$X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $1$ and...












0












$begingroup$


Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$



My input



$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $



$X_1sim Exp(1)=e^{-x}implies G(1,1)$



Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $



Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$



$X_2sim G(1,2)$



$P(X_2-X_1<0)$



I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If you fix $X_2$, can you find $P(X_1<X_2)$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:26










  • $begingroup$
    But won't it include $X_2$ how will I get final result ?
    $endgroup$
    – Daman deep
    Dec 13 '18 at 14:27






  • 3




    $begingroup$
    If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:29


















0












$begingroup$


Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$



My input



$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $



$X_1sim Exp(1)=e^{-x}implies G(1,1)$



Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $



Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$



$X_2sim G(1,2)$



$P(X_2-X_1<0)$



I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If you fix $X_2$, can you find $P(X_1<X_2)$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:26










  • $begingroup$
    But won't it include $X_2$ how will I get final result ?
    $endgroup$
    – Daman deep
    Dec 13 '18 at 14:27






  • 3




    $begingroup$
    If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:29
















0












0








0





$begingroup$


Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$



My input



$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $



$X_1sim Exp(1)=e^{-x}implies G(1,1)$



Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $



Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$



$X_2sim G(1,2)$



$P(X_2-X_1<0)$



I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.










share|cite|improve this question









$endgroup$




Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$



My input



$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $



$X_1sim Exp(1)=e^{-x}implies G(1,1)$



Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $



Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$



$X_2sim G(1,2)$



$P(X_2-X_1<0)$



I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.







probability statistics probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 14:21









Daman deepDaman deep

750418




750418








  • 1




    $begingroup$
    If you fix $X_2$, can you find $P(X_1<X_2)$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:26










  • $begingroup$
    But won't it include $X_2$ how will I get final result ?
    $endgroup$
    – Daman deep
    Dec 13 '18 at 14:27






  • 3




    $begingroup$
    If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:29
















  • 1




    $begingroup$
    If you fix $X_2$, can you find $P(X_1<X_2)$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:26










  • $begingroup$
    But won't it include $X_2$ how will I get final result ?
    $endgroup$
    – Daman deep
    Dec 13 '18 at 14:27






  • 3




    $begingroup$
    If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:29










1




1




$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26




$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26












$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27




$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27




3




3




$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29






$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29












1 Answer
1






active

oldest

votes


















1












$begingroup$

Smiley is right.



Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$



(and leave $X_2-X_1$ for what it is!)






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%2f3038060%2fx-1-be-an-exponential-random-variable-with-mean-1-and-x-2-be-a-gamma-rando%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









    1












    $begingroup$

    Smiley is right.



    Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$



    (and leave $X_2-X_1$ for what it is!)






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Smiley is right.



      Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$



      (and leave $X_2-X_1$ for what it is!)






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Smiley is right.



        Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$



        (and leave $X_2-X_1$ for what it is!)






        share|cite|improve this answer









        $endgroup$



        Smiley is right.



        Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$



        (and leave $X_2-X_1$ for what it is!)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 14:42









        drhabdrhab

        101k545136




        101k545136






























            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%2f3038060%2fx-1-be-an-exponential-random-variable-with-mean-1-and-x-2-be-a-gamma-rando%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Aardman Animations

            Are they similar matrix