# Integral defined as zero implies function is zero

• coolnessitself
In summary, the author is trying to show that a family of pdfs is complete. He is unsure of how to do this for his specific function, but believes it will require that g(x) be continuous.
coolnessitself

## Homework Statement

$$\int\limits_{\theta}^{\infty} f(x) g(x) dx = 0$$
$$\theta > 0$$
$$f(x) = ke^{-k(x-\theta)}$$
Show g(x) is identically 0.

## 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?

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

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

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?)

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?

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.

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## 1. What does it mean for an integral to be defined as zero?

An integral being defined as zero means that the area under the curve of a function is equal to zero. This can happen when the function crosses the x-axis and has both positive and negative values canceling each other out.

## 2. How does an integral defined as zero relate to the function being zero?

If an integral is defined as zero, it means that the function has no area under the curve. This can only happen if the function is equal to zero at every point on the interval of integration, meaning the function is identically zero.

## 3. Can an integral defined as zero have non-zero values for the function?

No, an integral defined as zero implies that the function is equal to zero at every point on the interval of integration. If the function has non-zero values, the integral will also have a non-zero value.

## 4. Can a function be non-zero if its integral is defined as zero?

No, if an integral is defined as zero, it means the function is equal to zero at every point on the interval of integration. This means the function is identically zero and cannot have non-zero values.

## 5. What are some real-life applications of an integral defined as zero?

An integral defined as zero can be used in physics to calculate the work done by a conservative force, or in economics to calculate the net profit of a company. It can also be used to find the average value of a function over an interval if the function is known to have a mean value of zero.

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