- #1
Watts
- 38
- 0
Assume I have a data set that I am trying to find a distribution that describes how the data is distributed. Assume I have found a function say [itex] f(x) = e^{ - x^2 } [/itex] that describes the distribution of data. Statistics tells me that my first move in doing so is to normalize this function so that [itex]\int\limits_{ - \infty }^\infty {P(x)dx} = 1 [/itex]. The common approach is to integrate [itex]\int\limits_{ - \infty }^\infty {e^{ - x^2 } dx} = \sqrt \pi [/itex] and multiply the function times the reciprocal of that integral [itex]\frac{1}{{\sqrt \pi }} \cdot \int\limits_{ - \infty }^\infty {e^{ - x^2 } dx} = 1[/itex]. But what if I can normalize it a different way say integrate the function [itex]\int\limits_{ - \infty }^\infty {e^{ - x^2 } dx} = \sqrt \pi [/itex] and place the result of that integral in the parentheses beside the variable [itex]P(x) = e^{ - (\sqrt \pi \cdot x)^2 } = e^{ - \pi \cdot x^2 }[/itex] instead of in front of the function. If you now integrate the function [itex]\int\limits_{ - \infty }^\infty {e^{ - \pi \cdot x^2 } dx} = 1[/itex] it still is equal to one. So my question is which PDF do I use? I have normalized the same function two different ways. Any thoughts on this?