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I have two (related), ungraded homework problems. I am hoping to receive advice on how to proceed, or to be alerted if I'm on the wrong path (or, alternatively, to have the problems solved for me).

Let $P\sim Q$ be equivalent probability measures.


Prove that $dQ/dP > 0 \,\,\,\,\,\text{ P-a.s. (and hence Q-a.s.)}$


$\displaystyle \ \ \int_\Omega dQ = 1 $

If $P\sim Q$ and $dQ/dP < 0$ then:

$1 = \int_\Omega \frac{dQ}{dP}dP = E^P[\frac{dQ}{dP}]\leq 0$

Which is a contradiction, therefore $\frac{dQ}{dP} > 0 \,\,\,\text{P-a.s. (and hence Q-a.s.)}$.


Prove that $dQ/dP = (dP/dQ)^{-1} \,\,\,\text{P-a.s. (and hence Q-a.s.)}$


By the definition of equivalent probability measures, we have that:

$\displaystyle \ \ \int_\Omega dP = \int_\Omega \frac{dP}{dQ}\frac{dQ}{dP}dP = \int_\Omega \frac{dP}{dQ}dQ$

Therefore the equality is true.

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In question, why would the expectation with respect to $P$ of $\frac{dQ}{dP}$ would be non-positive? – Davide Giraudo Nov 24 '12 at 12:19
@DavideGiraudo I've revised to make this clear. – Jase Nov 24 '12 at 12:19
The negation of "$f>0$ almost everywhere" is not "$f\leqslant 0$ almost everywhere", it's rather "there is a set of non-zero measure on which $f\leqslant 0$". Now integrate on this set. – Davide Giraudo Nov 24 '12 at 12:21
@DavideGiraudo I'm afraid I don't follow. – Jase Nov 24 '12 at 12:32
@DavideGiraudo What you mean by integrating on a set of non-zero measure on which $f \leq 0$. All of that. – Jase Nov 24 '12 at 12:51
  1. Let $f:=\frac{dQ}{dP}$, and assume that "$f>0$ $P$-a.e. is not true". That means that we can find $S\subset\Omega$ such that $P(S)>0$ and $f(x)\leqslant 0$ for all $x\in S$. This gives $$Q(S)=\int_{S}f(x)dP(x)\leqslant 0,$$ so $Q(S)=0$. It's a contradiction as $P\ll Q$.

  2. What you did would be correct if you replace $\Omega$ by an arbitrary measurable set $A$.

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