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Interesting density function

  1. Nov 18, 2008 #1
    I am trying to make a function which is exponential for a while, and then turns gaussian:
    f(l,d) = \lambda e^{-\lambda d} , 0 < d < l

    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
    (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?

  2. jcsd
  3. Nov 19, 2008 #2


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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?
  4. Nov 19, 2008 #3
    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.
  5. Nov 19, 2008 #4
    But the integral of the second function over the domain in question is not 1.
  6. Nov 19, 2008 #5
    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 dont think?
  7. Nov 19, 2008 #6
    It is 1/2, to be exact.

    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.

    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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