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Math Help - recognising uniform distibution

  1. #1
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    recognising uniform distibution

    I am going through an example where one needs to find score and information for a random sample of size n from the logistic distribution with a density function

    f_Y(y;\mu)=\frac{e^{y-\mu}}{[1+e^{(e-\mu)}]^2}

    and the score is s_Y(Y;\mu)=\Sigma_{i=1}^n(-1+2F_Y(y_i;\mu))

    then the explanation goes, 'by recognising that F_Y(Y;\mu)\sim Unif[0;1] we can calculate information... and they use the formula for Var of a Uniform distribution.

    Can someone help me to see that F_Y(Y;\mu)\sim Unif[0;1]?
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  2. #2
    Moo
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    Hello,

    That's a standard property, if Fy has an inverse (it's the case in particular if the rv has a continuous density).

    Indeed : P(F_Y(Y)\leq x)=P(Y\leq F_Y^{-1}(x))=F_Y(F_Y^{-1}(x))=x, for x\in[0,1] since F_Y is a probability and is hence taking values between 0 and 1.
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  3. #3
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    Quote Originally Posted by Volga View Post
    I am going through an example where one needs to find score and information for a random sample of size n from the logistic distribution with a density function

    f_Y(y;\mu)=\frac{e^{y-\mu}}{[1+e^{(e-\mu)}]^2}

    and the score is s_Y(Y;\mu)=\Sigma_{i=1}^n(-1+2F_Y(y_i;\mu))

    then the explanation goes, 'by recognising that F_Y(Y;\mu)\sim Unif[0;1] we can calculate information... and they use the formula for Var of a Uniform distribution.

    Can someone help me to see that F_Y(Y;\mu)\sim Unif[0;1]?
    There are a number of ways of looking at this, one is the usual transformation method where if $$U has pdf $$f_U(u) we determine the distribution of $$V=h(U) where $$h(.) is an increasing function.

    f_V(v)=\dfrac{f_U(u)}{h'(u)}

    But if h(u)=F_U(u) then h'(u)=f_U(u) so f_V(v)=1 in the range of $$h(u) (which is $$[0,1]).

    CB
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  4. #4
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    Nice one, I didn't know that, thank you!
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