Convolution of weighted uniformly distributed random variables

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I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.



Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...



Thanks in advance!







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    up vote
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    down vote

    favorite












    I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.



    Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...



    Thanks in advance!







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.



      Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...



      Thanks in advance!







      share|cite|improve this question












      I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.



      Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...



      Thanks in advance!









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 20 at 12:46









      P3rs3rk3r

      385




      385




















          2 Answers
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          On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$



          Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get



          $$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$



          Solving the integral will give you the wanted result.






          share|cite|improve this answer



























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            Hint:



            For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.






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              2 Answers
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              2 Answers
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              up vote
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              On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$



              Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get



              $$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$



              Solving the integral will give you the wanted result.






              share|cite|improve this answer
























                up vote
                0
                down vote













                On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$



                Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get



                $$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$



                Solving the integral will give you the wanted result.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$



                  Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get



                  $$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$



                  Solving the integral will give you the wanted result.






                  share|cite|improve this answer












                  On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$



                  Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get



                  $$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$



                  Solving the integral will give you the wanted result.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 20 at 13:06









                  Gono

                  3,404316




                  3,404316




















                      up vote
                      0
                      down vote













                      Hint:



                      For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Hint:



                        For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Hint:



                          For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.






                          share|cite|improve this answer












                          Hint:



                          For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 20 at 13:36









                          drhab

                          87.8k541119




                          87.8k541119






















                               

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