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







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  • 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
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$begingroup$

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 $g(z)=0$.







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    $begingroup$

    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






























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