Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p(x)$.

Thanks in advance!

share|improve this question
    
+1 => >1000. Congrats :) –  Jacob Feb 26 '13 at 16:51
    
Are you sure that y(x)=\int(t*p(x,t)dt) / p(x) –  Occupy Gezi Mar 1 '13 at 21:55
    
@AnilBaseski Sure. I think that part is correct. –  Robert Smith Mar 2 '13 at 0:15
    
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
show 1 more comment

1 Answer

up vote 2 down vote accepted
+50

Despite appearances, this is really a one-dimensional variational problem. The unknown $y$ is only a function of $x$, so we can treat the integral in $t$ as a "black box", that is, just some ordinary function $F(x,y) := \int L(t,y)p(x,t)\,\mathrm dt$. Then your functional becomes $$E[L]=\iint L\big(t,y(x)\big)p(x,t)\,\mathrm dx\,\mathrm dt=\int F\big(x,y(x)\big)\,\mathrm dx,$$ and you can apply the usual Euler-Lagrange equation in $x$ and $y(x)$.

In this specific case, we can write $F$ out explicitly. Observe that $$L(t,y)p(x,t)=(y-t)^2p(x,t)=y^2p(x,t)-2typ(x,t)+t^2p(x,t),$$ so $$F(x,y)=\int L(t,y)p(x,t)\,\mathrm dt=y^2\int p(x,t)\,\mathrm dt-2y\int tp(x,t)\,\mathrm dt+\int t^2p(x,t)\,\mathrm dt\\=y^2q(x)-2yr(x)+s(x),$$ where $q(x)=\int p(x,t)\,\mathrm dt$, $r(x)=\int tp(x,t)\,\mathrm dt$, and $s(x)=\int t^2p(x,t)\,\mathrm dt$.


Heck, $F\big(x,y(x)\big)$ doesn't involve any derivatives of $y$, so we don't even need any variational calculus at all. The value of $F\big(x,y(x)\big)$ at any point $x$ is independent of the value of $y$ at any other point, so you can just choose $y(x)$ at each $x$ independently to minimize $F\big(x,y(x)\big)$ at that point. That is to say, when you find $y(1)$ to minimize $F\big(1,y(1)\big)$, you don't have to worry about what $y(2)$ or $y(1.1)$ or $y(42)$ is going to be. That makes things ridiculously easy: $$\frac{\partial}{\partial y}F(x,y)=2yq(x)-2r(x)=0\implies y=\frac{r(x)}{q(x)}=\frac{\int tp(x,t)\,\mathrm dt}{\int p(x,t)\,\mathrm dt}.$$


Anyway, I guess what you really want to understand is what the original objective $E[L]=\iint \big(y(x)-t\big)^2p(x,t)\,\mathrm dx\,\mathrm dt$ is all about. Well, clearly the term $L(t,y)=\big(y(x)-t\big)^2$ says that you want $y(x)-t$ to be small, that is, you want $y(x)$ to be close to $t$. But $t$ could be anything, while $y$ only depends on $x$! So how can we ever hope to get $y(x)$ to be close to $t$, whatever $t$ might be? The key is that for any given $x$, some values of $t$ are more likely than others — that's what $p(x,t)$ tells you. So if you know the value of $x$, then you know what values of $t$ are more likely at that $x$, and you can pick $y(x)$ to be somewhere in the middle of those values.

That's all that's really going on here. Don't think of the $p(x,t)$'s as "cross terms", think of them as weights in a weighted average (or a weighted regression, maybe). After all, the optimal value of $y(x)$ turns out to be nothing but the expected value of $t$ conditioned on $x$.

share|improve this answer
    
Great answer, thanks! Regarding the other question, I was referring to the fact that $E[L] = \sum_{i} \sum_{j} (y(x_{i}) - t_{j})^{2}p(x_{i},t_{j})$ contains terms such as $(y(x_{1}) - t_{2})^{2}p(x_{1},t_{2})$ which seems very unusual in a regression, so I'd like to get an explanation about it. –  Robert Smith Mar 7 '13 at 4:38
    
@Robert: See my update. –  Rahul Mar 7 '13 at 15:25
    
Excellent. Now it makes much more sense :-) –  Robert Smith Mar 7 '13 at 23:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.