Finding a distribution function of random variable sum












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Let $xi_1, xi_2, xi_3$ independent random variables in $(Omega, mathcal{F},mathbb{P}).$ Also, they are evenly distributed in $[0,1]$. I need to find a distribution function of sum $xi_1+ xi_2+ xi_3$.



So, I know that $F_{xi}(x)=mathbb{P}(omega : xi(omega leq x)=mathbb{P}(xi leq x)=mathbb{P}(xi^{-1}(-infty;x])$. Also I know that $mathbb{P}(xi_1,xi_2,xi_3)=mathbb{P}(xi_1)* mathbb{P}(xi_1)* mathbb{P}(xi_1)$.



But how to use it and find a distribution function?










share|cite|improve this question









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    1












    $begingroup$


    Let $xi_1, xi_2, xi_3$ independent random variables in $(Omega, mathcal{F},mathbb{P}).$ Also, they are evenly distributed in $[0,1]$. I need to find a distribution function of sum $xi_1+ xi_2+ xi_3$.



    So, I know that $F_{xi}(x)=mathbb{P}(omega : xi(omega leq x)=mathbb{P}(xi leq x)=mathbb{P}(xi^{-1}(-infty;x])$. Also I know that $mathbb{P}(xi_1,xi_2,xi_3)=mathbb{P}(xi_1)* mathbb{P}(xi_1)* mathbb{P}(xi_1)$.



    But how to use it and find a distribution function?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $xi_1, xi_2, xi_3$ independent random variables in $(Omega, mathcal{F},mathbb{P}).$ Also, they are evenly distributed in $[0,1]$. I need to find a distribution function of sum $xi_1+ xi_2+ xi_3$.



      So, I know that $F_{xi}(x)=mathbb{P}(omega : xi(omega leq x)=mathbb{P}(xi leq x)=mathbb{P}(xi^{-1}(-infty;x])$. Also I know that $mathbb{P}(xi_1,xi_2,xi_3)=mathbb{P}(xi_1)* mathbb{P}(xi_1)* mathbb{P}(xi_1)$.



      But how to use it and find a distribution function?










      share|cite|improve this question









      $endgroup$




      Let $xi_1, xi_2, xi_3$ independent random variables in $(Omega, mathcal{F},mathbb{P}).$ Also, they are evenly distributed in $[0,1]$. I need to find a distribution function of sum $xi_1+ xi_2+ xi_3$.



      So, I know that $F_{xi}(x)=mathbb{P}(omega : xi(omega leq x)=mathbb{P}(xi leq x)=mathbb{P}(xi^{-1}(-infty;x])$. Also I know that $mathbb{P}(xi_1,xi_2,xi_3)=mathbb{P}(xi_1)* mathbb{P}(xi_1)* mathbb{P}(xi_1)$.



      But how to use it and find a distribution function?







      probability probability-theory probability-distributions random-variables






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      asked Dec 10 '18 at 6:50









      AtstovasAtstovas

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          Convolution of two distributions.



          $t_{x_0} = 0$



          $t_{x_1} = 1$



          $t_{h_0} = 0$



          $t_{h_1} = 1$



          Thus $$f_Y(t) = 0, t le t_{x_0}+t_{h_0} ,$$



          $$f_Y(t) = int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(tau)f_H(t-tau)dtau, text{ } t_{x_0}+t_{h_0} le t le t_{x_1}+t_{h_1},$$



          $$f_Y(t) = 0, t ge t_{x_1}+t_{h_1} ,$$



          THese translate to the following solution



          First convolve two uniform distributions



          $X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$



          $$Y(t) = x(t).h(t) = int_{-infty}^{infty} x(tau)h(t-tau)dtau$$
          The above convolution reduces to



          $y(t) = 0, tlt 0$



          $y(t) = int_{max(0,t-1)}^{min(1,t)} dtau , 0lt t lt 2$,



          $y(t) = 0 , t gt 2$



          The middle one will have to split into two intervals, namely $0lt t lt 1$ and $1lt t lt 2$



          $y(t) = 0, tlt 0$



          $y(t) = int_{0}^{t} dtau =t, 0lt tlt 1$,



          $y(t) = int_{t-1}^{1} dtau = 2-t, 1lt tlt 2$,



          $y(t) = 0 , t gt 2$



          Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$



          For $0lt tlt 1$, the bounds are



          $t_{s_0} = 0$



          $t_{s_1} = 1$



          $t_{y_0} = 0$



          $t_{y_1} = 1$



          $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau = int_{0}^{t}tau dtau = frac{t^2}{2}, 0lt tlt 1$$,



          For $1lt tlt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are



          $t_{s_0} = 0$



          $t_{s_1} = 1$



          $t_{y_0} = 0$



          $t_{y_1} = 1$



          and for the interval $(1,2)$ the bounds are



          $t_{s_0} = 0$



          $t_{s_1} = 1$



          $t_{y_0} = 1$



          $t_{y_1} = 2$



          Thus $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau + int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau$$ $$ = int_{t-1}^{1}tau dtau + int_{1}^{t}(2-tau) dtau$$ $$ = -frac{1}{2}(2t^2-6t+3), 1lt tlt 2$$,



          For $2lt tlt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are



          $t_{s_0} = 0$



          $t_{s_1} = 1$



          $t_{y_0} = 1$



          $t_{y_1} = 2$



          Thus $$W(t) = int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau $$ $$ = int_{t-1}^{2} (2- tau) dtau$$ $$ = frac{(t-3)^2}{2}, 2lt t lt 3$$



          and finally $W(t) = 0, tgt 3$



          Thus the $W(t)$ is defined by



          $W(t) = 0 , tlt 0$



          $W(t) = frac{t^2}{2}, 0lt t lt 1$



          $W(t) = -t^2+3t-frac{3}{2}, 1lt t lt 2$



          $W(t) = frac{(t-3)^2}{2}, 2lt t lt 3$



          $W(t) = 0 , tgt 3$






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            One way is (I used $Z$ instead of $xi$)$$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{min(z,,1)}int_{z_2=0}^{min(z-z_3,,1)}int_{z_1=0}^{min(z-z_2-z_3,,1)}dz_1,dz_2,dz_3$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How to integrate this?
              $endgroup$
              – Atstovas
              Dec 10 '18 at 9:00










            • $begingroup$
              I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
              $endgroup$
              – BlackMath
              Dec 10 '18 at 9:24












            • $begingroup$
              A similar questions is found here math.stackexchange.com/questions/2631501/…
              $endgroup$
              – BlackMath
              Dec 10 '18 at 9:32










            • $begingroup$
              can you show your solution when $z<1$?
              $endgroup$
              – Atstovas
              Dec 11 '18 at 17:40










            • $begingroup$
              It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
              $endgroup$
              – BlackMath
              Dec 11 '18 at 23:52













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            2 Answers
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            2 Answers
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            1












            $begingroup$

            Convolution of two distributions.



            $t_{x_0} = 0$



            $t_{x_1} = 1$



            $t_{h_0} = 0$



            $t_{h_1} = 1$



            Thus $$f_Y(t) = 0, t le t_{x_0}+t_{h_0} ,$$



            $$f_Y(t) = int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(tau)f_H(t-tau)dtau, text{ } t_{x_0}+t_{h_0} le t le t_{x_1}+t_{h_1},$$



            $$f_Y(t) = 0, t ge t_{x_1}+t_{h_1} ,$$



            THese translate to the following solution



            First convolve two uniform distributions



            $X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$



            $$Y(t) = x(t).h(t) = int_{-infty}^{infty} x(tau)h(t-tau)dtau$$
            The above convolution reduces to



            $y(t) = 0, tlt 0$



            $y(t) = int_{max(0,t-1)}^{min(1,t)} dtau , 0lt t lt 2$,



            $y(t) = 0 , t gt 2$



            The middle one will have to split into two intervals, namely $0lt t lt 1$ and $1lt t lt 2$



            $y(t) = 0, tlt 0$



            $y(t) = int_{0}^{t} dtau =t, 0lt tlt 1$,



            $y(t) = int_{t-1}^{1} dtau = 2-t, 1lt tlt 2$,



            $y(t) = 0 , t gt 2$



            Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$



            For $0lt tlt 1$, the bounds are



            $t_{s_0} = 0$



            $t_{s_1} = 1$



            $t_{y_0} = 0$



            $t_{y_1} = 1$



            $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau = int_{0}^{t}tau dtau = frac{t^2}{2}, 0lt tlt 1$$,



            For $1lt tlt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are



            $t_{s_0} = 0$



            $t_{s_1} = 1$



            $t_{y_0} = 0$



            $t_{y_1} = 1$



            and for the interval $(1,2)$ the bounds are



            $t_{s_0} = 0$



            $t_{s_1} = 1$



            $t_{y_0} = 1$



            $t_{y_1} = 2$



            Thus $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau + int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau$$ $$ = int_{t-1}^{1}tau dtau + int_{1}^{t}(2-tau) dtau$$ $$ = -frac{1}{2}(2t^2-6t+3), 1lt tlt 2$$,



            For $2lt tlt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are



            $t_{s_0} = 0$



            $t_{s_1} = 1$



            $t_{y_0} = 1$



            $t_{y_1} = 2$



            Thus $$W(t) = int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau $$ $$ = int_{t-1}^{2} (2- tau) dtau$$ $$ = frac{(t-3)^2}{2}, 2lt t lt 3$$



            and finally $W(t) = 0, tgt 3$



            Thus the $W(t)$ is defined by



            $W(t) = 0 , tlt 0$



            $W(t) = frac{t^2}{2}, 0lt t lt 1$



            $W(t) = -t^2+3t-frac{3}{2}, 1lt t lt 2$



            $W(t) = frac{(t-3)^2}{2}, 2lt t lt 3$



            $W(t) = 0 , tgt 3$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Convolution of two distributions.



              $t_{x_0} = 0$



              $t_{x_1} = 1$



              $t_{h_0} = 0$



              $t_{h_1} = 1$



              Thus $$f_Y(t) = 0, t le t_{x_0}+t_{h_0} ,$$



              $$f_Y(t) = int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(tau)f_H(t-tau)dtau, text{ } t_{x_0}+t_{h_0} le t le t_{x_1}+t_{h_1},$$



              $$f_Y(t) = 0, t ge t_{x_1}+t_{h_1} ,$$



              THese translate to the following solution



              First convolve two uniform distributions



              $X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$



              $$Y(t) = x(t).h(t) = int_{-infty}^{infty} x(tau)h(t-tau)dtau$$
              The above convolution reduces to



              $y(t) = 0, tlt 0$



              $y(t) = int_{max(0,t-1)}^{min(1,t)} dtau , 0lt t lt 2$,



              $y(t) = 0 , t gt 2$



              The middle one will have to split into two intervals, namely $0lt t lt 1$ and $1lt t lt 2$



              $y(t) = 0, tlt 0$



              $y(t) = int_{0}^{t} dtau =t, 0lt tlt 1$,



              $y(t) = int_{t-1}^{1} dtau = 2-t, 1lt tlt 2$,



              $y(t) = 0 , t gt 2$



              Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$



              For $0lt tlt 1$, the bounds are



              $t_{s_0} = 0$



              $t_{s_1} = 1$



              $t_{y_0} = 0$



              $t_{y_1} = 1$



              $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau = int_{0}^{t}tau dtau = frac{t^2}{2}, 0lt tlt 1$$,



              For $1lt tlt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are



              $t_{s_0} = 0$



              $t_{s_1} = 1$



              $t_{y_0} = 0$



              $t_{y_1} = 1$



              and for the interval $(1,2)$ the bounds are



              $t_{s_0} = 0$



              $t_{s_1} = 1$



              $t_{y_0} = 1$



              $t_{y_1} = 2$



              Thus $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau + int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau$$ $$ = int_{t-1}^{1}tau dtau + int_{1}^{t}(2-tau) dtau$$ $$ = -frac{1}{2}(2t^2-6t+3), 1lt tlt 2$$,



              For $2lt tlt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are



              $t_{s_0} = 0$



              $t_{s_1} = 1$



              $t_{y_0} = 1$



              $t_{y_1} = 2$



              Thus $$W(t) = int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau $$ $$ = int_{t-1}^{2} (2- tau) dtau$$ $$ = frac{(t-3)^2}{2}, 2lt t lt 3$$



              and finally $W(t) = 0, tgt 3$



              Thus the $W(t)$ is defined by



              $W(t) = 0 , tlt 0$



              $W(t) = frac{t^2}{2}, 0lt t lt 1$



              $W(t) = -t^2+3t-frac{3}{2}, 1lt t lt 2$



              $W(t) = frac{(t-3)^2}{2}, 2lt t lt 3$



              $W(t) = 0 , tgt 3$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Convolution of two distributions.



                $t_{x_0} = 0$



                $t_{x_1} = 1$



                $t_{h_0} = 0$



                $t_{h_1} = 1$



                Thus $$f_Y(t) = 0, t le t_{x_0}+t_{h_0} ,$$



                $$f_Y(t) = int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(tau)f_H(t-tau)dtau, text{ } t_{x_0}+t_{h_0} le t le t_{x_1}+t_{h_1},$$



                $$f_Y(t) = 0, t ge t_{x_1}+t_{h_1} ,$$



                THese translate to the following solution



                First convolve two uniform distributions



                $X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$



                $$Y(t) = x(t).h(t) = int_{-infty}^{infty} x(tau)h(t-tau)dtau$$
                The above convolution reduces to



                $y(t) = 0, tlt 0$



                $y(t) = int_{max(0,t-1)}^{min(1,t)} dtau , 0lt t lt 2$,



                $y(t) = 0 , t gt 2$



                The middle one will have to split into two intervals, namely $0lt t lt 1$ and $1lt t lt 2$



                $y(t) = 0, tlt 0$



                $y(t) = int_{0}^{t} dtau =t, 0lt tlt 1$,



                $y(t) = int_{t-1}^{1} dtau = 2-t, 1lt tlt 2$,



                $y(t) = 0 , t gt 2$



                Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$



                For $0lt tlt 1$, the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 0$



                $t_{y_1} = 1$



                $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau = int_{0}^{t}tau dtau = frac{t^2}{2}, 0lt tlt 1$$,



                For $1lt tlt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 0$



                $t_{y_1} = 1$



                and for the interval $(1,2)$ the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 1$



                $t_{y_1} = 2$



                Thus $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau + int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau$$ $$ = int_{t-1}^{1}tau dtau + int_{1}^{t}(2-tau) dtau$$ $$ = -frac{1}{2}(2t^2-6t+3), 1lt tlt 2$$,



                For $2lt tlt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 1$



                $t_{y_1} = 2$



                Thus $$W(t) = int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau $$ $$ = int_{t-1}^{2} (2- tau) dtau$$ $$ = frac{(t-3)^2}{2}, 2lt t lt 3$$



                and finally $W(t) = 0, tgt 3$



                Thus the $W(t)$ is defined by



                $W(t) = 0 , tlt 0$



                $W(t) = frac{t^2}{2}, 0lt t lt 1$



                $W(t) = -t^2+3t-frac{3}{2}, 1lt t lt 2$



                $W(t) = frac{(t-3)^2}{2}, 2lt t lt 3$



                $W(t) = 0 , tgt 3$






                share|cite|improve this answer











                $endgroup$



                Convolution of two distributions.



                $t_{x_0} = 0$



                $t_{x_1} = 1$



                $t_{h_0} = 0$



                $t_{h_1} = 1$



                Thus $$f_Y(t) = 0, t le t_{x_0}+t_{h_0} ,$$



                $$f_Y(t) = int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(tau)f_H(t-tau)dtau, text{ } t_{x_0}+t_{h_0} le t le t_{x_1}+t_{h_1},$$



                $$f_Y(t) = 0, t ge t_{x_1}+t_{h_1} ,$$



                THese translate to the following solution



                First convolve two uniform distributions



                $X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$



                $$Y(t) = x(t).h(t) = int_{-infty}^{infty} x(tau)h(t-tau)dtau$$
                The above convolution reduces to



                $y(t) = 0, tlt 0$



                $y(t) = int_{max(0,t-1)}^{min(1,t)} dtau , 0lt t lt 2$,



                $y(t) = 0 , t gt 2$



                The middle one will have to split into two intervals, namely $0lt t lt 1$ and $1lt t lt 2$



                $y(t) = 0, tlt 0$



                $y(t) = int_{0}^{t} dtau =t, 0lt tlt 1$,



                $y(t) = int_{t-1}^{1} dtau = 2-t, 1lt tlt 2$,



                $y(t) = 0 , t gt 2$



                Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$



                For $0lt tlt 1$, the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 0$



                $t_{y_1} = 1$



                $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau = int_{0}^{t}tau dtau = frac{t^2}{2}, 0lt tlt 1$$,



                For $1lt tlt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 0$



                $t_{y_1} = 1$



                and for the interval $(1,2)$ the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 1$



                $t_{y_1} = 2$



                Thus $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau + int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau$$ $$ = int_{t-1}^{1}tau dtau + int_{1}^{t}(2-tau) dtau$$ $$ = -frac{1}{2}(2t^2-6t+3), 1lt tlt 2$$,



                For $2lt tlt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are



                $t_{s_0} = 0$



                $t_{s_1} = 1$



                $t_{y_0} = 1$



                $t_{y_1} = 2$



                Thus $$W(t) = int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau $$ $$ = int_{t-1}^{2} (2- tau) dtau$$ $$ = frac{(t-3)^2}{2}, 2lt t lt 3$$



                and finally $W(t) = 0, tgt 3$



                Thus the $W(t)$ is defined by



                $W(t) = 0 , tlt 0$



                $W(t) = frac{t^2}{2}, 0lt t lt 1$



                $W(t) = -t^2+3t-frac{3}{2}, 1lt t lt 2$



                $W(t) = frac{(t-3)^2}{2}, 2lt t lt 3$



                $W(t) = 0 , tgt 3$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 11 '18 at 10:24

























                answered Dec 11 '18 at 5:04









                Satish RamanathanSatish Ramanathan

                9,66531323




                9,66531323























                    2












                    $begingroup$

                    One way is (I used $Z$ instead of $xi$)$$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{min(z,,1)}int_{z_2=0}^{min(z-z_3,,1)}int_{z_1=0}^{min(z-z_2-z_3,,1)}dz_1,dz_2,dz_3$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      How to integrate this?
                      $endgroup$
                      – Atstovas
                      Dec 10 '18 at 9:00










                    • $begingroup$
                      I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:24












                    • $begingroup$
                      A similar questions is found here math.stackexchange.com/questions/2631501/…
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:32










                    • $begingroup$
                      can you show your solution when $z<1$?
                      $endgroup$
                      – Atstovas
                      Dec 11 '18 at 17:40










                    • $begingroup$
                      It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
                      $endgroup$
                      – BlackMath
                      Dec 11 '18 at 23:52


















                    2












                    $begingroup$

                    One way is (I used $Z$ instead of $xi$)$$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{min(z,,1)}int_{z_2=0}^{min(z-z_3,,1)}int_{z_1=0}^{min(z-z_2-z_3,,1)}dz_1,dz_2,dz_3$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      How to integrate this?
                      $endgroup$
                      – Atstovas
                      Dec 10 '18 at 9:00










                    • $begingroup$
                      I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:24












                    • $begingroup$
                      A similar questions is found here math.stackexchange.com/questions/2631501/…
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:32










                    • $begingroup$
                      can you show your solution when $z<1$?
                      $endgroup$
                      – Atstovas
                      Dec 11 '18 at 17:40










                    • $begingroup$
                      It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
                      $endgroup$
                      – BlackMath
                      Dec 11 '18 at 23:52
















                    2












                    2








                    2





                    $begingroup$

                    One way is (I used $Z$ instead of $xi$)$$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{min(z,,1)}int_{z_2=0}^{min(z-z_3,,1)}int_{z_1=0}^{min(z-z_2-z_3,,1)}dz_1,dz_2,dz_3$$






                    share|cite|improve this answer









                    $endgroup$



                    One way is (I used $Z$ instead of $xi$)$$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{min(z,,1)}int_{z_2=0}^{min(z-z_3,,1)}int_{z_1=0}^{min(z-z_2-z_3,,1)}dz_1,dz_2,dz_3$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 10 '18 at 7:10









                    BlackMathBlackMath

                    30518




                    30518












                    • $begingroup$
                      How to integrate this?
                      $endgroup$
                      – Atstovas
                      Dec 10 '18 at 9:00










                    • $begingroup$
                      I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:24












                    • $begingroup$
                      A similar questions is found here math.stackexchange.com/questions/2631501/…
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:32










                    • $begingroup$
                      can you show your solution when $z<1$?
                      $endgroup$
                      – Atstovas
                      Dec 11 '18 at 17:40










                    • $begingroup$
                      It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
                      $endgroup$
                      – BlackMath
                      Dec 11 '18 at 23:52




















                    • $begingroup$
                      How to integrate this?
                      $endgroup$
                      – Atstovas
                      Dec 10 '18 at 9:00










                    • $begingroup$
                      I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:24












                    • $begingroup$
                      A similar questions is found here math.stackexchange.com/questions/2631501/…
                      $endgroup$
                      – BlackMath
                      Dec 10 '18 at 9:32










                    • $begingroup$
                      can you show your solution when $z<1$?
                      $endgroup$
                      – Atstovas
                      Dec 11 '18 at 17:40










                    • $begingroup$
                      It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
                      $endgroup$
                      – BlackMath
                      Dec 11 '18 at 23:52


















                    $begingroup$
                    How to integrate this?
                    $endgroup$
                    – Atstovas
                    Dec 10 '18 at 9:00




                    $begingroup$
                    How to integrate this?
                    $endgroup$
                    – Atstovas
                    Dec 10 '18 at 9:00












                    $begingroup$
                    I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
                    $endgroup$
                    – BlackMath
                    Dec 10 '18 at 9:24






                    $begingroup$
                    I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, ,z-z_3,,z-z_2-z_3$. For $zgeq 1$ there are more cases.
                    $endgroup$
                    – BlackMath
                    Dec 10 '18 at 9:24














                    $begingroup$
                    A similar questions is found here math.stackexchange.com/questions/2631501/…
                    $endgroup$
                    – BlackMath
                    Dec 10 '18 at 9:32




                    $begingroup$
                    A similar questions is found here math.stackexchange.com/questions/2631501/…
                    $endgroup$
                    – BlackMath
                    Dec 10 '18 at 9:32












                    $begingroup$
                    can you show your solution when $z<1$?
                    $endgroup$
                    – Atstovas
                    Dec 11 '18 at 17:40




                    $begingroup$
                    can you show your solution when $z<1$?
                    $endgroup$
                    – Atstovas
                    Dec 11 '18 at 17:40












                    $begingroup$
                    It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
                    $endgroup$
                    – BlackMath
                    Dec 11 '18 at 23:52






                    $begingroup$
                    It will be $$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{z}int_{z_2=0}^{z-z_3}int_{z_1=0}^{z-z_2-z_3}dz_1,dz_2,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral.
                    $endgroup$
                    – BlackMath
                    Dec 11 '18 at 23:52




















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