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Math Help - Estimating lognormal parameters from a truncated empirical distribution

  1. #1
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    Estimating lognormal parameters from a truncated empirical distribution

    I have a set of 50 values that I assume are part of a longnormal distribution. However, the values I have are a biased sample, in the sense that they are all under 1,000. (In other words, I do not have any information on how many values could or should exceed 1,000, or what those values may be.)

    I know that if I had a "full sample" I could calculate the mean and variance of my sample and back into the relevant mu and sigma parameters, but is there a way to estimate the full distribution's mu and sigma parameters using some kind of limited expected value and limited variance function?

    Thanks in advance!

    - Steve J
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Steve_J View Post
    I have a set of 50 values that I assume are part of a longnormal distribution. However, the values I have are a biased sample, in the sense that they are all under 1,000. (In other words, I do not have any information on how many values could or should exceed 1,000, or what those values may be.)

    I know that if I had a "full sample" I could calculate the mean and variance of my sample and back into the relevant mu and sigma parameters, but is there a way to estimate the full distribution's mu and sigma parameters using some kind of limited expected value and limited variance function?

    Thanks in advance!

    - Steve J

    Do you know how the data were/are censored?

    CB
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    All the values above the threshold were just completely ignored.
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    Quote Originally Posted by Steve_J View Post
    All the values above the threshold were just completely ignored.
    It should be possible to compute the conditional mean and variance of censored log-normal data, thogh it might be easier to do this numerically rather than analytically, but that should still be sufficient to do the fit (though this is not a routine task)

    The mean of the truncated distribution is:

    \mu^*=\frac{\int_0^{1000} x p(x)\;dx}{\int_0^{1000} p(x)\;dx}

    where p(x) is the density (in this case the log-normal density) of the un-truncated distribution

    and the variance of the truncated distribution is:

     <br />
\sigma^*=\frac{\int_0^{1000} (x-\mu^*)^2 p(x)\;dx}{\int_0^{1000} p(x)\;dx}<br />

    and the task is to find the parameters of the underlying log-normal that when truncated give the same mean and variance as your actual data.

    CB
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    Thanks, CB, but I think I'm still not 100% sure of what to do next. I believe the formulas you've outlined below are broadly applicable across all distributions; I'm not sure what the specific form of the equations will look like when the lognormal distribution is censored as described, nor exactly how to derive it.

    Again, I appreciate your help, and hope you can help get me the rest of the way there.

    - Steve J
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  6. #6
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    Quote Originally Posted by Steve_J View Post
    Thanks, CB, but I think I'm still not 100% sure of what to do next. I believe the formulas you've outlined below are broadly applicable across all distributions; I'm not sure what the specific form of the equations will look like when the lognormal distribution is censored as described, nor exactly how to derive it.

    Again, I appreciate your help, and hope you can help get me the rest of the way there.

    - Steve J
    Do you have the mean and variance of your sample?

    The density of the log-normal distribution is:

    p(x,\mu,\sigma^2)=\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{(\ln(x)-\mu)^2}{2\sigma^2}}

    So now we can (numerically at least) determin the mean and variance of the truncated log-normal using the formulas I posted earlier.

    An alternative is to compute the mean and variance og the log of your data and fit a truncated normal distribution.

    In fact there are many ways of going about this (most will produce slightly different answers but all should be pretty similar).

    CB
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