How to find pdf of X+Y given X and Y are dependent.












0












$begingroup$


The joint pdf is f(x,y) = $$frac{2}{5}(2x+3y)$$ for $0leq x leq 1,0leq y leq 1$
Normally if the random variables are independent, you can apply the convolution definition Z = X + Y which looks like $g(y) = int_{infty}^{infty}f_1(y-z)f_2(z)dz$. So I believe you can remove the dependence of one random variable on another. But I don't really know how to proceed at that point. For reference the answer is $g(y) = {z}^2$ for $0leq z leq 1$ and $z(2-z)$ $1leq z leq 2$






integration probability-theory probability-distributions random-variables







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For dependent, $g(y)=int_{-infty}^infty f(y-z,z)dz$.
    $endgroup$
    – NCh
    Dec 11 '18 at 2:41










  • $begingroup$
    But plugging that into the joint pdf for $0leq zleq1$ does not give ${z}^2$. What am I missing?
    $endgroup$
    – Sir lethian
    Dec 11 '18 at 9:11












  • $begingroup$
    Please show the calculations.
    $endgroup$
    – NCh
    Dec 11 '18 at 15:28
















0












$begingroup$


The joint pdf is f(x,y) = $$frac{2}{5}(2x+3y)$$ for $0leq x leq 1,0leq y leq 1$
Normally if the random variables are independent, you can apply the convolution definition Z = X + Y which looks like $g(y) = int_{infty}^{infty}f_1(y-z)f_2(z)dz$. So I believe you can remove the dependence of one random variable on another. But I don't really know how to proceed at that point. For reference the answer is $g(y) = {z}^2$ for $0leq z leq 1$ and $z(2-z)$ $1leq z leq 2$






integration probability-theory probability-distributions random-variables







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For dependent, $g(y)=int_{-infty}^infty f(y-z,z)dz$.
    $endgroup$
    – NCh
    Dec 11 '18 at 2:41










  • $begingroup$
    But plugging that into the joint pdf for $0leq zleq1$ does not give ${z}^2$. What am I missing?
    $endgroup$
    – Sir lethian
    Dec 11 '18 at 9:11












  • $begingroup$
    Please show the calculations.
    $endgroup$
    – NCh
    Dec 11 '18 at 15:28














0












0








0





$begingroup$


The joint pdf is f(x,y) = $$frac{2}{5}(2x+3y)$$ for $0leq x leq 1,0leq y leq 1$
Normally if the random variables are independent, you can apply the convolution definition Z = X + Y which looks like $g(y) = int_{infty}^{infty}f_1(y-z)f_2(z)dz$. So I believe you can remove the dependence of one random variable on another. But I don't really know how to proceed at that point. For reference the answer is $g(y) = {z}^2$ for $0leq z leq 1$ and $z(2-z)$ $1leq z leq 2$






integration probability-theory probability-distributions random-variables







share|cite|improve this question











$endgroup$




The joint pdf is f(x,y) = $$frac{2}{5}(2x+3y)$$ for $0leq x leq 1,0leq y leq 1$
Normally if the random variables are independent, you can apply the convolution definition Z = X + Y which looks like $g(y) = int_{infty}^{infty}f_1(y-z)f_2(z)dz$. So I believe you can remove the dependence of one random variable on another. But I don't really know how to proceed at that point. For reference the answer is $g(y) = {z}^2$ for $0leq z leq 1$ and $z(2-z)$ $1leq z leq 2$






integration probability-theory probability-distributions random-variables




integration probability-theory probability-distributions random-variables






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 1:18







Sir lethian

















asked Dec 11 '18 at 1:07









Sir lethianSir lethian

163




163








  • 1




    $begingroup$
    For dependent, $g(y)=int_{-infty}^infty f(y-z,z)dz$.
    $endgroup$
    – NCh
    Dec 11 '18 at 2:41










  • $begingroup$
    But plugging that into the joint pdf for $0leq zleq1$ does not give ${z}^2$. What am I missing?
    $endgroup$
    – Sir lethian
    Dec 11 '18 at 9:11












  • $begingroup$
    Please show the calculations.
    $endgroup$
    – NCh
    Dec 11 '18 at 15:28














  • 1




    $begingroup$
    For dependent, $g(y)=int_{-infty}^infty f(y-z,z)dz$.
    $endgroup$
    – NCh
    Dec 11 '18 at 2:41










  • $begingroup$
    But plugging that into the joint pdf for $0leq zleq1$ does not give ${z}^2$. What am I missing?
    $endgroup$
    – Sir lethian
    Dec 11 '18 at 9:11












  • $begingroup$
    Please show the calculations.
    $endgroup$
    – NCh
    Dec 11 '18 at 15:28








1




1




$begingroup$
For dependent, $g(y)=int_{-infty}^infty f(y-z,z)dz$.
$endgroup$
– NCh
Dec 11 '18 at 2:41




$begingroup$
For dependent, $g(y)=int_{-infty}^infty f(y-z,z)dz$.
$endgroup$
– NCh
Dec 11 '18 at 2:41












$begingroup$
But plugging that into the joint pdf for $0leq zleq1$ does not give ${z}^2$. What am I missing?
$endgroup$
– Sir lethian
Dec 11 '18 at 9:11






$begingroup$
But plugging that into the joint pdf for $0leq zleq1$ does not give ${z}^2$. What am I missing?
$endgroup$
– Sir lethian
Dec 11 '18 at 9:11














$begingroup$
Please show the calculations.
$endgroup$
– NCh
Dec 11 '18 at 15:28




$begingroup$
Please show the calculations.
$endgroup$
– NCh
Dec 11 '18 at 15:28










1 Answer
1






active

oldest

votes


















0












$begingroup$

For dependent $X$ and $Y$ the convolution formula for the density of $Z:=X+Y$ is
$$
g(z):=int_{-infty}^infty f(z-x,x),dx.tag1
$$
In this case the integrand in (1) evaluates to $f(z-x,x)=frac25left(2(z-x)+3(x)right)$, which simplifies to $frac25(2z+x)$, provided
$$
0le z-xle1qquadrm{and}qquad 0le xle 1tag2
$$
and equals zero otherwise. So to evaluate (1), which is an integral over $x$, you need to determine, for each fixed value of $z$, the range of $x$ values where the conditions in (2) are satisfied. Draw a picture to see the limits of integration:





From the picture it is clear that you need to argue by cases:




  1. If $0le zle1$, the limits run from $x=0$ to $x=z$ and therefore $g(z)=int_{x=0}^zfrac25(2z+x),dx$.


  2. If $1le zle 2$, the limits run from $x=z-1$ to $x=1$ and therefore $g(z)=int_{x=z-1}^1frac25(2z+x),dx$.


  3. For $z$ outside these two ranges the integrand is zero so $g(z)=0$.







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%2f3034731%2fhow-to-find-pdf-of-xy-given-x-and-y-are-dependent%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












    $begingroup$

    For dependent $X$ and $Y$ the convolution formula for the density of $Z:=X+Y$ is
    $$
    g(z):=int_{-infty}^infty f(z-x,x),dx.tag1
    $$
    In this case the integrand in (1) evaluates to $f(z-x,x)=frac25left(2(z-x)+3(x)right)$, which simplifies to $frac25(2z+x)$, provided
    $$
    0le z-xle1qquadrm{and}qquad 0le xle 1tag2
    $$
    and equals zero otherwise. So to evaluate (1), which is an integral over $x$, you need to determine, for each fixed value of $z$, the range of $x$ values where the conditions in (2) are satisfied. Draw a picture to see the limits of integration:





    From the picture it is clear that you need to argue by cases:




    1. If $0le zle1$, the limits run from $x=0$ to $x=z$ and therefore $g(z)=int_{x=0}^zfrac25(2z+x),dx$.


    2. If $1le zle 2$, the limits run from $x=z-1$ to $x=1$ and therefore $g(z)=int_{x=z-1}^1frac25(2z+x),dx$.


    3. For $z$ outside these two ranges the integrand is zero so $g(z)=0$.







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For dependent $X$ and $Y$ the convolution formula for the density of $Z:=X+Y$ is
      $$
      g(z):=int_{-infty}^infty f(z-x,x),dx.tag1
      $$
      In this case the integrand in (1) evaluates to $f(z-x,x)=frac25left(2(z-x)+3(x)right)$, which simplifies to $frac25(2z+x)$, provided
      $$
      0le z-xle1qquadrm{and}qquad 0le xle 1tag2
      $$
      and equals zero otherwise. So to evaluate (1), which is an integral over $x$, you need to determine, for each fixed value of $z$, the range of $x$ values where the conditions in (2) are satisfied. Draw a picture to see the limits of integration:





      From the picture it is clear that you need to argue by cases:




      1. If $0le zle1$, the limits run from $x=0$ to $x=z$ and therefore $g(z)=int_{x=0}^zfrac25(2z+x),dx$.


      2. If $1le zle 2$, the limits run from $x=z-1$ to $x=1$ and therefore $g(z)=int_{x=z-1}^1frac25(2z+x),dx$.


      3. For $z$ outside these two ranges the integrand is zero so $g(z)=0$.







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For dependent $X$ and $Y$ the convolution formula for the density of $Z:=X+Y$ is
        $$
        g(z):=int_{-infty}^infty f(z-x,x),dx.tag1
        $$
        In this case the integrand in (1) evaluates to $f(z-x,x)=frac25left(2(z-x)+3(x)right)$, which simplifies to $frac25(2z+x)$, provided
        $$
        0le z-xle1qquadrm{and}qquad 0le xle 1tag2
        $$
        and equals zero otherwise. So to evaluate (1), which is an integral over $x$, you need to determine, for each fixed value of $z$, the range of $x$ values where the conditions in (2) are satisfied. Draw a picture to see the limits of integration:





        From the picture it is clear that you need to argue by cases:




        1. If $0le zle1$, the limits run from $x=0$ to $x=z$ and therefore $g(z)=int_{x=0}^zfrac25(2z+x),dx$.


        2. If $1le zle 2$, the limits run from $x=z-1$ to $x=1$ and therefore $g(z)=int_{x=z-1}^1frac25(2z+x),dx$.


        3. For $z$ outside these two ranges the integrand is zero so $g(z)=0$.







        share|cite|improve this answer









        $endgroup$



        For dependent $X$ and $Y$ the convolution formula for the density of $Z:=X+Y$ is
        $$
        g(z):=int_{-infty}^infty f(z-x,x),dx.tag1
        $$
        In this case the integrand in (1) evaluates to $f(z-x,x)=frac25left(2(z-x)+3(x)right)$, which simplifies to $frac25(2z+x)$, provided
        $$
        0le z-xle1qquadrm{and}qquad 0le xle 1tag2
        $$
        and equals zero otherwise. So to evaluate (1), which is an integral over $x$, you need to determine, for each fixed value of $z$, the range of $x$ values where the conditions in (2) are satisfied. Draw a picture to see the limits of integration:





        From the picture it is clear that you need to argue by cases:




        1. If $0le zle1$, the limits run from $x=0$ to $x=z$ and therefore $g(z)=int_{x=0}^zfrac25(2z+x),dx$.


        2. If $1le zle 2$, the limits run from $x=z-1$ to $x=1$ and therefore $g(z)=int_{x=z-1}^1frac25(2z+x),dx$.


        3. For $z$ outside these two ranges the integrand is zero so $g(z)=0$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 2:22









        grand_chatgrand_chat

        20.3k11326




        20.3k11326






























            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%2f3034731%2fhow-to-find-pdf-of-xy-given-x-and-y-are-dependent%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
            Read more