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Math Help - problem of calculus of variations

  1. #1
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    problem of calculus of variations

    Hello everyone

    I have the following problem. I need to find a function, say f(x), that maximizes
    \int_{a}^{b}u(x)f(x)dx, where u(x) is the square of the cumulative distribution function of the standard normal minus one half, i.e. u(x)=(\phi(x)-1/2)^2.
    The maximisation problem does not have prespecified boundary values for a and b, and is subject to the constraint \int_{a}^{b}f(x)dx = 1

    any help is really precious

    thanks

    Lorenzo
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  2. #2
    A Plied Mathematician
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    Hmm. Typically, a problem with an integral constraint is dealt with using the Lagrange multiplier method. That is, instead of setting

    L=u(x)\,f(x), and solving the DE

    \displaystyle\frac{d}{dx}\,\frac{\partial L}{\partial f'}-\frac{\partial L}{\partial f}=0,

    you do the following. Set L=u(x)\,f(x)+\lambda\,f(x), and set

    \displaystyle\frac{d}{dx}\,\frac{\partial L}{\partial f'}-\frac{\partial L}{\partial f}=0.

    The problem is that in all these cases,

    \dfrac{\partial L}{\partial f'}=0,

    which makes all the resulting equations much too constrained. Even if you let \lambda=\lambda(x), you still have problems. You end up with \lambda=-u(x), or \lambda(x)=-u(x), an absurdity.

    There's something I'm missing here, or something is really whacked out. Can you please double-check your problem statement, and make sure it's correct? In particular, I'd expect to see an f'(x) floating around somewhere, at least if the calculus of variations is going to be the right tool for the job.
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