Finding expectation of joint uniform continuous distribution without integrating
$begingroup$
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
asked Nov 26 '18 at 21:28
agbltagblt
17914
$endgroup$
add a comment |
$begingroup$
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
asked Nov 26 '18 at 21:28
agbltagblt
17914
$endgroup$
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
add a comment |
$begingroup$
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
asked Nov 26 '18 at 21:28
agbltagblt
17914
$endgroup$
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
asked Nov 26 '18 at 21:28
agbltagblt
17914
asked Nov 26 '18 at 21:28
agbltagblt
17914
asked Nov 26 '18 at 21:28
agbltagblt
17914
asked Nov 26 '18 at 21:28
agbltagblt
17914
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $5$.
$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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 $5$.
$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $5$.
$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $5$.
$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
$endgroup$
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 $5$.
$$mathsf E(Xmid (X,Y){in}triangle(0,0)(10,0)(5,5))=5$$
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
answered Nov 27 '18 at 0:02
Graham KempGraham Kemp
85.1k43378
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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