Compute $P(X+Y>Z)$ where $X, Y$ and $Z$ are independent, uniform random variables in the interval $[0,1]$.












1












$begingroup$


I know that if $W=X+Y$ then
$$f_{W}(t) = tmathbf{1}_{[0,1]}(t)+(2-t)mathbf{1}_{(1,2]}(t).$$
Thus, we want to find $P(W>Z)=P(W-Z>0).$ Using Bayes Theorem we get that
$$P(W-Z>0) = P(W-Z>0|Wleq 1)P(Wleq 1)+P(W-Z>0|W> 1)P(W> 1)$$
$$=int_{0}^{1} int_{0}^{w} dz cdot P(Wleq 1)+P(Zleq 1)P(W>1)$$
$$=frac{1}{2}cdot frac{1}{2}+1cdot frac{1}{2}=frac{3}{4}.$$



Is this calculation correct?



Edit



Based on the suggestion made in the comment we consider the following:



$$P(W>Z)=int_{0}^{2}P(Z<w)f_{W}(w)dw$$
$$=int_{0}^{1}w^2cdot dw+int_{1}^{2}(2-w)cdot dw=frac{1}{3}+2-frac{3}{2}=5/6.$$
Does this make sense?






probability-theory







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  • $begingroup$
    No this is not correct. To compute probabilities, you need to condition on convenient things. It is not convenient to condition on the event ${Wleq 1}$ because it is not precise enough (there are too many scenarios to consider, each depending on the specific value of $W$ given that $Wleq 1$). So, why not make life easy and condition on all possible values of $W$? $$P[Z<W] = int_{-infty}^{infty} P[Z<W|W=w]f_W(w)dw=int_{0}^{2} P[Z<W|W=w]f_W(w)dw$$
    $endgroup$
    – Michael
    Dec 3 '18 at 16:43












  • $begingroup$
    @Michael I made an edit. Is this correct now?
    $endgroup$
    – Hello_World
    Dec 3 '18 at 16:50










  • $begingroup$
    Yes that looks good now. If you want, you can write that answer yourself in the answer field below and give it "best answer" (which is good practice when solving a problem based on hints).
    $endgroup$
    – Michael
    Dec 3 '18 at 16:51


















1












$begingroup$


I know that if $W=X+Y$ then
$$f_{W}(t) = tmathbf{1}_{[0,1]}(t)+(2-t)mathbf{1}_{(1,2]}(t).$$
Thus, we want to find $P(W>Z)=P(W-Z>0).$ Using Bayes Theorem we get that
$$P(W-Z>0) = P(W-Z>0|Wleq 1)P(Wleq 1)+P(W-Z>0|W> 1)P(W> 1)$$
$$=int_{0}^{1} int_{0}^{w} dz cdot P(Wleq 1)+P(Zleq 1)P(W>1)$$
$$=frac{1}{2}cdot frac{1}{2}+1cdot frac{1}{2}=frac{3}{4}.$$



Is this calculation correct?



Edit



Based on the suggestion made in the comment we consider the following:



$$P(W>Z)=int_{0}^{2}P(Z<w)f_{W}(w)dw$$
$$=int_{0}^{1}w^2cdot dw+int_{1}^{2}(2-w)cdot dw=frac{1}{3}+2-frac{3}{2}=5/6.$$
Does this make sense?






probability-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    No this is not correct. To compute probabilities, you need to condition on convenient things. It is not convenient to condition on the event ${Wleq 1}$ because it is not precise enough (there are too many scenarios to consider, each depending on the specific value of $W$ given that $Wleq 1$). So, why not make life easy and condition on all possible values of $W$? $$P[Z<W] = int_{-infty}^{infty} P[Z<W|W=w]f_W(w)dw=int_{0}^{2} P[Z<W|W=w]f_W(w)dw$$
    $endgroup$
    – Michael
    Dec 3 '18 at 16:43












  • $begingroup$
    @Michael I made an edit. Is this correct now?
    $endgroup$
    – Hello_World
    Dec 3 '18 at 16:50










  • $begingroup$
    Yes that looks good now. If you want, you can write that answer yourself in the answer field below and give it "best answer" (which is good practice when solving a problem based on hints).
    $endgroup$
    – Michael
    Dec 3 '18 at 16:51
















1












1








1





$begingroup$


I know that if $W=X+Y$ then
$$f_{W}(t) = tmathbf{1}_{[0,1]}(t)+(2-t)mathbf{1}_{(1,2]}(t).$$
Thus, we want to find $P(W>Z)=P(W-Z>0).$ Using Bayes Theorem we get that
$$P(W-Z>0) = P(W-Z>0|Wleq 1)P(Wleq 1)+P(W-Z>0|W> 1)P(W> 1)$$
$$=int_{0}^{1} int_{0}^{w} dz cdot P(Wleq 1)+P(Zleq 1)P(W>1)$$
$$=frac{1}{2}cdot frac{1}{2}+1cdot frac{1}{2}=frac{3}{4}.$$



Is this calculation correct?



Edit



Based on the suggestion made in the comment we consider the following:



$$P(W>Z)=int_{0}^{2}P(Z<w)f_{W}(w)dw$$
$$=int_{0}^{1}w^2cdot dw+int_{1}^{2}(2-w)cdot dw=frac{1}{3}+2-frac{3}{2}=5/6.$$
Does this make sense?






probability-theory







share|cite|improve this question











$endgroup$




I know that if $W=X+Y$ then
$$f_{W}(t) = tmathbf{1}_{[0,1]}(t)+(2-t)mathbf{1}_{(1,2]}(t).$$
Thus, we want to find $P(W>Z)=P(W-Z>0).$ Using Bayes Theorem we get that
$$P(W-Z>0) = P(W-Z>0|Wleq 1)P(Wleq 1)+P(W-Z>0|W> 1)P(W> 1)$$
$$=int_{0}^{1} int_{0}^{w} dz cdot P(Wleq 1)+P(Zleq 1)P(W>1)$$
$$=frac{1}{2}cdot frac{1}{2}+1cdot frac{1}{2}=frac{3}{4}.$$



Is this calculation correct?



Edit



Based on the suggestion made in the comment we consider the following:



$$P(W>Z)=int_{0}^{2}P(Z<w)f_{W}(w)dw$$
$$=int_{0}^{1}w^2cdot dw+int_{1}^{2}(2-w)cdot dw=frac{1}{3}+2-frac{3}{2}=5/6.$$
Does this make sense?






probability-theory




probability-theory






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 16:49







Hello_World

















asked Dec 3 '18 at 16:36









Hello_WorldHello_World

4,11621731




4,11621731












  • $begingroup$
    No this is not correct. To compute probabilities, you need to condition on convenient things. It is not convenient to condition on the event ${Wleq 1}$ because it is not precise enough (there are too many scenarios to consider, each depending on the specific value of $W$ given that $Wleq 1$). So, why not make life easy and condition on all possible values of $W$? $$P[Z<W] = int_{-infty}^{infty} P[Z<W|W=w]f_W(w)dw=int_{0}^{2} P[Z<W|W=w]f_W(w)dw$$
    $endgroup$
    – Michael
    Dec 3 '18 at 16:43












  • $begingroup$
    @Michael I made an edit. Is this correct now?
    $endgroup$
    – Hello_World
    Dec 3 '18 at 16:50










  • $begingroup$
    Yes that looks good now. If you want, you can write that answer yourself in the answer field below and give it "best answer" (which is good practice when solving a problem based on hints).
    $endgroup$
    – Michael
    Dec 3 '18 at 16:51




















  • $begingroup$
    No this is not correct. To compute probabilities, you need to condition on convenient things. It is not convenient to condition on the event ${Wleq 1}$ because it is not precise enough (there are too many scenarios to consider, each depending on the specific value of $W$ given that $Wleq 1$). So, why not make life easy and condition on all possible values of $W$? $$P[Z<W] = int_{-infty}^{infty} P[Z<W|W=w]f_W(w)dw=int_{0}^{2} P[Z<W|W=w]f_W(w)dw$$
    $endgroup$
    – Michael
    Dec 3 '18 at 16:43












  • $begingroup$
    @Michael I made an edit. Is this correct now?
    $endgroup$
    – Hello_World
    Dec 3 '18 at 16:50










  • $begingroup$
    Yes that looks good now. If you want, you can write that answer yourself in the answer field below and give it "best answer" (which is good practice when solving a problem based on hints).
    $endgroup$
    – Michael
    Dec 3 '18 at 16:51


















$begingroup$
No this is not correct. To compute probabilities, you need to condition on convenient things. It is not convenient to condition on the event ${Wleq 1}$ because it is not precise enough (there are too many scenarios to consider, each depending on the specific value of $W$ given that $Wleq 1$). So, why not make life easy and condition on all possible values of $W$? $$P[Z<W] = int_{-infty}^{infty} P[Z<W|W=w]f_W(w)dw=int_{0}^{2} P[Z<W|W=w]f_W(w)dw$$
$endgroup$
– Michael
Dec 3 '18 at 16:43






$begingroup$
No this is not correct. To compute probabilities, you need to condition on convenient things. It is not convenient to condition on the event ${Wleq 1}$ because it is not precise enough (there are too many scenarios to consider, each depending on the specific value of $W$ given that $Wleq 1$). So, why not make life easy and condition on all possible values of $W$? $$P[Z<W] = int_{-infty}^{infty} P[Z<W|W=w]f_W(w)dw=int_{0}^{2} P[Z<W|W=w]f_W(w)dw$$
$endgroup$
– Michael
Dec 3 '18 at 16:43














$begingroup$
@Michael I made an edit. Is this correct now?
$endgroup$
– Hello_World
Dec 3 '18 at 16:50




$begingroup$
@Michael I made an edit. Is this correct now?
$endgroup$
– Hello_World
Dec 3 '18 at 16:50












$begingroup$
Yes that looks good now. If you want, you can write that answer yourself in the answer field below and give it "best answer" (which is good practice when solving a problem based on hints).
$endgroup$
– Michael
Dec 3 '18 at 16:51






$begingroup$
Yes that looks good now. If you want, you can write that answer yourself in the answer field below and give it "best answer" (which is good practice when solving a problem based on hints).
$endgroup$
– Michael
Dec 3 '18 at 16:51












2 Answers
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$begingroup$

$x+y > z\
0le z le 1$



Either $x +y > 1$ and we can ignore $z$ or $x+yle 1$ and $z < x+y$



$int_0^1int_0^{1-x}int_0^{x+y} dz dy dx + int_0^1int_{1-x}^{1} dy dx\
frac 13 + frac 12$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Using the Irwin–Hall distribution



    $$W = X + Y,qquad f_W(w)= w 1_{(0,1]}(w) + (2 - w)1_{(1,2)}(w)$$



    then you get



    $$
    begin{align}
    P(W > Z)
    &= P(W > Zmid W > 1)P(W > 1)+ P(W > Z, W leq 1) \
    &= 1 cdot int_1^2(2-w)dw +
    int_0^1int_0^wwdzdw \
    &= frac{1}{2} + frac{1}{3}
    end{align}$$



    as Doug M points out. You forgot a $w$ and $dw$ in $int_{0}^{1} int_{0}^{w} dz$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      $x+y > z\
      0le z le 1$



      Either $x +y > 1$ and we can ignore $z$ or $x+yle 1$ and $z < x+y$



      $int_0^1int_0^{1-x}int_0^{x+y} dz dy dx + int_0^1int_{1-x}^{1} dy dx\
      frac 13 + frac 12$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $x+y > z\
        0le z le 1$



        Either $x +y > 1$ and we can ignore $z$ or $x+yle 1$ and $z < x+y$



        $int_0^1int_0^{1-x}int_0^{x+y} dz dy dx + int_0^1int_{1-x}^{1} dy dx\
        frac 13 + frac 12$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $x+y > z\
          0le z le 1$



          Either $x +y > 1$ and we can ignore $z$ or $x+yle 1$ and $z < x+y$



          $int_0^1int_0^{1-x}int_0^{x+y} dz dy dx + int_0^1int_{1-x}^{1} dy dx\
          frac 13 + frac 12$






          share|cite|improve this answer









          $endgroup$



          $x+y > z\
          0le z le 1$



          Either $x +y > 1$ and we can ignore $z$ or $x+yle 1$ and $z < x+y$



          $int_0^1int_0^{1-x}int_0^{x+y} dz dy dx + int_0^1int_{1-x}^{1} dy dx\
          frac 13 + frac 12$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 17:00









          Doug MDoug M

          44.4k31854




          44.4k31854























              0












              $begingroup$

              Using the Irwin–Hall distribution



              $$W = X + Y,qquad f_W(w)= w 1_{(0,1]}(w) + (2 - w)1_{(1,2)}(w)$$



              then you get



              $$
              begin{align}
              P(W > Z)
              &= P(W > Zmid W > 1)P(W > 1)+ P(W > Z, W leq 1) \
              &= 1 cdot int_1^2(2-w)dw +
              int_0^1int_0^wwdzdw \
              &= frac{1}{2} + frac{1}{3}
              end{align}$$



              as Doug M points out. You forgot a $w$ and $dw$ in $int_{0}^{1} int_{0}^{w} dz$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Using the Irwin–Hall distribution



                $$W = X + Y,qquad f_W(w)= w 1_{(0,1]}(w) + (2 - w)1_{(1,2)}(w)$$



                then you get



                $$
                begin{align}
                P(W > Z)
                &= P(W > Zmid W > 1)P(W > 1)+ P(W > Z, W leq 1) \
                &= 1 cdot int_1^2(2-w)dw +
                int_0^1int_0^wwdzdw \
                &= frac{1}{2} + frac{1}{3}
                end{align}$$



                as Doug M points out. You forgot a $w$ and $dw$ in $int_{0}^{1} int_{0}^{w} dz$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Using the Irwin–Hall distribution



                  $$W = X + Y,qquad f_W(w)= w 1_{(0,1]}(w) + (2 - w)1_{(1,2)}(w)$$



                  then you get



                  $$
                  begin{align}
                  P(W > Z)
                  &= P(W > Zmid W > 1)P(W > 1)+ P(W > Z, W leq 1) \
                  &= 1 cdot int_1^2(2-w)dw +
                  int_0^1int_0^wwdzdw \
                  &= frac{1}{2} + frac{1}{3}
                  end{align}$$



                  as Doug M points out. You forgot a $w$ and $dw$ in $int_{0}^{1} int_{0}^{w} dz$.






                  share|cite|improve this answer









                  $endgroup$



                  Using the Irwin–Hall distribution



                  $$W = X + Y,qquad f_W(w)= w 1_{(0,1]}(w) + (2 - w)1_{(1,2)}(w)$$



                  then you get



                  $$
                  begin{align}
                  P(W > Z)
                  &= P(W > Zmid W > 1)P(W > 1)+ P(W > Z, W leq 1) \
                  &= 1 cdot int_1^2(2-w)dw +
                  int_0^1int_0^wwdzdw \
                  &= frac{1}{2} + frac{1}{3}
                  end{align}$$



                  as Doug M points out. You forgot a $w$ and $dw$ in $int_{0}^{1} int_{0}^{w} dz$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 22:31









                  Benjamin ChristoffersenBenjamin Christoffersen

                  2009




                  2009






























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