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Thread: second order stochastic dominance for convex functions

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    second order stochastic dominance for convex functions




    Hi,

    To gain better understanding of second-order stochastic dominance I am trying to replicate some basic proofs. The proof of `standard' SSD (relating to a concave utility function) is okay but that for convex utility functions seems to be a lot harder.
    Recall that SSD holds iff:
    \int^{y}_{-\infty} (G(t) - F(t)) dt \geq 0 for any y.

    For the convex version of SSD, often referred to as Descending SSD, the condition above becomes:
    \int_{x}^{\infty} (G(t) - F(t)) dt \geq 0 for any x.

    I have found a paper (Levy & Wiener (1998) that gives two hints but does not provide the proof itself. The first hint is that if u(x) is convex <-> -u(-x) is concave. I have managed to show that SSD holds when -u(-x) is concave (the pitfall there is to not mess up any of the minus-signs for the first and second derivative...), but the second hint remains unclear to me. They suggest that x DSSD y <-> -y SSD -x, so that plugging in -y SSD -x into the first condition above, this immediately leads to x DSSD y.

    Can anyone show me some intermediate steps of the derivation? I have the gut feeling that a survival function should show up somewhere (i.e. (1-F(x)) ) but I cannot seem to understand where or when in the derivation.

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    Re: second order stochastic dominance for convex functions


    Do you mean show something like this? (the theorem 8 in page 46, page 4 of the pdf)

    http://ocw.mit.edu/courses/economics...10_notes05.pdf

    It is stating and proving that the expected utility is larger for every weakly increasing concave utility function;
    and you want to show the given condition is equivalent to this, in the descending SSD version (with convex one)?

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