Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2exp(-lambda y)$.












0












$begingroup$



Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.




I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.



Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.



However, I struggle with two things:




  1. How to determine the range of $g(s,t)$ function?

  2. How to determine the lower and upper boundry pf the integral?






probability-theory statistics probability-distributions definite-integrals marginal-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you need the range of the density?
    $endgroup$
    – orange
    Dec 14 '18 at 22:02










  • $begingroup$
    @orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
    $endgroup$
    – user1607
    Dec 14 '18 at 22:04












  • $begingroup$
    Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
    $endgroup$
    – orange
    Dec 14 '18 at 22:18












  • $begingroup$
    But again, you integrate over the domain and not the range.
    $endgroup$
    – orange
    Dec 14 '18 at 22:25






  • 1




    $begingroup$
    "is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
    $endgroup$
    – Did
    Dec 15 '18 at 10:15
















0












$begingroup$



Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.




I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.



Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.



However, I struggle with two things:




  1. How to determine the range of $g(s,t)$ function?

  2. How to determine the lower and upper boundry pf the integral?






probability-theory statistics probability-distributions definite-integrals marginal-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you need the range of the density?
    $endgroup$
    – orange
    Dec 14 '18 at 22:02










  • $begingroup$
    @orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
    $endgroup$
    – user1607
    Dec 14 '18 at 22:04












  • $begingroup$
    Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
    $endgroup$
    – orange
    Dec 14 '18 at 22:18












  • $begingroup$
    But again, you integrate over the domain and not the range.
    $endgroup$
    – orange
    Dec 14 '18 at 22:25






  • 1




    $begingroup$
    "is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
    $endgroup$
    – Did
    Dec 15 '18 at 10:15














0












0








0


0



$begingroup$



Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.




I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.



Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.



However, I struggle with two things:




  1. How to determine the range of $g(s,t)$ function?

  2. How to determine the lower and upper boundry pf the integral?






probability-theory statistics probability-distributions definite-integrals marginal-distribution







share|cite|improve this question











$endgroup$





Find the density function $X+Y$, where $X$ and $Y$ have a joint density $f(x,y)=lambda^2e^{-lambda y}$ for $0leq x leq y, lambda >0 $.




I used the Density transformation theorem to find the joint density of $g(s,t)= lambda^2 e^{-lambda t}$.



Then I find the marginal density $g(s)= int g(s,t) dt$ where $s= x + y$, and $t = y$.



However, I struggle with two things:




  1. How to determine the range of $g(s,t)$ function?

  2. How to determine the lower and upper boundry pf the integral?






probability-theory statistics probability-distributions definite-integrals marginal-distribution




probability-theory statistics probability-distributions definite-integrals marginal-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 10:04









Davide Giraudo

127k16153268




127k16153268










asked Dec 14 '18 at 21:57









user1607user1607

1718




1718












  • $begingroup$
    Why do you need the range of the density?
    $endgroup$
    – orange
    Dec 14 '18 at 22:02










  • $begingroup$
    @orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
    $endgroup$
    – user1607
    Dec 14 '18 at 22:04












  • $begingroup$
    Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
    $endgroup$
    – orange
    Dec 14 '18 at 22:18












  • $begingroup$
    But again, you integrate over the domain and not the range.
    $endgroup$
    – orange
    Dec 14 '18 at 22:25






  • 1




    $begingroup$
    "is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
    $endgroup$
    – Did
    Dec 15 '18 at 10:15


















  • $begingroup$
    Why do you need the range of the density?
    $endgroup$
    – orange
    Dec 14 '18 at 22:02










  • $begingroup$
    @orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
    $endgroup$
    – user1607
    Dec 14 '18 at 22:04












  • $begingroup$
    Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
    $endgroup$
    – orange
    Dec 14 '18 at 22:18












  • $begingroup$
    But again, you integrate over the domain and not the range.
    $endgroup$
    – orange
    Dec 14 '18 at 22:25






  • 1




    $begingroup$
    "is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
    $endgroup$
    – Did
    Dec 15 '18 at 10:15
















$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02




$begingroup$
Why do you need the range of the density?
$endgroup$
– orange
Dec 14 '18 at 22:02












$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04






$begingroup$
@orange is that not the "right way" to always write a density together with its range? Also, if i am not mistaken, the two question are related. The boundaries of the integral can be determined from the range of the distribution? Of course, I am asking this questions because i dont know any better
$endgroup$
– user1607
Dec 14 '18 at 22:04














$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18






$begingroup$
Well, for the range, the function is independent of $x$ and is continuous. So get the limit for $yto 0$ and $yto infty$.
$endgroup$
– orange
Dec 14 '18 at 22:18














$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25




$begingroup$
But again, you integrate over the domain and not the range.
$endgroup$
– orange
Dec 14 '18 at 22:25




1




1




$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15




$begingroup$
"is that not the "right way" to always write a density together with its range?" Indeed, and this is why the writing in your title, say, is misleading. You should get used to write PDFs correctly, that is, their range included, here, $$f(x,y)=lambda^2e^{-lambda y}mathbf 1_{0<x<y}$$ and the rest of your questions becomes automatically trivial...
$endgroup$
– Did
Dec 15 '18 at 10:15










1 Answer
1






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0












$begingroup$

The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.



First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.






share|cite|improve this answer









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    $begingroup$

    The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.



    First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.



      First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.



        First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.






        share|cite|improve this answer









        $endgroup$



        The pdf of $Z:=X+Y$ is $$int_{0le xle y}lambda^2 e^{-lambda y}delta (x+y-z)dxdy=int_{z/2}^zlambda^2 e^{-lambda y}dy=lambda (e^{-lambda z/2}-e^{-lambda z})$$for $z=0$. (Note this is clearly a pdf.) I'll explain the steps below.



        First we write down an integral using the $delta$ function; then we integrate out $x$, which only retains value of $y$ for which $x+y-z$ is negative when $x=0$ but positive when $x=y$, so that the integral is around the delta peak. These constraints can be stated as $yle zle 2y$, or equivalently $y/2le zle y$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 10:15









        J.G.J.G.

        28.8k22845




        28.8k22845






























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