Let F1 and F2 denote distribution functions for absolutely continuous distributions on the real line and let p1 and p2 denote the corresponding density functions. is p1^a p2^(1-a) where 0 ≤ a ≤ 1, density function? how do I prove that?
There are two requirements - non-negativity for all x and ∫=1
First part is easy, since exponentiation preserves negativity, p1a > 0 and p21-a > 0 ∴ p = p1a x p21-a > 0 for all x
The integral part is only valid if the two ps are identical (or a convex combination), if not the integral will not evaluate to 1. Easy to check by letting p1 be a standard normal density function and p2 be a uniform density function, then for any a the function f = p1a x p21-a is neither a normal d.f. nor a uniform d.f. and you can't prove that the ∫=1 and therefore can't prove that f is a density function
Hello, the short answer is no, p1^a p2^(1-a) where 0 ≤ a ≤ 1 is not always a density function, because there exist a two functions that don't satisfy it.
If you take for example two uniform distributions with separate ranges. this production will be zero in all IR, Thus the integral ∫p1^a p2^(1-a) is zero. and it should be 1 if it's a density function.
If F1and F2 are distribution functions for absolutely continuous distributions, then the function f(x)=F2′(x)−F1′(x)f(x) = F2'(x) - F1'(x)f(x)=F2′(x)−F1′(x) (assuming the derivatives exist) would be a valid density function if it satisfies the properties of a probability density function: it must be non-negative and integrate to 1 over the entire domain. In other words, the function needs to be integrable and represent the rate of change of the cumulative distribution functions.
- Badges
- Science topic
- Similar topics
- Engineering
- Electrical Engineering