Verification of result

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I've recently I've come across this and am questioning it's validity but can't fully convince myself of (or against) it. Here are my intuitions why it may be true but I can't fully but I can't find a counter-example of prove it....


Claim: $$ E_{\mathbb{P}}[Y|\mathscr{F}_t]=E_{\mathbb{Q}}[Y\frac{Z_t}{Z_T}|\mathscr{F}_t]. . $$


Thoughts:

Suppose that $X_t$ is an integrable $\mathscr{F}$-adapted process, $\mathbb{Q}\sim\mathbb{P}$ are equivalent measures and $Y$ is an $\mathscr{F}_T$-measurable random-variable for some $T>t$. Let $Z_t$ be the corresponding density process $$ Z_t \triangleq \frac{d\mathbb{Q}}{d\mathbb{P}}|_{\mathscr{F}_t}. $$ It is known, for example in Theorem 3.8 of Jacod and Shiryaev's book, that $$ E_{\mathbb{Q}}[Y|\mathscr{F}_t]=E_{\mathbb{P}}[Y\frac{Z_T}{Z_t}|\mathscr{F}_t]. $$ Moreover, in Corollary 15.2.2 of Cohen and Elliot's book, it is known that $(Z_t)^{-1}$ is also a $(\mathbb{Q},\mathscr{F}_t)$-martingale since both measures are equivalent.

This seems wrong, but the Cohen result makes me believe otherwise.... Can someone confirm this?

So the claim seems to be a mis-reading of the Jacod and Shiryaev result but on second glance maybe the Samuel and Cohen result justifies it...Just don't see why/not..

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Here are my two cents. I think the claim is true at least if the condition (c) below holds true (sorry for the first posts where I mixed up the algebra).

First let's recall things a bit following your post :

$Z \triangleq \frac{d\mathbb{Q}}{d\mathbb{P}}$ and $Z^{-1} \triangleq \frac{d\mathbb{P}}{d\mathbb{Q}}$

and we have $Z^{-1}=\frac{1}{Z}$ as $\mathbb{Q}\sim\mathbb{P}$ by hypothesis (this is easy to show there posts on the forum on this I think).

Now the condition (c)

Let's redefine following your post for any $\forall t>0$ :

$$Z_t = E_{\mathbb{P}}[Z|\mathscr{F}_t] \triangleq \frac{d\mathbb{Q}}{d\mathbb{P}}|_{\mathscr{F}_t} $$

and

$$Z_t^{-1} = E_{\mathbb{Q}}[Z^{-1} |\mathscr{F}_t] \triangleq \frac{d\mathbb{P}}{d\mathbb{Q}}|_{\mathscr{F}_t} $$

If we have in this setting :

$$Z_t^{-1}=\frac{1}{Z_t} ~~~~(c)$$

Your claim is true under condition (c). Proof : Starting with the Jacod and Shiryaev theorem but switching $\mathbb{P}$ and $\mathbb{Q}$ we get :

$$E_{\mathbb{P}}[Y|\mathscr{F}_t]=E_{\mathbb{Q}}[Y\frac{Z_T^{-1}}{Z_t^{-1}}|\mathscr{F}_t]$$

then using the property (c) above we get your claim because :

$$E_{\mathbb{Q}}[Y\frac{Z_T^{-1}}{Z_t^{-1}}|\mathscr{F}_t]=E_{\mathbb{Q}}[Y\frac{Z_t}{Z_T}|\mathscr{F}_t]$$

So the condition above is sufficient for your claim to be true and it remains to find suitable conditions for which it holds true, but I think it holds in quite a general framework, the question to know if the condition above is also necessary is unknown me, but note that I didn't use the theorem of Cohen and Elliott's book.

Note that condition (c) can be rewritten :

$$E_{\mathbb{Q}}[Z^{-1} |\mathscr{F}_t] =\frac{1}{E_{\mathbb{P}}[Z|\mathscr{F}_t]}$$ or

$$E_{\mathbb{P}}[Z|\mathscr{F}_t].E_{\mathbb{Q}}[Z^{-1} |\mathscr{F}_t] = 1$$

Unless mistaken this is the case when Girsanov's theorem applies because it holds true that $Z_t$ and $Z^{-1}_t$ can be expressed as :

$Z_t=exp(X_t-\frac{1}{2}<X>_t)$ and $Z^{-1}_t=exp(-X_t+\frac{1}{2}<X>_t)$ For some martingale $X_t$, but I don't know if for strict local martingale this is true.

Here is another one line argument :

Let's take Jacod and Shiryaev theorem (JST) with $Y'=Y.\frac{Z_t}{Z_T}$, note that $Y'$ is $\mathcal{F}_T$-measurable so that :

$$E_{\mathbb{Q}}[Y.\frac{Z_t}{Z_T}|\mathscr{F}_t]=\underbrace{E_{\mathbb{Q}}[Y'|\mathscr{F}_t]=E_{\mathbb{P}}[Y'\frac{Z_T}{Z_t}|\mathscr{F}_t]}_\text{by Jacod and Shiryaev's theorem applied to $Y'$}=E_{\mathbb{P}}[Y.\frac{Z_t}{Z_T}\frac{Z_T}{Z_t}|\mathscr{F}_t]=E_{\mathbb{P}}[Y|\mathscr{F}_t]$$ QED