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Math Help - Y=1/X is normal distribution, how to solve X?

  1. #1
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    Y=1/X is normal distribution, how to solve X?

    I have a distribution X, whose reciprocal Y=1/X is a standard normal distribution Y~N(0,1).

    the PDF I achieved for X is 1/(x^2*(2*pi)^0.5)*exp(-1/2*(1/x)^2).

    The mean I achieved is 0, by using μ=xf(x)dx

    I tried to solve variance of X by using σ2=x^2f(x)dx, however, something wrong, I can not solve it.

    Can somebody help to check my calculations?

    Thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Show me your work and I'll take a look at it.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ceseng View Post
    I have a distribution X, whose reciprocal Y=1/X is a standard normal distribution Y~N(0,1).

    the PDF I achieved for X is 1/(x^2*(2*pi)^0.5)*exp(-1/2*(1/x)^2).

    The mean I achieved is 0, by using μ=xf(x)dx

    I tried to solve variance of X by using σ2=x^2f(x)dx, however, something wrong, I can not solve it.

    Can somebody help to check my calculations?

    Thanks.
    Assuming the density is correct:

    The mean is wrong it has no mean, the integral for the mean is an improper integral that does not converge, it also has no variance.

    (which sounds about right from the reciprical of a standard normal rv)

    CB
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