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Let the probability measure Q be

$Q(A)=E[X^p1_A]/E[X^p]$

Then the expectation of Z with respect to probability measure Q is

$E_Q[Z]=E[ZX^p]/E[X^p]$

Why is this? Shouldn't taking the expectation involve integrating over the support of Z? Is that happening here? Why is Z just being put into the expectation and why did the dummy variable $1_A$ just disappear?

JJ chips
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1 Answers1

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To prove things like this, sometimes you need to go back to basics. Recall that expectation is defined in stages:

  • The expected value of a simple random variable $Z=\sum_{i=1}^n a_i{\bf 1}(E_i)$ is defined as $\sum_{i=1}^n a_i P(E_i)$.
  • When $Z$ is a positive random variable, $E[Z]$ is defined as $\sup \{E[W]:W\text{ is simple },W\le Z\}$. This is known to be equal to $\lim_{n} E[Z_n]$, where $Z_n\nearrow Z$ is a sequence of simple functions increasing to $Z$ (that is, $Z_1\le Z_2\le \dots$, and $\lim_n Z_n=Z$ almost surely).
  • In general, writing $Z=Z^+-Z^-$, we define $E[Z]=E[Z^+]-E[Z^-]$.

In your case, prove that $E_Q[Z]=E[ZX^p]/E[X^p]$ in stages. First, prove it assuming $Z$ is simple. Then, assume $Z$ is positive, let $Z_n\nearrow Z$ for $Z_n$ simple, and use the fact that $E_Q[Z]=\lim_n E_Q[Z_n]=\lim_n E[Z_nX^p]/E[X^p]$, along with the dominated convergence theorem. Finally, extend this to general $Z$ using $E[Z]=E[Z^+]-E[Z^-]$ (this is always the easiest part).

As a side note, by definition, the Radon Nikodym derivative of $Q$ with respect to $P$ is $$ \frac{dQ}{dP}=\frac{X^p}{E[X^p]}. $$ The notation for this derivative is convenient in that the formal equality $dQ=\frac{X^p}{E[X^p]}dP$ holds, which proves your statement immediately: $$ E_Q[Z]=\int Z\,dQ=\int Z \frac{X^p}{E[X^p]}dP=\frac{E[ZX^p]}{E[X^p]}. $$

Mike Earnest
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