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

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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.