Conditional Expectations under a Stock Inversion Measure Change

Summary

The document considers a Black–Scholes stock and a transformed payoff based on inverting the stock price around a fixed level. It introduces a measure associated with a power of the stock and states a pricing identity under that measure. The question is whether the transformed expectation can instead be written under the original risk-neutral measure when the inverted process has the same distribution as the original stock under that measure.

The central issue is conditional expectation: equality in distribution of two processes under different measures alone does not establish equality of conditional expectations with respect to the same information. A sound argument must specify the conditioning sigma-algebra and show that the relevant conditional law of the inverted process under the changed measure matches the stock’s conditional law under the risk-neutral measure. The document supplies the setup and claimed distributional correspondence but no resolution or proof, so the measure-change identity and its precise conditions remain open.

Key ideas

• A stock inversion is used to relate an original payoff to a transformed payoff.
• The pricing identity involves a measure defined through a power of the stock.
• Equality of distributions does not by itself establish equality of conditional expectations across measures.
• The argument requires matching conditional laws relative to the same information set.
• The document poses the proof question without resolving its assumptions.

Tags

# Is $E_t^{Q}(g(Y))=E_t^{Q^Z}(g(Y))$?


# Is $E_t^{Q}(g(Y))=E_t^{Q^Z}(g(Y))$?












Consider $$Z(t)=\left(\frac{S(t)}{H}\right)^p$$where $S$ has a standard Black-scholes Dynamics for a stock, $H$ is a postive constant and $p =1 - \frac{2r}{\sigma^2}$

and a simple claim with a pay-off at time T specified by pay-off function g. The arbitrate-free time t value is $$\pi^g(t)=e^{-r(T-t)}E^{\mathbb{Q}}_t(g(S(T)))=e^{-r(T-t)}f(S(t),t)$$, where $\tilde{g}(x)=\frac{x}{H}g(\frac{H^2}{x})$

Define $\mathbb{Q}^{Z}$ measure such $\frac{d\mathbb{Q}^{Z}}{d\mathbb{Q}}$.

I have shown that $$\pi^{\tilde{g}(t)}=e^{-r(T-t)}\left(\frac{S(t)}{H}\right) ^p E^{\mathbb{Q}^Z}_t\left(g\left(\frac{H^2}{S(T)}\right)\right)$$

and considering $Y(t)=\frac{H^2}{S(t)}$ I showed that $dY(t)=rY(t)dt+\sigma Y(t)(-dW^{\mathbb{Q}^Z(t)})$

Problem: Show that $$\pi^{\tilde{g}(t)}=e^{-r(T-t)}\left(\frac{S(t)}{H}\right) ^p E^{\mathbb{Q}}_t\left(g\left(\frac{H^2}{S(T)}\right)\right)$$, given the fact that Y under $\mathbb{Q}^Z$ is the same as the distribution of $S$ under $\mathbb{Q}$.

For me, it is straightforward to see that if g is measurable than $E_t^{Q}(g(Y))=E_t^{Q^Z}(g(Y))$ so that replacing on the identity would yield the same result. However I do not know if this equality holds since I am dealing with conditional expectations

Question:

How should I prove the expression of $\pi^{\tilde{g}(t)}$ is the same for both measures?

Thanks in advance!

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.