Sum of absolutely continuous independent random variables












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I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).



Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
$mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$



How did we get this last step?






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    $begingroup$


    I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).



    Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
    $mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$



    How did we get this last step?






    random-variables







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).



      Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
      $mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$



      How did we get this last step?






      random-variables







      share|cite|improve this question









      $endgroup$




      I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).



      Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
      $mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$



      How did we get this last step?






      random-variables




      random-variables






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 2 at 9:47









      Baffo rastaBaffo rasta

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          $$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.






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            $$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.






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              $begingroup$

              $$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.






              share|cite|improve this answer









              $endgroup$
















                1












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                $begingroup$

                $$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.






                share|cite|improve this answer









                $endgroup$



                $$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 12:14









                Kavi Rama MurthyKavi Rama Murthy

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