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

    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

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

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

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

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

    The problem is that in all these cases,

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

    which makes all the resulting equations much too constrained. Even if you let $\displaystyle \lambda=\lambda(x),$ you still have problems. You end up with $\displaystyle \lambda=-u(x),$ or $\displaystyle \lambda(x)=-u(x),$ 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 $\displaystyle f'(x)$ floating around somewhere, at least if the calculus of variations is going to be the right tool for the job.
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