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To prove this Lemma :

$$ E^Q \left[ Y|F(s) \right] = \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$$

with $P$ and $Q$ two equivalent probability measures and $Z(t)$ is the expectation of the Radon Nikodym derivative $Z(t) = E^P \left[\frac{dQ}{dP}|F(t) \right]$, and $Y$ is $F(t)-$ measurable.

I found that in Shreve's Stochastic Calculus for Finance II, the result is concluded only by checking the partial averaging property :

$$ \int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A E^Q \left[ Y|F(s) \right] dP$$ My question is why is it sufficient to check the partial averaging property to conclude the result? Why don't we just do without taking the expecation of both sides? Thank you

JasBeck
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1 Answers1

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You have a question to this link I gave you in your other question…

  1. why don't you ask there if you have a question?
  2. why don't you link to the answer you have a question to?

Nevertheless let's answer your question: You are wrong, in "Shreve's Stochastic Calculus for Finance II" they don't check $$\int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A E^Q \left[ Y|F(s) \right] dP$$ but they check if $$\int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A Y dQ$$ holds for $A \in F(s)$. Or if you write it with expectations instead of the integral form $$E_Q\left[1_A\frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]\right] = E_Q\left[1_AY\right]$$ And this is because this had necessarily to hold if $$\frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$$ should be the conditional expectation of $Y$ w.r.t to $F(s)$ besides the $F(s)$ - mesurability... this those is obvious. Maybe it's clearer to you if we set $$X := \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$$ Then $X$ is the $Q$-conditional expectation of $Y$ w.r.t. $F(s)$ iff

  • $X$ is $F(s)$ mb
  • $E_Q[X1_A] = E_Q[Y1_A]$ for all $A \in F(s)$

And this is exactly what is checked by: $$E_Q[X1_A] = \int_A X dQ = \int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A Y dQ = E_Q[Y1_A]$$

Gono
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