How can I use this density function as a likelihood function?

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Discussion Overview

The discussion revolves around the construction and interpretation of a piecewise density function that transitions from an exponential form to a Gaussian form. Participants explore its validity as a likelihood function, particularly in the context of defining it as a function of the parameter l given the variable d. The conversation includes technical reasoning about the properties of the density function and its integrals.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant, Dave, presents a piecewise function combining an exponential and a Gaussian form, questioning its definition as a likelihood function of l given d.
  • Another participant asks whether the proposed function is a valid density function and if its integral over (0, ∞) equals 1.
  • Some participants assert that the combination of the two functions is designed to ensure the total area under the density function equals 1.
  • Concerns are raised regarding the integral of the second function over its domain, with one participant noting it does not equal 1.
  • Another participant specifies that the integral of the second function from l to infinity is less than 1, suggesting it needs to be scaled to account for the area used by the first function.
  • Further clarification is provided that the scaling factor is necessary to ensure the total integrates to 1, with a specific mention that the integral equals 1/2 and requires adjustment.
  • Discussion includes the potential challenges in using the piecewise density as a likelihood function, particularly regarding continuity and differentiability, which may complicate maximum likelihood estimation for l.

Areas of Agreement / Disagreement

Participants generally agree that the proposed function can be interpreted as a density function, but there is disagreement regarding the specifics of its integrals and the implications for its use as a likelihood function. The discussion remains unresolved regarding the exact scaling needed and the continuity of the function.

Contextual Notes

Participants note that the piecewise function's continuity and differentiability depend on the choice of variance parameters, which may affect the maximum likelihood estimation process.

daviddoria
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I am trying to make a function which is exponential for a while, and then turns gaussian:
[tex] f(l,d) = \lambda e^{-\lambda d} , 0 < d < l[/tex]

and
[tex] f(l,d) = (1-\int_0^l \lambda e^{-\lambda d} dd) \frac{1}{\sigma \sqrt{2 \pi}} e^{-(d-l)^2/(2\sigma^2)} , l < d < \infty[/tex]
(That is supposed to be a piecewise function!)

You can see that as a function of d, the function is exponential until l and then gives the remaining weight (ie 1 - the area accumulated so far) to the gaussian.

The problem is, interpreted as a function of l (the likelihood of l given d instead of the probability of d given l), I don't understand if it is defined, since the piecewise region depends on l.

Does that make sense?

Thanks,
Dave
 
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For any given l, is f(l,d) is a density function at all ? Does its integral over (0,inf) become 1?
 
yea its a density function. They way the two functions are combined is exactly to make it a density function. It looks at the integral from 0 to l of the first function and then gives (1-that area) weight to the second function, so the total is now 1.
 
daviddoria said:
yea its a density function. They way the two functions are combined is exactly to make it a density function. It looks at the integral from 0 to l of the first function and then gives (1-that area) weight to the second function, so the total is now 1.

But the integral of the second function over the domain in question is not 1.
 
the integral of the second function from l to infinity would be some value < 1, and then it is scaled to instead be the remaining portion of "1" that has not been "used" by the first function. Maybe I did that scaling a bit wrong, but it doesn't change the point really I don't think?
 
daviddoria said:
the integral of the second function from l to infinity would be some value < 1,

It is 1/2, to be exact.

daviddoria said:
and then it is scaled to instead be the remaining portion of "1" that has not been "used" by the first function.

There's no "instead:" it's simply scaled by 1-<other integral>, and so it integrates to half of that scaling factor. You need to multiply by 2.

daviddoria said:
Maybe I did that scaling a bit wrong, but it doesn't change the point really I don't think?

The point had to do with using this density as a likelihood function, right? The likelihood function seems to me to be well-defined, but that doesn't mean it's going to be easy to work with. The proposed piecewise density function is not continuous unless you choose the variance parameters in a certain way and, even then, it's not continuously differentiable. Which is to say that the ML estimate for l may be quite difficult to work out.
 

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