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Math Help - Compare the expectation of two functions of same r.v.

  1. #1
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    Compare the expectation of two functions of same r.v.

    Hello everybody,

    does anyone have an idea how to prove the following inequality

    E(\log(1+c_1(e^{Z+b}-1)))\geq E(\log(1+c_2(e^{Z+b}-1)))?

    where Z is normally distributed with a non-negative mean and c_1,c_2 are constants with c_1,c_2 \in [0,1],c_1\geq c_2
    b>0 is also a constant.

    Thanks in advance
    Last edited by Juju; April 24th 2013 at 07:23 AM.
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  2. #2
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    Re: Compare the expectation of two functions of same r.v.

    Hey Juju.

    Try showing that the function is monotonic increasing and using that, show that if CDF_X(p) > CDF_Y(p) for all p in [0,1] then E[X] > E[Y].
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  3. #3
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    Re: Compare the expectation of two functions of same r.v.

    Thanks!

    Unfortuanately, I do not completely understand your hint. The function log(1+c(e^{Y-b}+1))) is increasing and convex in Y.

    The expectation is given by
    \int log(1+c(e^{y-b}+1)))\frac{1}{\sqrt{2 \pi \sigma^2}}e^{-0.5\frac{(x-\mu)^2}{\sigma^2}} dx

    Or do you mean that I should consider the cdf of the function of r.v. log(1+c(e^{Y-b}+1)))?
    P(log(1+c_1(e^{Y-b}+1))) \leq y)=P(e^Y\leq \frac{e^y-1-c_1}{c_1})
    But now it does not hold
     \frac{e^y-1-c_1}{c_1}\leq  \frac{e^y-1-c_2}{c_2}\; \forall y
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  4. #4
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    Re: Compare the expectation of two functions of same r.v.

    Yes, consider the CDF of the log function of the random variable. That was my initial intent and idea for you,
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