# Minimization of definite integral reduces to the minimization of a part of the integrand, why?

This post was edited following a comment that rightly stated that the original question was unsensible. The edited version follows.

Why does the equation in the second line implicitly define the function specification of $a(x)$ which minimizes the integral in the first line? What is the formal rule that is used here?

$$\int_{x_1}^{x_2}{f(a(x),x) g(x) dx}$$

$$a(x) = \underset{a^*}{argmin} f(a^*,x)$$

It is indeed, as remarked by one of the commenters, the derrivation of the bayesian mean minimum squared error estimator where $x$ is the data, $a(x)$ the estimator, $g(x)$ the marginal probability density of the data and f(a(x),x) the conditional mean squared error of the estimator. $g(x) \geq 0, \forall x\in [x_1,x_2]$

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The equation makes no sense. On the right, $x$ is a free variable; on the left, it is bound under the integral sign. In general, $\arg\min_a f(a,x)$ will depend on the value of $x$, so how can it be identical to something independent of $x$? It would be better if you post the actual part of the proof that you're struggling with. – Rahul Jan 3 '13 at 17:18
What does the textbook say, exactly? – user1551 Jan 3 '13 at 17:22
I recommend you look at MIT's Open Courseware on the Estimation with Minimum Mean Square Error which derives the conditional MMSE and take a look at this Quora post as well for a good explaination. – GEL Jan 3 '13 at 18:08
Thanks lewellen and Rahul Narain for the insight that the original question was ill posed and the reference to the derivation of the MMSE. – Skifozoa Jan 3 '13 at 19:33

To minimize the integral, the function itself is minimized for every value of $x$. If you consider the integral as a sum then you're summing $$f(a(x),x) g(x)$$ You want that to be as small as possible for every $x$. As $g(x)$ is independent of $a$, you can only minimize the first term. That's just what your second line tells you: the $a(x)$ is a function that gives you the a value that minimizes $f(a,x)$ for every $x$. Suppose $x=x_0$. Then $a(x_0)$, according to your second line, is given as $$\operatorname*{arg \min}_{a^*} f(a^*,x_0)$$ so $a(x_0)$ is the value of a that minimizes $f(a,x_0)$. Plug that into your product of $f$ and $g$ and the product is minimal for $x_0$. Doing this for every $x$ minimizes the integral.
@Skifozoa Integrals are like sums. To minimize the sum $a+b+c$, assuming that you have freedom to pick $a,b,c$ independently of each other, you would pick the smallest allowed value for each. Same with the integral. – user53153 Mar 6 '13 at 3:54