# Minimizing Lagrangian with two functions

I read this problem where I have to minimize a functional $E[L]$ using calculus of variations, but I'm not sure what is the procedure to follow.

The functional is the expected loss:

$$E[L] = \int\int L(t, y(x))p(x,t)dxdt$$

and we want to choose a $y(x)$ to minimize $E[L]$ in order to get the following result with $L = (y(x) - t)^2$

$$y(x) = \frac{\int tp(x,t)dt}{p(x)}$$

but in doing so, I think it's ignoring the variation on $p(x,t)$. What is the proper Lagrangian for this functional and how to proceed with the minimization when there is an additional integration (in this case) in $t$.

By the way, I'm used to see Lagragians in the form $L(q(t), q'(t),t)$ where $t$ is the independent variable and we get something along the lines of (informally):

$$\delta \displaystyle \int L(q(t), q'(t),t) dt= \displaystyle \int\left(\displaystyle \frac{ \partial L}{\partial q}\delta q + \displaystyle \frac{\partial L }{\partial q'}\delta q'\right) dt$$

An additional question: Why are we using a double integral in the expected loss? Thinking about it in discrete terms, the equivalent would be:

$$E[L] = \sum_{i} \sum_{j} (y(x_{i}) - t_{j})^{2}p(x_{i},t_{j})$$

which seems a bit odd in the context of a regression since we are taking into account cross terms, right? So, I guess that at some point we are requiring a minimum distance, for example, from a point $y(x_{2})$ to $t_{1}$.

UPDATE:

I found this paper in which there is a minimization using a different functional, but it's not explained why they ignored $p(x)$.

• What exactly is your question? Do you just want to know how to minimize $E[L]$ in this case? It's easy, you just integrate in $t$ first and then you're left with a one-dimensional variational problem in $x$ and $y(x)$. – Rahul Mar 6 '13 at 22:16
• Also, you're missing the square on $(y(x_i)-t_j)^2$ in your discrete equation, and I don't know what you mean about cross terms. // In any case, it looks like $p(x,t)$ is a fixed function and not a variable that you're optimizing, is that right? The variation is for figuring out how the value of the functional changes if you change the optimization variables, like $y(x)$. If you don't get to change $p(x,t)$ at all, there's no point in taking its variation. – Rahul Mar 6 '13 at 22:23