Let's say I have some function continuous over all real numbers such that .
How can I make it so that we know is a density curve?
Imposing the condition that is even comes to mind. Does this work and what else can be said?
Printable View
Let's say I have some function continuous over all real numbers such that .
How can I make it so that we know is a density curve?
Imposing the condition that is even comes to mind. Does this work and what else can be said?
You need to define what you mean by density in this context.
Sorry, it means to say a function in which the area under the entire curve is 1.
Which, I guess, is the same as saying...
.
I don't know if this is the answer to your question (I don't understand very well the question) but you can say that function f(x) is a density function if f(x) is pairs (simmetric to the y axes)...because from what I understand you know that the area to 0 to infinity is 1/2 :)
Hmm, I thought as much. Thanks!
I honestly don't know what that is, though I have the suspicion that you think I'm disguising a homework question? If so, then I disagree. Since the time it was asked, I realized it doesn't make much sense. The only concrete thing that one can say is that f(x) is even.Quote:
Originally Posted by Markus Hanke
By the way, if it's not odd to ask, why was the original post's TEX code edited?
No, it is just that the integral you gave is similar in form to the normalisation conditin of a quantum wave function :
Wave function - Wikipedia, the free encyclopedia
I took the liberty since the syntax of the original code was incorrect, so it didn't display at all. Hope you don't mind. I didn't change anything on the actual content, just on the syntax.Quote:
By the way, if it's not odd to ask, why was the original post's TEX code edited?