Finding expectation of joint uniform continuous distribution without integrating
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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 at 21:28
agblt
12413
add a comment |
up vote
3
down vote
favorite
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 at 21:28
agblt
12413
1
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
– Henry
Nov 26 at 23:50
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 at 21:28
agblt
12413
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 at 21:28
agblt
12413
asked Nov 26 at 21:28
agblt
12413
asked Nov 26 at 21:28
agblt
12413
asked Nov 26 at 21:28
agblt
12413
asked Nov 26 at 21:28
agblt
12413
12413
1
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
– Henry
Nov 26 at 23:50
add a comment |
1
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
– Henry
Nov 26 at 23:50
1
1
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
– Henry
Nov 26 at 23:50
Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
– Henry
Nov 26 at 23:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
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 at 0:02
Graham Kemp
84.6k43378
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 $1over50$?
– agblt
Nov 27 at 1:02
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 at 0:02
Graham Kemp
84.6k43378
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 $1over50$?
– agblt
Nov 27 at 1:02
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
add a comment |
up vote
1
down vote
accepted
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 at 0:02
Graham Kemp
84.6k43378
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 $1over50$?
– agblt
Nov 27 at 1:02
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 at 0:02
Graham Kemp
84.6k43378
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 at 0:02
Graham Kemp
84.6k43378
answered Nov 27 at 0:02
Graham Kemp
84.6k43378
answered Nov 27 at 0:02
Graham Kemp
84.6k43378
answered Nov 27 at 0:02
Graham Kemp
84.6k43378
84.6k43378
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 $1over50$?
– agblt
Nov 27 at 1:02
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
add a comment |
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 $1over50$?
– agblt
Nov 27 at 1:02
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
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 $1over50$?
– agblt
Nov 27 at 1:02
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 $1over50$?
– agblt
Nov 27 at 1:02
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
@abbot well, we don't need to know it's value to find the expectation.
– Graham Kemp
Nov 27 at 1:31
add a comment |
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Because the line from $(5,5)$ to $(5,0)$ is a median of the triangle and so shows the mean of the triangle
– Henry
Nov 26 at 23:50