Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

QUESTION 1

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

Attempt:

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

QUESTION 2

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

Attempt:

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.

share|improve this question
    
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
1  
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
add comment

1 Answer

  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 $A$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.