Z=X+Y pdf, there is a part I can't understand.












1












$begingroup$


Two independent random variable X,Y which is U(0,1)



what is the pdf of Z=X+Y?



$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}



I understand I have to differential $F_z(z)$



The thing I can't understand is the highlight part.



Why does it starts with z-1?






statistics random-variables







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
    $endgroup$
    – J.G.
    Jan 5 at 14:25










  • $begingroup$
    @J.G. Thank you I edited it.
    $endgroup$
    – yonghankwon0
    Jan 6 at 5:01
















1












$begingroup$


Two independent random variable X,Y which is U(0,1)



what is the pdf of Z=X+Y?



$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}



I understand I have to differential $F_z(z)$



The thing I can't understand is the highlight part.



Why does it starts with z-1?






statistics random-variables







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
    $endgroup$
    – J.G.
    Jan 5 at 14:25










  • $begingroup$
    @J.G. Thank you I edited it.
    $endgroup$
    – yonghankwon0
    Jan 6 at 5:01














1












1








1





$begingroup$


Two independent random variable X,Y which is U(0,1)



what is the pdf of Z=X+Y?



$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}



I understand I have to differential $F_z(z)$



The thing I can't understand is the highlight part.



Why does it starts with z-1?






statistics random-variables







share|cite|improve this question











$endgroup$




Two independent random variable X,Y which is U(0,1)



what is the pdf of Z=X+Y?



$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}



I understand I have to differential $F_z(z)$



The thing I can't understand is the highlight part.



Why does it starts with z-1?






statistics random-variables




statistics random-variables






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 5:00







yonghankwon0

















asked Jan 5 at 13:37









yonghankwon0yonghankwon0

155




155








  • 1




    $begingroup$
    By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
    $endgroup$
    – J.G.
    Jan 5 at 14:25










  • $begingroup$
    @J.G. Thank you I edited it.
    $endgroup$
    – yonghankwon0
    Jan 6 at 5:01














  • 1




    $begingroup$
    By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
    $endgroup$
    – J.G.
    Jan 5 at 14:25










  • $begingroup$
    @J.G. Thank you I edited it.
    $endgroup$
    – yonghankwon0
    Jan 6 at 5:01








1




1




$begingroup$
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
$endgroup$
– J.G.
Jan 5 at 14:25




$begingroup$
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
$endgroup$
– J.G.
Jan 5 at 14:25












$begingroup$
@J.G. Thank you I edited it.
$endgroup$
– yonghankwon0
Jan 6 at 5:01




$begingroup$
@J.G. Thank you I edited it.
$endgroup$
– yonghankwon0
Jan 6 at 5:01










1 Answer
1






active

oldest

votes


















1












$begingroup$

Convolution of probability densities



Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
Define $Z equiv X + Y$.
Note that
$$
F_Z(z)
= mathbb{P}(Z leq z)
= mathbb{P}(X + Y leq z)
= int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
= int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
$$
Differentiating,
$$
f_Z(z)
= int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
$$




Convolution of i.i.d. uniform distributions



If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
Plugging this into the integral of the previous section,
$$
f_Z(z)
= int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
$$
Proceeding by cases,
$$
f_Z(z)
= begin{cases}
int_0^z dx = z & text{if } 0 leq z leq 1 \
int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
end{cases}
$$






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Convolution of probability densities



    Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
    Define $Z equiv X + Y$.
    Note that
    $$
    F_Z(z)
    = mathbb{P}(Z leq z)
    = mathbb{P}(X + Y leq z)
    = int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
    = int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
    $$
    Differentiating,
    $$
    f_Z(z)
    = int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
    $$




    Convolution of i.i.d. uniform distributions



    If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
    Plugging this into the integral of the previous section,
    $$
    f_Z(z)
    = int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
    $$
    Proceeding by cases,
    $$
    f_Z(z)
    = begin{cases}
    int_0^z dx = z & text{if } 0 leq z leq 1 \
    int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
    end{cases}
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Convolution of probability densities



      Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
      Define $Z equiv X + Y$.
      Note that
      $$
      F_Z(z)
      = mathbb{P}(Z leq z)
      = mathbb{P}(X + Y leq z)
      = int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
      = int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
      $$
      Differentiating,
      $$
      f_Z(z)
      = int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
      $$




      Convolution of i.i.d. uniform distributions



      If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
      Plugging this into the integral of the previous section,
      $$
      f_Z(z)
      = int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
      $$
      Proceeding by cases,
      $$
      f_Z(z)
      = begin{cases}
      int_0^z dx = z & text{if } 0 leq z leq 1 \
      int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
      end{cases}
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Convolution of probability densities



        Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
        Define $Z equiv X + Y$.
        Note that
        $$
        F_Z(z)
        = mathbb{P}(Z leq z)
        = mathbb{P}(X + Y leq z)
        = int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
        = int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
        $$
        Differentiating,
        $$
        f_Z(z)
        = int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
        $$




        Convolution of i.i.d. uniform distributions



        If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
        Plugging this into the integral of the previous section,
        $$
        f_Z(z)
        = int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
        $$
        Proceeding by cases,
        $$
        f_Z(z)
        = begin{cases}
        int_0^z dx = z & text{if } 0 leq z leq 1 \
        int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
        end{cases}
        $$






        share|cite|improve this answer









        $endgroup$



        Convolution of probability densities



        Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
        Define $Z equiv X + Y$.
        Note that
        $$
        F_Z(z)
        = mathbb{P}(Z leq z)
        = mathbb{P}(X + Y leq z)
        = int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
        = int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
        $$
        Differentiating,
        $$
        f_Z(z)
        = int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
        $$




        Convolution of i.i.d. uniform distributions



        If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
        Plugging this into the integral of the previous section,
        $$
        f_Z(z)
        = int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
        $$
        Proceeding by cases,
        $$
        f_Z(z)
        = begin{cases}
        int_0^z dx = z & text{if } 0 leq z leq 1 \
        int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
        end{cases}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 5:25









        parsiadparsiad

        18.5k32453




        18.5k32453






























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