Finding expectation of joint uniform continuous distribution without integrating












3












$begingroup$


From SOA sample 138:




A machine consists of two components, whose lifetimes have the joint density function



$$f(x,y)=
begin{cases}
{1over50}, & text{for }x>0,y>0,x+y<10 \
0, & text{otherwise}
end{cases}$$



The machine operates until both components fail.
Calculate the expected operational time of the machine.




I was able to get the solution by deciding on case $Y>X$, drawing a triangle with vertices at $(0,0)$, $(0,10)$, and $(5,5)$ and then integrating $int_0^5 int_{x}^{10-x}{yover50} dy dx$ to get $2.5$, and then doubling by symmetry for the case of $X>Y$ to get $5$.



However the SOA solution does it a different way that I am trying to understand:




Suppose the component represented by the random variable $X$ fails last. This is represented by
the triangle with vertices at $(0, 0)$, $(10, 0)$ and $(5, 5)$. Because the density is uniform over this
region, the mean value of $X$ and thus the expected operational time of the machine is 5. By
symmetry, if the component represented by the random variable $Y$ fails last, the expected
operational time of the machine is also $5$. Thus, the unconditional expected operational time of
the machine must be $5$ as well.




I bolded the part that I do not understand. Where do they get $5$ just by looking at the triangle, which is $25$ in area?






probability statistics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
    $endgroup$
    – Henry
    Nov 26 '18 at 23:50
















3












$begingroup$


From SOA sample 138:




A machine consists of two components, whose lifetimes have the joint density function



$$f(x,y)=
begin{cases}
{1over50}, & text{for }x>0,y>0,x+y<10 \
0, & text{otherwise}
end{cases}$$



The machine operates until both components fail.
Calculate the expected operational time of the machine.




I was able to get the solution by deciding on case $Y>X$, drawing a triangle with vertices at $(0,0)$, $(0,10)$, and $(5,5)$ and then integrating $int_0^5 int_{x}^{10-x}{yover50} dy dx$ to get $2.5$, and then doubling by symmetry for the case of $X>Y$ to get $5$.



However the SOA solution does it a different way that I am trying to understand:




Suppose the component represented by the random variable $X$ fails last. This is represented by
the triangle with vertices at $(0, 0)$, $(10, 0)$ and $(5, 5)$. Because the density is uniform over this
region, the mean value of $X$ and thus the expected operational time of the machine is 5. By
symmetry, if the component represented by the random variable $Y$ fails last, the expected
operational time of the machine is also $5$. Thus, the unconditional expected operational time of
the machine must be $5$ as well.




I bolded the part that I do not understand. Where do they get $5$ just by looking at the triangle, which is $25$ in area?






probability statistics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
    $endgroup$
    – Henry
    Nov 26 '18 at 23:50














3












3








3





$begingroup$


From SOA sample 138:




A machine consists of two components, whose lifetimes have the joint density function



$$f(x,y)=
begin{cases}
{1over50}, & text{for }x>0,y>0,x+y<10 \
0, & text{otherwise}
end{cases}$$



The machine operates until both components fail.
Calculate the expected operational time of the machine.




I was able to get the solution by deciding on case $Y>X$, drawing a triangle with vertices at $(0,0)$, $(0,10)$, and $(5,5)$ and then integrating $int_0^5 int_{x}^{10-x}{yover50} dy dx$ to get $2.5$, and then doubling by symmetry for the case of $X>Y$ to get $5$.



However the SOA solution does it a different way that I am trying to understand:




Suppose the component represented by the random variable $X$ fails last. This is represented by
the triangle with vertices at $(0, 0)$, $(10, 0)$ and $(5, 5)$. Because the density is uniform over this
region, the mean value of $X$ and thus the expected operational time of the machine is 5. By
symmetry, if the component represented by the random variable $Y$ fails last, the expected
operational time of the machine is also $5$. Thus, the unconditional expected operational time of
the machine must be $5$ as well.




I bolded the part that I do not understand. Where do they get $5$ just by looking at the triangle, which is $25$ in area?






probability statistics







share|cite|improve this question









$endgroup$




From SOA sample 138:




A machine consists of two components, whose lifetimes have the joint density function



$$f(x,y)=
begin{cases}
{1over50}, & text{for }x>0,y>0,x+y<10 \
0, & text{otherwise}
end{cases}$$



The machine operates until both components fail.
Calculate the expected operational time of the machine.




I was able to get the solution by deciding on case $Y>X$, drawing a triangle with vertices at $(0,0)$, $(0,10)$, and $(5,5)$ and then integrating $int_0^5 int_{x}^{10-x}{yover50} dy dx$ to get $2.5$, and then doubling by symmetry for the case of $X>Y$ to get $5$.



However the SOA solution does it a different way that I am trying to understand:




Suppose the component represented by the random variable $X$ fails last. This is represented by
the triangle with vertices at $(0, 0)$, $(10, 0)$ and $(5, 5)$. Because the density is uniform over this
region, the mean value of $X$ and thus the expected operational time of the machine is 5. By
symmetry, if the component represented by the random variable $Y$ fails last, the expected
operational time of the machine is also $5$. Thus, the unconditional expected operational time of
the machine must be $5$ as well.




I bolded the part that I do not understand. Where do they get $5$ just by looking at the triangle, which is $25$ in area?






probability statistics




probability statistics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '18 at 21:28









agbltagblt

17914




17914








  • 1




    $begingroup$
    Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
    $endgroup$
    – Henry
    Nov 26 '18 at 23:50














  • 1




    $begingroup$
    Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
    $endgroup$
    – Henry
    Nov 26 '18 at 23:50








1




1




$begingroup$
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
$endgroup$
– Henry
Nov 26 '18 at 23:50




$begingroup$
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
$endgroup$
– Henry
Nov 26 '18 at 23:50










1 Answer
1






active

oldest

votes


















1












$begingroup$

It has naught to do with the area.



The isosceles triangle, $triangle(0,0)(10,0)(5,5)$, is symmetrical about the vertical line through $overline{~(5,0)(5,5)~}$, and the density inside that shape is uniform.   Therefore the (conditional) mean value of $X$ given it is inside that shape is $5$.



$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
    $endgroup$
    – agblt
    Nov 27 '18 at 1:02












  • $begingroup$
    @abbot well, we don't need to know it's value to find the expectation.
    $endgroup$
    – Graham Kemp
    Nov 27 '18 at 1:31











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1 Answer
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1












$begingroup$

It has naught to do with the area.



The isosceles triangle, $triangle(0,0)(10,0)(5,5)$, is symmetrical about the vertical line through $overline{~(5,0)(5,5)~}$, and the density inside that shape is uniform.   Therefore the (conditional) mean value of $X$ given it is inside that shape is $5$.



$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
    $endgroup$
    – agblt
    Nov 27 '18 at 1:02












  • $begingroup$
    @abbot well, we don't need to know it's value to find the expectation.
    $endgroup$
    – Graham Kemp
    Nov 27 '18 at 1:31
















1












$begingroup$

It has naught to do with the area.



The isosceles triangle, $triangle(0,0)(10,0)(5,5)$, is symmetrical about the vertical line through $overline{~(5,0)(5,5)~}$, and the density inside that shape is uniform.   Therefore the (conditional) mean value of $X$ given it is inside that shape is $5$.



$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
    $endgroup$
    – agblt
    Nov 27 '18 at 1:02












  • $begingroup$
    @abbot well, we don't need to know it's value to find the expectation.
    $endgroup$
    – Graham Kemp
    Nov 27 '18 at 1:31














1












1








1





$begingroup$

It has naught to do with the area.



The isosceles triangle, $triangle(0,0)(10,0)(5,5)$, is symmetrical about the vertical line through $overline{~(5,0)(5,5)~}$, and the density inside that shape is uniform.   Therefore the (conditional) mean value of $X$ given it is inside that shape is $5$.



$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$






share|cite|improve this answer









$endgroup$



It has naught to do with the area.



The isosceles triangle, $triangle(0,0)(10,0)(5,5)$, is symmetrical about the vertical line through $overline{~(5,0)(5,5)~}$, and the density inside that shape is uniform.   Therefore the (conditional) mean value of $X$ given it is inside that shape is $5$.



$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 0:02









Graham KempGraham Kemp

85.1k43378




85.1k43378












  • $begingroup$
    So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
    $endgroup$
    – agblt
    Nov 27 '18 at 1:02












  • $begingroup$
    @abbot well, we don't need to know it's value to find the expectation.
    $endgroup$
    – Graham Kemp
    Nov 27 '18 at 1:31


















  • $begingroup$
    So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
    $endgroup$
    – agblt
    Nov 27 '18 at 1:02












  • $begingroup$
    @abbot well, we don't need to know it's value to find the expectation.
    $endgroup$
    – Graham Kemp
    Nov 27 '18 at 1:31
















$begingroup$
So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
$endgroup$
– agblt
Nov 27 '18 at 1:02






$begingroup$
So once we can make this triangle, and the density is uniform, we don't care about what the joint density specifically is, meaning it could just as well be $1over1000$ as $1over50$?
$endgroup$
– agblt
Nov 27 '18 at 1:02














$begingroup$
@abbot well, we don't need to know it's value to find the expectation.
$endgroup$
– Graham Kemp
Nov 27 '18 at 1:31




$begingroup$
@abbot well, we don't need to know it's value to find the expectation.
$endgroup$
– Graham Kemp
Nov 27 '18 at 1:31


















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