a $Z$.

  • Another thing I did: Well, I did think of Radon-Nikodym (duh)

$$E^{\mathbb Q}[X] = E^{\mathbb P}[X \frac{d \mathbb Q}{d \mathbb P}]$$.

I guess $Z = \frac{d \mathbb Q}{d \mathbb P}$ otherwise not sure how that's relevant but I guess since $$\mathbb Q(z) = \int_z \frac{d \mathbb Q}{d \mathbb P} d \mathbb P \ \forall z \in \sigma(Z) \subseteq 2^{\Omega}$$,

$\frac{d \mathbb Q}{d \mathbb P}$ is a version of $E[\frac{d \mathbb Q}{d \mathbb P} | Z]$

  • Gee how informative. Well, I think $\sigma(Z)$ can be only either $\{\emptyset, \Omega\}$, in which case the $Z$ is any (almost surely?) constant random variable or $2^{\Omega} = \sigma(X) = \sigma(S_1)$, in which case $q_u = 1_{u}$ which would make sense iff $q_u$ is degenerate, which I guess violates equivalence assumption.

  • For $(3)$,

I guess $Z=S_1$? I'm not sure what that says. I was kinda expecting (lol) that real world probabilities $E[1_A]$ and risk-neutral probabilities $E[1_A | B] = \frac{E[1_A1_B]}{E[1_B]}$ would be like, respectively, prior $P(A)$ and posterior $P(A|B)$ probabilities.

Edit: $$\frac{d \mathbb Q}{d \mathbb P} = \frac{q_u}{p_u}1_{u} + \frac{q_d}{p_d}1_{d}$$ ?

Based on Section 4.5 of Etheridge's A Course in Financial Calculus, I guess

$$\frac{d \mathbb Q}{d \mathbb P} = (\frac{q_u}{p_u})^u(\frac{q_d}{p_d})^{1-u}$$

This avoids indicator functions in favour of exponents as in the binomial theorem, binomial model or binomial distribution.

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2 Answers 2


Risk neutral pricing is not really related to the real probability $\mathbb P$. Instead you could say the price at time $k$ is the conditional expectation of the future payoff given the history up to time $k$: so $V_k=E(V_n\mid S_1,\dots, S_k)$ (in the zero interest case).

Risk-neutral pricing boils down to the following.

In the one-period binomial model, it is guaranteed that the value of the option $V_1$ at time 1 will be a linear function of the value of the stock at time 1, $S_1$. That's just because between any two points $(S_1(H),V_1(H))$ and $(S_1(T),V_1(T))$ in the plane there can be drawn a straight line. So $$V_1=\alpha S_1+\beta$$ for constant $\alpha$, $\beta$.

Now we assume that the stated price of the stock at time 0, $S_0$, is the correct one. (So we're not pondering the possibility that maybe the stock is ill-priced.)

Then by the principle that

the value now of two cars tomorrow is twice the value now of one car tomorrow,

(so we're of course just trading these cars, not driving them... i.e. not associating any utility with them) we should have $$V_0=\alpha S_0+\beta$$ (well, in the interest zero case). That's it -- now you have priced the option and found the price to be $V_0$.

Since the above was just a cold-blooded analysis of the correct price, we're not showing any positive or negative attitude towards risk ... so it's risk-neutral.

  • $\begingroup$ So what's the relationship with conditional expectation please and thanks? $\endgroup$
    – BCLC
    Commented Jan 9, 2018 at 5:56
  • 1
    $\begingroup$ thanks Bjørn Kjos-Hanssen 1. include in answer? (i'll think about it more later) 2. any comments here pls? (comment first then downvote! hehe) $\endgroup$
    – BCLC
    Commented Jan 11, 2018 at 23:22
  • $\begingroup$ ? quant.stackexchange.com/questions/37548/… $\endgroup$
    – BCLC
    Commented Sep 20, 2021 at 21:55

Trying to answer my question.

3 incoherently poorly presented goals:

Goal 1: Why $E^{\mathbb Q}[X]$ and not $E^{\mathbb P}[X]$ in computing price, in terms of conditional expectation?

Consider the conditional expectation $E^{\mathbb P}[X|\mathscr G]$ where $\mathscr G \subseteq \mathscr F$. Then, we have 2 cases:

  1. $\mathscr G = \mathscr G_0 := \{\emptyset, \Omega\} \to E^{\mathbb Q}[E^{\mathbb P}[X|\mathscr G]] = E^{\mathbb P}[X]$
  2. $\mathscr G = \mathscr F \to E^{\mathbb Q}[E^{\mathbb P}[X|\mathscr G]] = E^{\mathbb Q}[X]$

We observe that $E^{\mathbb P}[X]$ is double expectation with $\mathscr G = \mathscr G_0$, which we must not use because...idk.

Goal 2: What's $\frac{d \mathbb Q}{d \mathbb P}$?

It seems possible candidates are $\frac{d \mathbb Q}{d \mathbb P} = A1_u + B1_d$ that satisfy $E[X \frac{d \mathbb Q}{d \mathbb P}] = X_uq_u+X_dq_d$. Writing $X=X_u1_u+X_d1_d$, I think we can try $A = \frac{q_u}{p_u}$ and $B = \frac{q_d}{p_d}$

Goal 3: Relate $\frac{d \mathbb Q}{d \mathbb P}$ to conditional expectation:

Not sure. $\frac{d \mathbb Q}{d \mathbb P}$ is supposed to be a version of some conditional expectation. Here's what I got so far:

  • $X$ is a version of $E[X|\mathscr F]$
  • $E[X]$ is a version of $E[X|\mathscr F_0]$
  • $E^{\mathbb Q}[X]$ is a version of $E^{\mathbb Q}[X|\mathscr F_0]$

But is $\frac{d \mathbb Q}{d \mathbb P}$ a version of something?

Besides $E[\frac{d \mathbb Q}{d \mathbb P}|\mathscr F]$ and $E^{\mathbb Q}[\frac{d \mathbb Q}{d \mathbb P}|\mathscr F]$ i guess

  • 3
    $\begingroup$ It may be best to keep your question and answer together as just one focused question... that fits the StackExchange Q&A format best. People are more interested in "answering a question" than "grading an essay" :) $\endgroup$ Commented Jan 12, 2018 at 0:00
  • $\begingroup$ @BjørnKjos-Hanssen do you know Cunningham's Law? $\endgroup$
    – BCLC
    Commented Sep 20, 2021 at 21:55
  • 1
    $\begingroup$ Thanks, I've noticed somethnig similar without knowing the name. Often the best way to find how to do something is to find a question where the asker already knows that, but wants to do more (and they're often not asking a good question or getting good answers!). $\endgroup$ Commented Sep 21, 2021 at 19:09

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