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Thread: distribution of ln(X) when X is binodal normal

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    distribution of ln(X) when X is binodal normal




    Hi everyone.
    I am trying to find the distribution of ln(X) when i know that X is distributed as a binodal normal distribution, in other words:


    +
    which I suppose also can be written as


    How do I then go about finding the distribution of ln(X)? I am quite stuck here so I dont have that much "effort" to show, I assume i need to do some sort of transformation.
    Any input on how to start here is more than welcome!

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    Re: distribution of ln(X) when X is binodal normal

    That would be undefined for X<=0.
    I don't have emotions and sometimes that makes me very sad.

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    Re: distribution of ln(X) when X is binodal normal

    I am trying to find the distribution of ln(X)
    I suppose you want to find the probability density function of \ln(X)

    Seemingly you are talking about the mixtures of normal distribution, and its pdf will just be the weighted sum of normal pdfs. In such case the pdf of the resulting random variable after transformation should be also equal to the weighted sum of the individual pdfs after transformation. As long as you know how to find the pdf for a continuous transformation it should be fine.

    Another issue here is that the support of normal (mixture) random variable is including the negative real line, and therefore the log transformation will be invalid there.

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    Re: distribution of ln(X) when X is binodal normal

    Yes, X is between (0,1) btw.

    So, I write and i get


    I cannot get the + to work in latex so i wrote plus, sorry.
    I already tried this but it does not look like im getting anywhere.
    Is there any hope of getting this into the familiar normal form?

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    Re: distribution of ln(X) when X is binodal normal

    If X is between 0 and 1 then X is not normally distributed. Are you saying each piece of the mixture has a truncated normal (truncated betwee 0 and 1).
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    Re: distribution of ln(X) when X is binodal normal

    or, well i am doing an approximation here, since I cannot logit it will be some error in the model but the real parameter is only between 0 and 1, but ofcourse I will have some issues in the ends here...

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    Re: distribution of ln(X) when X is binodal normal

    But if i close my eyes and assume that this is ok, is ln(x) with the right kind of assumptions still normal? How do i go about finding the mean and variance?

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    Re: distribution of ln(X) when X is binodal normal


    Sorry I am kinda confused here.

    Is there any hope of getting this into the familiar normal form?
    If the original pdf is weighted sum then the resulting one is also a weighted sum.

    since I cannot logit ...
    Logit should be another transformation which is not the natural logarithm. Are you talking about first transforming the mixture normal by logit onto the (0,1) interval and then transform again by log?

    How do i go about finding the mean and variance?
    Well if this is your ultimate goal, there are some approximation method. Finding the pdf first usually cannot help much. It is still possible to have a closed-form solution but again you need to state your transformation and the underlying distribution assumption again clearly before we further discuss on it.

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