Find Distribution of Sum of Random Variables given only Joint Distribution












0












$begingroup$


So if we have two random variables $X, Y$, with unknown distributions. Then we have a random variable $Z$, such that $Z=X+Y$.



Firstly, how do we find the CDF of $Z$, i.e. $F_Z(z)$, given the joint PDF of $X, Y$;



$$f_{XY}(x,y) for x, y > 0$$



Intuitively, as far as I can tell, the expression to find this is (hopefully):



$$F_Z(z) = int_{0}^{z} int_{0}^{z-x} f_{XY}(x,y) dy dx $$



Except I wasn't certain about whether this is the case, so I wanted to prove it. Therefore my real question is how to show this expression is correct?



Perhaps something along the lines of ?:



begin{align}
F_Z(z) & = P(Z le z) \
& = P(X + Y le z) \
...??? \
...??? \
& = P(Y le z -x and X le z)
end{align}



Any help would be great!






statistics probability-distributions bivariate-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your formula for the CDF of $Z$ is correct. You wrote $F_Z(z) = P(X+Y leq z)$. Can you express the last quantity in terms of the density?
    $endgroup$
    – Michh
    Dec 29 '18 at 16:34










  • $begingroup$
    @Michh I'm sorry, honestly, I've no clue. Any help?
    $endgroup$
    – ptolemy0
    Jan 1 at 10:08
















0












$begingroup$


So if we have two random variables $X, Y$, with unknown distributions. Then we have a random variable $Z$, such that $Z=X+Y$.



Firstly, how do we find the CDF of $Z$, i.e. $F_Z(z)$, given the joint PDF of $X, Y$;



$$f_{XY}(x,y) for x, y > 0$$



Intuitively, as far as I can tell, the expression to find this is (hopefully):



$$F_Z(z) = int_{0}^{z} int_{0}^{z-x} f_{XY}(x,y) dy dx $$



Except I wasn't certain about whether this is the case, so I wanted to prove it. Therefore my real question is how to show this expression is correct?



Perhaps something along the lines of ?:



begin{align}
F_Z(z) & = P(Z le z) \
& = P(X + Y le z) \
...??? \
...??? \
& = P(Y le z -x and X le z)
end{align}



Any help would be great!






statistics probability-distributions bivariate-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your formula for the CDF of $Z$ is correct. You wrote $F_Z(z) = P(X+Y leq z)$. Can you express the last quantity in terms of the density?
    $endgroup$
    – Michh
    Dec 29 '18 at 16:34










  • $begingroup$
    @Michh I'm sorry, honestly, I've no clue. Any help?
    $endgroup$
    – ptolemy0
    Jan 1 at 10:08














0












0








0





$begingroup$


So if we have two random variables $X, Y$, with unknown distributions. Then we have a random variable $Z$, such that $Z=X+Y$.



Firstly, how do we find the CDF of $Z$, i.e. $F_Z(z)$, given the joint PDF of $X, Y$;



$$f_{XY}(x,y) for x, y > 0$$



Intuitively, as far as I can tell, the expression to find this is (hopefully):



$$F_Z(z) = int_{0}^{z} int_{0}^{z-x} f_{XY}(x,y) dy dx $$



Except I wasn't certain about whether this is the case, so I wanted to prove it. Therefore my real question is how to show this expression is correct?



Perhaps something along the lines of ?:



begin{align}
F_Z(z) & = P(Z le z) \
& = P(X + Y le z) \
...??? \
...??? \
& = P(Y le z -x and X le z)
end{align}



Any help would be great!






statistics probability-distributions bivariate-distributions







share|cite|improve this question









$endgroup$




So if we have two random variables $X, Y$, with unknown distributions. Then we have a random variable $Z$, such that $Z=X+Y$.



Firstly, how do we find the CDF of $Z$, i.e. $F_Z(z)$, given the joint PDF of $X, Y$;



$$f_{XY}(x,y) for x, y > 0$$



Intuitively, as far as I can tell, the expression to find this is (hopefully):



$$F_Z(z) = int_{0}^{z} int_{0}^{z-x} f_{XY}(x,y) dy dx $$



Except I wasn't certain about whether this is the case, so I wanted to prove it. Therefore my real question is how to show this expression is correct?



Perhaps something along the lines of ?:



begin{align}
F_Z(z) & = P(Z le z) \
& = P(X + Y le z) \
...??? \
...??? \
& = P(Y le z -x and X le z)
end{align}



Any help would be great!






statistics probability-distributions bivariate-distributions




statistics probability-distributions bivariate-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 16:13









ptolemy0ptolemy0

154




154












  • $begingroup$
    Your formula for the CDF of $Z$ is correct. You wrote $F_Z(z) = P(X+Y leq z)$. Can you express the last quantity in terms of the density?
    $endgroup$
    – Michh
    Dec 29 '18 at 16:34










  • $begingroup$
    @Michh I'm sorry, honestly, I've no clue. Any help?
    $endgroup$
    – ptolemy0
    Jan 1 at 10:08


















  • $begingroup$
    Your formula for the CDF of $Z$ is correct. You wrote $F_Z(z) = P(X+Y leq z)$. Can you express the last quantity in terms of the density?
    $endgroup$
    – Michh
    Dec 29 '18 at 16:34










  • $begingroup$
    @Michh I'm sorry, honestly, I've no clue. Any help?
    $endgroup$
    – ptolemy0
    Jan 1 at 10:08
















$begingroup$
Your formula for the CDF of $Z$ is correct. You wrote $F_Z(z) = P(X+Y leq z)$. Can you express the last quantity in terms of the density?
$endgroup$
– Michh
Dec 29 '18 at 16:34




$begingroup$
Your formula for the CDF of $Z$ is correct. You wrote $F_Z(z) = P(X+Y leq z)$. Can you express the last quantity in terms of the density?
$endgroup$
– Michh
Dec 29 '18 at 16:34












$begingroup$
@Michh I'm sorry, honestly, I've no clue. Any help?
$endgroup$
– ptolemy0
Jan 1 at 10:08




$begingroup$
@Michh I'm sorry, honestly, I've no clue. Any help?
$endgroup$
– ptolemy0
Jan 1 at 10:08










1 Answer
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0












$begingroup$

We have
$$begin{align*} F_Z(z) &= P(X+Y leq z) = E[mathbf{1}_{X+Y leq z}]\
&= int_{[0,infty) times [0,infty)} mathbf{1}_{x+y leq z} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y
end{align*}$$
using the following property of the density: $E[g(X,Y)] = int_{[0,infty) times [0,infty)} g(x,y), f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y$. Now notice that $mathbf{1}_{x+y leq z} = mathbf{1}_{x leq z} mathbf{1}_{y leq z - x}$, that is $x+y leq z$ if and only if $x leq z$ and $y leq z - x$. So using Fubini's theorem, we get
$$
begin{align*} F_Z(z) = int_{[0,infty) times [0,infty)} mathbf{1}_{x leq z} mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y &= int_{[0,infty)} mathbf{1}_{x leq z} left(int_{[0,infty)}mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x \
&= int_0^z left(int_0^{z-x} f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x
end{align*}$$
which proves the result. Normally you would not have to go into as much detail and from the first equation above you could directly obtain the result.






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    1 Answer
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    $begingroup$

    We have
    $$begin{align*} F_Z(z) &= P(X+Y leq z) = E[mathbf{1}_{X+Y leq z}]\
    &= int_{[0,infty) times [0,infty)} mathbf{1}_{x+y leq z} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y
    end{align*}$$
    using the following property of the density: $E[g(X,Y)] = int_{[0,infty) times [0,infty)} g(x,y), f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y$. Now notice that $mathbf{1}_{x+y leq z} = mathbf{1}_{x leq z} mathbf{1}_{y leq z - x}$, that is $x+y leq z$ if and only if $x leq z$ and $y leq z - x$. So using Fubini's theorem, we get
    $$
    begin{align*} F_Z(z) = int_{[0,infty) times [0,infty)} mathbf{1}_{x leq z} mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y &= int_{[0,infty)} mathbf{1}_{x leq z} left(int_{[0,infty)}mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x \
    &= int_0^z left(int_0^{z-x} f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x
    end{align*}$$
    which proves the result. Normally you would not have to go into as much detail and from the first equation above you could directly obtain the result.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We have
      $$begin{align*} F_Z(z) &= P(X+Y leq z) = E[mathbf{1}_{X+Y leq z}]\
      &= int_{[0,infty) times [0,infty)} mathbf{1}_{x+y leq z} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y
      end{align*}$$
      using the following property of the density: $E[g(X,Y)] = int_{[0,infty) times [0,infty)} g(x,y), f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y$. Now notice that $mathbf{1}_{x+y leq z} = mathbf{1}_{x leq z} mathbf{1}_{y leq z - x}$, that is $x+y leq z$ if and only if $x leq z$ and $y leq z - x$. So using Fubini's theorem, we get
      $$
      begin{align*} F_Z(z) = int_{[0,infty) times [0,infty)} mathbf{1}_{x leq z} mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y &= int_{[0,infty)} mathbf{1}_{x leq z} left(int_{[0,infty)}mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x \
      &= int_0^z left(int_0^{z-x} f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x
      end{align*}$$
      which proves the result. Normally you would not have to go into as much detail and from the first equation above you could directly obtain the result.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We have
        $$begin{align*} F_Z(z) &= P(X+Y leq z) = E[mathbf{1}_{X+Y leq z}]\
        &= int_{[0,infty) times [0,infty)} mathbf{1}_{x+y leq z} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y
        end{align*}$$
        using the following property of the density: $E[g(X,Y)] = int_{[0,infty) times [0,infty)} g(x,y), f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y$. Now notice that $mathbf{1}_{x+y leq z} = mathbf{1}_{x leq z} mathbf{1}_{y leq z - x}$, that is $x+y leq z$ if and only if $x leq z$ and $y leq z - x$. So using Fubini's theorem, we get
        $$
        begin{align*} F_Z(z) = int_{[0,infty) times [0,infty)} mathbf{1}_{x leq z} mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y &= int_{[0,infty)} mathbf{1}_{x leq z} left(int_{[0,infty)}mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x \
        &= int_0^z left(int_0^{z-x} f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x
        end{align*}$$
        which proves the result. Normally you would not have to go into as much detail and from the first equation above you could directly obtain the result.






        share|cite|improve this answer









        $endgroup$



        We have
        $$begin{align*} F_Z(z) &= P(X+Y leq z) = E[mathbf{1}_{X+Y leq z}]\
        &= int_{[0,infty) times [0,infty)} mathbf{1}_{x+y leq z} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y
        end{align*}$$
        using the following property of the density: $E[g(X,Y)] = int_{[0,infty) times [0,infty)} g(x,y), f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y$. Now notice that $mathbf{1}_{x+y leq z} = mathbf{1}_{x leq z} mathbf{1}_{y leq z - x}$, that is $x+y leq z$ if and only if $x leq z$ and $y leq z - x$. So using Fubini's theorem, we get
        $$
        begin{align*} F_Z(z) = int_{[0,infty) times [0,infty)} mathbf{1}_{x leq z} mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}x mathrm{d}y &= int_{[0,infty)} mathbf{1}_{x leq z} left(int_{[0,infty)}mathbf{1}_{y leq z - x} ; f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x \
        &= int_0^z left(int_0^{z-x} f_{X,Y}(x,y) , mathrm{d}y right)mathrm{d}x
        end{align*}$$
        which proves the result. Normally you would not have to go into as much detail and from the first equation above you could directly obtain the result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 10:46









        MichhMichh

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