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Consider any feasible $p:[0,1]^2\to [0,1]$ that allows discontinuities and the problem

$$\min_{p(.)} \int_0^1\int_0^1 p(x,y)^2 dF(x) dG(y)$$

s.t. $$\int_0^1 p(x,y)dG(y)=k\phantom{0} for \phantom{0} a.e.\phantom{0} x\in [0,1]$$

where $F,G$ are distribution functions and $1/2<k<1$.

How could I solve the problem analytically?

user91360
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2 Answers2

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By your condition on $p(x,y)$, we see that $p(x,y)=p(y)$ a.e.. Hence you just need to find $$ \min_p \int_0^1 p(y)^2 dG(y) $$ s.t. $$ \int_0^1 p(y) dG(y) =k $$ To minimize the functional, we perform a variation according to Lagrange multipliers. Take $p \to p + h \xi$ then we see first order terms in $h$ correspond to $$ \delta^{(1)} =\int_0^1 [2p(y) - \lambda ]\xi(y) dG(y)$$ To get a critical point, we need $p(y)=\lambda/2$ a.e. since we need $\delta^{(1)} =0 $ for all test functions $\xi$. But the constraint gives us that the solution is $$ p(x,y) = k \quad a.e. $$ For reference : Calculus of variations: Lagrange multipliers

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Jeb
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  • Hi,thanks for the answer. I think the constraint on the marginal does not imply p(x,y)=p(y) a.e.because there are other classes of infinite many solutions satisfying the constraint but depend essentially on both x and y. – user91360 Jul 10 '15 at 17:06
  • The example for the multiplicity of solutions is here : http://math.stackexchange.com/questions/1131684/finding-measure-given-constant-margins – user91360 Jul 11 '15 at 13:28
  • @user91360 because you're minimizing p, if enough to show it in just x by your constraint...sorry if that wasn't clear. – Jeb Jul 13 '15 at 18:05
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Could I modify the above reasoning by first considering the relaxed problem,

$$ \min_p \int_0^1\int_0^1 p(x,y)^2 dG(y)dF(x) $$ s.t. $$ \int_0^1 \int_0^1 p(x,y) dG(y)dF(x) =k $$

Take $p \to p + h \xi$ then we see first order terms in $h$ correspond to $$ \delta^{(1)} =\int_0^1 \int_0^1 [2p(x,y) - \lambda ]\xi(x,y) dG(y)dF(x)$$ To get a critical point, we need $p(x,y)=\lambda/2$ a.e. Then,

$$ p(x,y) = k \quad a.e. $$

Since the solution is also feasible for the unrelaxed problem, we conclude the solution to the original problem is also the constant.

user91360
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