Solution: Equivalent Measures and Lebesgue's Decomposition

Exercise: Equivalent Measures and Lebesgue's Decomposition

Part 1

$Z = \exp(\theta X - \theta^2/2) > 0$ everywhere on $\mathbb{R}$ since the exponential is positive. Hence $\mathbb{Q}(A) = \mathbb{E}^{\mathbb{P}}[Z\mathbf{1}_A] = 0$ iff $\mathbb{P}(A) = 0$ (as $Z > 0$ $\mathbb{P}$ -a.s.). So $\mathbb{P}$ and $\mathbb{Q}$ have the same null sets: $\mathbb{P} \sim \mathbb{Q}$ .

$d\mathbb{P}/d\mathbb{Q} = 1/Z = \exp(-\theta X + \theta^2/2)$ . Under $\mathbb{Q}$ , $X \sim \mathcal{N}(\theta, 1)$ :

\mathbb{E}^{\mathbb{Q}}[1/Z] = \mathbb{E}^{\mathbb{Q}}[e^{-\theta X + \theta^2/2}] = e^{\theta^2/2}\cdot \mathbb{E}^{\mathbb{Q}}[e^{-\theta X}] = e^{\theta^2/2}\cdot e^{-\theta\cdot\theta + \theta^2/2} = e^{\theta^2/2 - \theta^2/2} = 1. \quad \checkmark

(Used $\mathbb{E}^{\mathbb{Q}}[e^{-\theta X}] = e^{-\theta\mu_\mathbb{Q} + \theta^2\sigma_\mathbb{Q}^2/2} = e^{-\theta^2 + \theta^2/2} = e^{-\theta^2/2}$ .)

Part 2

$\mathbb{P} = \text{Uniform}[0, 1]$ — supported on $[0, 1]$ . $\mathbb{Q} = \text{Uniform}[0.5, 1.5]$ — supported on $[0.5, 1.5]$ .

$\mathbb{Q} \ll \mathbb{P}$ ? Take $A = \{1.3\}$ (just a singleton outside $[0, 1]$ , or more substantively $A = (1, 1.5]$ — support of $\mathbb{Q}$ but null under $\mathbb{P}$ ). $\mathbb{P}(A) = 0$ but $\mathbb{Q}(A) = 0.5 \ne 0$ . So $\mathbb{Q}$ is not absolutely continuous w.r.t. $\mathbb{P}$ .
$\mathbb{P} \ll \mathbb{Q}$ ? Take $A = [0, 0.5)$ . $\mathbb{Q}(A) = 0$ but $\mathbb{P}(A) = 0.5 \ne 0$ . So $\mathbb{P}$ is not absolutely continuous w.r.t. $\mathbb{Q}$ .

Neither is absolutely continuous with respect to the other, even though both measures have a common "region of overlap" $[0.5, 1]$ .

Part 3

Lebesgue decomposition of $\mathbb{Q}$ w.r.t. $\mathbb{P}$ :

Absolutely continuous part: restrict $\mathbb{Q}$ to the common support $[0.5, 1]$ . Its density w.r.t. Lebesgue on $[0.5, 1]$ is $1$ (from $\mathbb{Q}$ ), and $\mathbb{P}$ has density $1$ on $[0, 1]$ . So on $[0.5, 1]$ , $d\mathbb{Q}_{\text{ac}}/d\mathbb{P} = 1$ . Total mass: $\mathbb{Q}([0.5, 1]) = 0.5$ .
Singular part: $\mathbb{Q}$ restricted to $(1, 1.5]$ , which is $\mathbb{P}$ -null. Total mass: $0.5$ .

So $\mathbb{Q} = \mathbb{Q}_{\text{ac}} + \mathbb{Q}_{\text{sing}}$ where:

\mathbb{Q}_{\text{ac}}(A) = \text{Lebesgue}(A \cap [0.5, 1]), \qquad \mathbb{Q}_{\text{sing}}(A) = \text{Lebesgue}(A \cap (1, 1.5]).

Each is a sub-probability measure (mass $0.5$ ); together they sum to $\mathbb{Q}$ .

Part 4

Financial interpretation of equivalence.

\mathbb{Q} \ll \mathbb{P}

but not

\mathbb{P} \ll \mathbb{Q}

, there exists an event

A

with

\mathbb{P}(A) > 0

but

\mathbb{Q}(A) = 0

. This means

\mathbb{Q}

assigns zero probability to an outcome that is possible in the real world. The option price under $\mathbb{Q}$ is then blind to all payoffs that occur on $A$ .

Example: if $\mathbb{Q}$ assigned zero probability to "stock goes below $50," then a put option struck at $60 would be priced as zero — but under $\mathbb{P}$ (real world) that outcome can happen, and the put holder would actually receive a positive payoff. The seller of that put would be caught dramatically short.

This is why equivalent martingale measures (EMMs) must be equivalent, not merely absolutely continuous. Both directions of null-set agreement must hold, so that the pricing model sees every possibility the real world sees. Equivalence is the no-arbitrage + full-market-coverage condition.

Takeaways

Equivalence = same null sets. $\mathbb{P} \sim \mathbb{Q}$ iff $Z = d\mathbb{Q}/d\mathbb{P}$ is strictly positive $\mathbb{P}$ -a.s. For exponential-form derivatives (normal shifts, Girsanov), strict positivity is automatic.
Lebesgue's decomposition always works: $\mathbb{Q} = \mathbb{Q}_{\text{ac}} + \mathbb{Q}_{\text{sing}}$ . If $\mathbb{Q}_{\text{sing}} = 0$ , then $\mathbb{Q} \ll \mathbb{P}$ .
Risk-neutral pricing requires equivalence, not just absolute continuity. Otherwise the pricing measure can be blind to real-world possibilities, yielding prices that miss payoffs.
Singular measures are the "pathological" case. In finance, if two probability models disagree on zero-probability events, either they are looking at disjoint worlds (Lebesgue's singular part), or the models are not properly calibrated to the same market.