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As the title states, my question is whether first order stochastic dominance is conserved under change of measure, for instance from the $\mathbb{P}$ measure to $\mathbb{Q}$ measure and change of numeraire to yet another measure say $\mathbb{Q}_N$?

I am not a probabilist, but I believe it is conserved since change of measure and/or numeraire involves multiplication by an exponential martingale which is a positive process, but if someone can give a formal proof that shows why it is or isn't invariant under change of measure that would be great.

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  • $\begingroup$ It is the covariance of the expotential martingale with the underlyings that matters, not the fact that it is positive. $\endgroup$
    – Arshdeep
    Jun 1 at 2:24
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Consider a coin independently tossed 10 times. Assume under the measure $P$, $Pr(H)$ > 0.5 but not equal to 1.

Let a risk neutral person be iteratively given the gamble between getting atleast $n$ heads versus atleast $n$ tails. Clearly the person always chooses getting atleast $n$ heads. Let $X$ represent number of heads and $Y$ the number of tails. Thus we have

$E(1_{X>=n})>E(1_{Y>=n})$ for all $n$, and thus $X$ FOSD the $Y$ under this probability measure.

Now change to a measure where $Pr'(H)< 0.5$ but not equal to 0. Specify event probabilities by retaining the independence assumption. This is equivalent to the previous measure, as all event sets having 0 or 1 probability in the former retain the same probability in the latter.

Now consider the same gamble given to the same person. Now, the signs get flipped! $E'(1_{X>=n})<E'(1_{Y>=n})$ for all $n$, so now the $Y$ FOSD $X$. So the assertion is false.

When does it preserve FOSD? A look at some special cases below:

  1. If the RV's are such that the minimum of $X$ exceeds the maximum of $Y$, then the stochastic dominance will be maintained under any change of measure, because the $Pr(w: X(w)>X_{min})=0$ and $Pr(w:Y(w)<Y_{max})=1$ under any equivalent measure, so the distributions again are non intersecting.

  2. For discrete RV's $X$ and $Y$, a sufficient condition for violation of FOSD order is for the means to change order. This is because the mean can be written as the sum across all $u$, of $Pr(X>=u)$.

  3. What can the Radon-Nikodym derivative (RND) look like so that FOSD is not violated?

By simple change of measure using the RND $Z$,

$E'(1_{X>=u})=E(1_{X>=u})+cov(Z,1_{X>=u})$

and

$E'(1_{Y>=u})=E(1_{Y>=u})+cov(Z,1_{Y>=u})$

and thus it suffices that: $cov(Z,1_{X>=u})>cov(Z,1_{Y>=u})$ for all $u$. Intuitively, $Z$ has more of $X$ in it than $Y$, or $X$ is a better predictor of $Z$ than $Y$. It also suffices of Z is independent of both X and Y.

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  • $\begingroup$ Great example! How to show this in more generality? Do you have a link to some source with a proof? $\endgroup$ Jun 1 at 5:05
  • $\begingroup$ Thanks! I don't have a link unfortunately. Which assertion's proof are you looking for please? $\endgroup$
    – Arshdeep
    Jun 1 at 10:51
  • $\begingroup$ What I'd be looking for is a theorem/proposition which states under what conditions FOSD is maintained under change of measure/numeraire and proof of this assertion. I have been googling but somehow haven't come across this (so I'm probably not searching for the right terms) $\endgroup$ Jun 1 at 13:34
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    $\begingroup$ Please look at the added cases and see if they are of any help. $\endgroup$
    – Arshdeep
    Jun 1 at 13:52

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