proof if X Weibull distributed then Y=aX^b (a,b>0) also Weibull distributed












0












$begingroup$


It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:



$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$



= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$



= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$



if k = s-1



= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$



So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0



I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.






statistics proof-verification probability-distributions







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  • $begingroup$
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    – StubbornAtom
    Dec 30 '18 at 10:03
















0












$begingroup$


It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:



$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$



= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$



= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$



if k = s-1



= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$



So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0



I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.






statistics proof-verification probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Here's a MathJax tutorial.
    $endgroup$
    – StubbornAtom
    Dec 30 '18 at 10:03














0












0








0





$begingroup$


It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:



$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$



= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$



= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$



if k = s-1



= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$



So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0



I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.






statistics proof-verification probability-distributions







share|cite|improve this question











$endgroup$




It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:



$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$



= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$



= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$



if k = s-1



= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$



So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0



I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.






statistics proof-verification probability-distributions




statistics proof-verification probability-distributions






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edited Dec 30 '18 at 9:41







dsamara

















asked Dec 30 '18 at 9:26









dsamaradsamara

12




12












  • $begingroup$
    Here's a MathJax tutorial.
    $endgroup$
    – StubbornAtom
    Dec 30 '18 at 10:03


















  • $begingroup$
    Here's a MathJax tutorial.
    $endgroup$
    – StubbornAtom
    Dec 30 '18 at 10:03
















$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03




$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03










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$begingroup$

Other approach:



Let $U$ have standard exponential distribution.



Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
parameters $theta$ and $k$.



Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.



Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
showing directly that $Y$ has Weibull distribution with parameters
$atheta^{b}$ and $kb^{-1}$.






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    0












    $begingroup$

    Other approach:



    Let $U$ have standard exponential distribution.



    Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
    showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
    parameters $theta$ and $k$.



    Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
    has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.



    Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
    showing directly that $Y$ has Weibull distribution with parameters
    $atheta^{b}$ and $kb^{-1}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Other approach:



      Let $U$ have standard exponential distribution.



      Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
      showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
      parameters $theta$ and $k$.



      Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
      has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.



      Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
      showing directly that $Y$ has Weibull distribution with parameters
      $atheta^{b}$ and $kb^{-1}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Other approach:



        Let $U$ have standard exponential distribution.



        Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
        showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
        parameters $theta$ and $k$.



        Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
        has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.



        Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
        showing directly that $Y$ has Weibull distribution with parameters
        $atheta^{b}$ and $kb^{-1}$.






        share|cite|improve this answer









        $endgroup$



        Other approach:



        Let $U$ have standard exponential distribution.



        Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
        showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
        parameters $theta$ and $k$.



        Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
        has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.



        Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
        showing directly that $Y$ has Weibull distribution with parameters
        $atheta^{b}$ and $kb^{-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 9:55









        drhabdrhab

        103k545136




        103k545136






























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