Thank you to those who have shared their insights here.
Let me rephrase one more time.
$$\int_{0}^{\infty}f(x)g(x)dx=\int_{0}^{\infty}f(x)h(x)dx$$
1) Can we "cancel" the ##f(x)## terms on each side, thereby now having
$$\int_{0}^{\infty}g(x)dx=\int_{0}^{\infty}h(x)dx$$
2) Can we also now say...
Thank you all for looking at this. I wish I had a more concrete example but I'm just trying to understand a certain concept here. So please allow me to rephrase my original question.
How about this? Let's say ## f = f(x,y) ## and also ## g = g(x,y) ##.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx...
Thank you Frabjous.
How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that ## g(a)=g(b) ##??
...or even that ## a = b ##?
Thanks again
Howdy all,
Let's say we have, in general an expression:
$$ \int f(x) g(x) dx $$
But in through some machinations, we have, for parameter ##a##,
$$ \int f(x) g(x) dx = \int f(x) g(a) dx $$
...can we conclude that ## g(x) = g(a) ## ????
Thanks
hutchphd and Stephen Tashi, thank you for your quick replies. Stephen I think you've already made a great point. But regardless here is my attempt to "tighten up" my question.
Ok, so let's say we have some data that ostensibly "comes from" the normal...
Ok, I'm sure I can find a smarter way to pose this question, and I will try to define the question more carefully in coming days. That having been said, consider this:
Let's say we have a random variable X (or whatever). I can calculate the moments of this variable with no problem. In fact let's...