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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Here's a MathJax tutorial.
    $endgroup$
    – 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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










1 Answer
1






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056643%2fproof-if-x-weibull-distributed-then-y-axb-a-b0-also-weibull-distributed%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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

        102k545136




        102k545136






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056643%2fproof-if-x-weibull-distributed-then-y-axb-a-b0-also-weibull-distributed%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more