# Integral defined as zero implies function is zero

1. Dec 4, 2009

### coolnessitself

1. The problem statement, all variables and given/known data
$$\int\limits_{\theta}^{\infty} f(x) g(x) dx = 0$$
$$\theta > 0$$
$$f(x) = ke^{-k(x-\theta)}$$
Show g(x) is identically 0.

2. Relevant equations

3. The attempt at a solution
$$f$$ is always >= 0 since it behaves exponentially in the region of interest.
From something like https://www.physicsforums.com/showthread.php?t=299145" I could say that IF $$f(x)g(x)\ge 0$$ in this region, then $$g(x)=0$$, but I don't know anything about g. I could say that assuming g is positive somewhere, then the integral wouldn't be zero, and that assuming g is negative somewhere, the integral wouldn't be zero, but what about the case where $$g(c_1) f(c_1) = k$$ and $$g(c_2) f(c_2) = -k$$. Then g is positive somewhere and negative somewhere else such that the product cancels out. Wouldn't that allow the integral to be zero?

Last edited by a moderator: Apr 24, 2017
2. Dec 4, 2009

### LCKurtz

What are your assumptions? g continuous? True for all k and $\theta$?

3. Dec 5, 2009

### HallsofIvy

Staff Emeritus
As it stands, this simply isn't true! Are you requiring that g be continuous?

4. Dec 8, 2009

### coolnessitself

I guess I should explain a bit more. I'm trying to show that a family of pdfs is complete. I no longer have access to my paper (for the time being), so I'll reproduce what I can remember. The notation seems standard, so I'll just type out a definition:
Let the random variable Z of either the continuous type or the discrete type have a pdf or pmf that is one member of the family $$\left\{h(z;\theta) : \theta \in \Omega\right\}$$. If the condition $$E[u(Z)]=0$$, for every $$\theta \in\Omega$$ requires that u(z) be zero except on a set of points that has probability zero for each $$h(z;\theta),\;\theta\in\Omega$$, then the family $$\left\{h(z;\theta) : \theta \in \Omega\right\}$$ is called a complete family of probability density or mass functions.

Here, my pdf is the f I typed in the first post, and I need to show it's complete. As far as I can tell, I know nothing more about g(x). So, being the naive little statistician that I am, I would try and set the integral of g*f equal to zero and show that g is identically zero.

(ps the book sucks so there's a good chance the authors performed a little hand waving, although they stress this example so I would guess the error is on my part, not theirs)
(pps the book is hogg mckean and craig)
(ppps is there a way to use dollar signs \$ instead of typing [ tex ] each time?)

5. Dec 8, 2009

### Dick

Do you mean f(x)=k*exp(-k*|x-theta|)? Then as k->infinity f(x) behaves like a multiple of the delta function centered at theta. Is that what they are getting at?

6. Dec 11, 2009

### Mute

The key to proving that is the fact that $\theta$ is arbitrary. The point is not that the integral is zero for some $\theta$, it's that it's zero for every $\theta$.

You can prove it for you specific function by, for example, taking a derivative with respect to $\theta$. Note, however, I think that it will require that g(x) be continuous over the integration region. I'm not sure if there is a more general proof for your function.

Last edited: Dec 11, 2009