Mixing
Grimmett and Stirzaker Ex 3.11.20 p85
Clash Royale CLAN TAG#URR8PPP
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Let R(p) be the reliability function of a network G, each edge is working with probability p.
(a) Show that $R(p_1p_2) leq R(p_1)R(p_2) $ if $0 leq p_1, p_2 leq 1$
Their proof is:
$P$(edge is blue) $= p_1$
$P$(edge is yellow) $= p_2$
$P$(edge is green) $= p_1p_2$
$p_1$ and $p_2$ are independent of each other and all other edges.
An edge is green if is is both yellow and blue,
If there is a working green connection, then there is also a blue and green connection.
Thus:
beginalign
P(textgreen connection) &leq P(textblue connection and yellow connection)\
&=P(textblue connection)P(textyellow connection)
endalign
Thing is this is not really a proof just a statement of the result?
Where does the $leq$ come from in the first line? Given that $p_1$ and $p_2$ are independent I would expect their to be an equality as shown in the second line? What have I missed?
probability
asked Jul 17 at 17:16
Bazman
365211
add a comment |Â
up vote
-1
down vote
favorite
Let R(p) be the reliability function of a network G, each edge is working with probability p.
(a) Show that $R(p_1p_2) leq R(p_1)R(p_2) $ if $0 leq p_1, p_2 leq 1$
Their proof is:
$P$(edge is blue) $= p_1$
$P$(edge is yellow) $= p_2$
$P$(edge is green) $= p_1p_2$
$p_1$ and $p_2$ are independent of each other and all other edges.
An edge is green if is is both yellow and blue,
If there is a working green connection, then there is also a blue and green connection.
Thus:
beginalign
P(textgreen connection) &leq P(textblue connection and yellow connection)\
&=P(textblue connection)P(textyellow connection)
endalign
Thing is this is not really a proof just a statement of the result?
Where does the $leq$ come from in the first line? Given that $p_1$ and $p_2$ are independent I would expect their to be an equality as shown in the second line? What have I missed?
probability
asked Jul 17 at 17:16
Bazman
365211
1
What exactly is meant by a connection? If it is a monochrome path between two nodes, then I guess what is meant is that if there is a green connection then there is certainly a blue connection and a yellow connection, (the same path) but there might be another path with all yellow edges and yet another path with all blue edges.
– saulspatz
Jul 17 at 17:46
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let R(p) be the reliability function of a network G, each edge is working with probability p.
(a) Show that $R(p_1p_2) leq R(p_1)R(p_2) $ if $0 leq p_1, p_2 leq 1$
Their proof is:
$P$(edge is blue) $= p_1$
$P$(edge is yellow) $= p_2$
$P$(edge is green) $= p_1p_2$
$p_1$ and $p_2$ are independent of each other and all other edges.
An edge is green if is is both yellow and blue,
If there is a working green connection, then there is also a blue and green connection.
Thus:
beginalign
P(textgreen connection) &leq P(textblue connection and yellow connection)\
&=P(textblue connection)P(textyellow connection)
endalign
Thing is this is not really a proof just a statement of the result?
Where does the $leq$ come from in the first line? Given that $p_1$ and $p_2$ are independent I would expect their to be an equality as shown in the second line? What have I missed?
probability
asked Jul 17 at 17:16
Bazman
365211
Let R(p) be the reliability function of a network G, each edge is working with probability p.
(a) Show that $R(p_1p_2) leq R(p_1)R(p_2) $ if $0 leq p_1, p_2 leq 1$
Their proof is:
$P$(edge is blue) $= p_1$
$P$(edge is yellow) $= p_2$
$P$(edge is green) $= p_1p_2$
$p_1$ and $p_2$ are independent of each other and all other edges.
An edge is green if is is both yellow and blue,
If there is a working green connection, then there is also a blue and green connection.
Thus:
beginalign
P(textgreen connection) &leq P(textblue connection and yellow connection)\
&=P(textblue connection)P(textyellow connection)
endalign
Thing is this is not really a proof just a statement of the result?
Where does the $leq$ come from in the first line? Given that $p_1$ and $p_2$ are independent I would expect their to be an equality as shown in the second line? What have I missed?
probability
asked Jul 17 at 17:16
Bazman
365211
asked Jul 17 at 17:16
Bazman
365211
asked Jul 17 at 17:16
Bazman
365211
asked Jul 17 at 17:16
Bazman
365211
365211
1
What exactly is meant by a connection? If it is a monochrome path between two nodes, then I guess what is meant is that if there is a green connection then there is certainly a blue connection and a yellow connection, (the same path) but there might be another path with all yellow edges and yet another path with all blue edges.
– saulspatz
Jul 17 at 17:46
add a comment |Â
1
What exactly is meant by a connection? If it is a monochrome path between two nodes, then I guess what is meant is that if there is a green connection then there is certainly a blue connection and a yellow connection, (the same path) but there might be another path with all yellow edges and yet another path with all blue edges.
– saulspatz
Jul 17 at 17:46
1
1
What exactly is meant by a connection? If it is a monochrome path between two nodes, then I guess what is meant is that if there is a green connection then there is certainly a blue connection and a yellow connection, (the same path) but there might be another path with all yellow edges and yet another path with all blue edges.
– saulspatz
Jul 17 at 17:46
What exactly is meant by a connection? If it is a monochrome path between two nodes, then I guess what is meant is that if there is a green connection then there is certainly a blue connection and a yellow connection, (the same path) but there might be another path with all yellow edges and yet another path with all blue edges.
– saulspatz
Jul 17 at 17:46
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
The proof is rather terse. In greater detail, imagine coloring each edge in the network blue with probability $p_1$, and also color each edge yellow with probability $p_2$, with all colorings independent. Identify an edge being blue with the edge working under the "blue" regime, and identify an edge being yellow with the edge working under the "yellow" regime. If the edge is both blue and yellow, it's considered working under the "green" regime; under the "green" regime, any given edge has probability $p_1p_2$ of working.
Recall the reliability function under any given regime is defined as the expectation of the indicator that a path of the given color exists between source and sink. The point here is that if a green path exists, then a blue path exists and a yellow path exists, but not conversely, because the blue and yellow paths may not cover the same edges. In symbols,
$$
I(textgreen path exists)le I(textblue path exists)I(textyellow path exists).
$$
Taking expectations and using independence yields the result.
answered Jul 17 at 17:51
grand_chat
17.9k11121
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The proof is rather terse. In greater detail, imagine coloring each edge in the network blue with probability $p_1$, and also color each edge yellow with probability $p_2$, with all colorings independent. Identify an edge being blue with the edge working under the "blue" regime, and identify an edge being yellow with the edge working under the "yellow" regime. If the edge is both blue and yellow, it's considered working under the "green" regime; under the "green" regime, any given edge has probability $p_1p_2$ of working.
Recall the reliability function under any given regime is defined as the expectation of the indicator that a path of the given color exists between source and sink. The point here is that if a green path exists, then a blue path exists and a yellow path exists, but not conversely, because the blue and yellow paths may not cover the same edges. In symbols,
$$
I(textgreen path exists)le I(textblue path exists)I(textyellow path exists).
$$
Taking expectations and using independence yields the result.
answered Jul 17 at 17:51
grand_chat
17.9k11121
add a comment |Â
up vote
1
down vote
accepted
The proof is rather terse. In greater detail, imagine coloring each edge in the network blue with probability $p_1$, and also color each edge yellow with probability $p_2$, with all colorings independent. Identify an edge being blue with the edge working under the "blue" regime, and identify an edge being yellow with the edge working under the "yellow" regime. If the edge is both blue and yellow, it's considered working under the "green" regime; under the "green" regime, any given edge has probability $p_1p_2$ of working.
Recall the reliability function under any given regime is defined as the expectation of the indicator that a path of the given color exists between source and sink. The point here is that if a green path exists, then a blue path exists and a yellow path exists, but not conversely, because the blue and yellow paths may not cover the same edges. In symbols,
$$
I(textgreen path exists)le I(textblue path exists)I(textyellow path exists).
$$
Taking expectations and using independence yields the result.
answered Jul 17 at 17:51
grand_chat
17.9k11121
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The proof is rather terse. In greater detail, imagine coloring each edge in the network blue with probability $p_1$, and also color each edge yellow with probability $p_2$, with all colorings independent. Identify an edge being blue with the edge working under the "blue" regime, and identify an edge being yellow with the edge working under the "yellow" regime. If the edge is both blue and yellow, it's considered working under the "green" regime; under the "green" regime, any given edge has probability $p_1p_2$ of working.
Recall the reliability function under any given regime is defined as the expectation of the indicator that a path of the given color exists between source and sink. The point here is that if a green path exists, then a blue path exists and a yellow path exists, but not conversely, because the blue and yellow paths may not cover the same edges. In symbols,
$$
I(textgreen path exists)le I(textblue path exists)I(textyellow path exists).
$$
Taking expectations and using independence yields the result.
answered Jul 17 at 17:51
grand_chat
17.9k11121
The proof is rather terse. In greater detail, imagine coloring each edge in the network blue with probability $p_1$, and also color each edge yellow with probability $p_2$, with all colorings independent. Identify an edge being blue with the edge working under the "blue" regime, and identify an edge being yellow with the edge working under the "yellow" regime. If the edge is both blue and yellow, it's considered working under the "green" regime; under the "green" regime, any given edge has probability $p_1p_2$ of working.
Recall the reliability function under any given regime is defined as the expectation of the indicator that a path of the given color exists between source and sink. The point here is that if a green path exists, then a blue path exists and a yellow path exists, but not conversely, because the blue and yellow paths may not cover the same edges. In symbols,
$$
I(textgreen path exists)le I(textblue path exists)I(textyellow path exists).
$$
Taking expectations and using independence yields the result.
answered Jul 17 at 17:51
grand_chat
17.9k11121
answered Jul 17 at 17:51
grand_chat
17.9k11121
answered Jul 17 at 17:51
grand_chat
17.9k11121
answered Jul 17 at 17:51
grand_chat
17.9k11121
17.9k11121
add a comment |Â
add a comment |Â
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1
What exactly is meant by a connection? If it is a monochrome path between two nodes, then I guess what is meant is that if there is a green connection then there is certainly a blue connection and a yellow connection, (the same path) but there might be another path with all yellow edges and yet another path with all blue edges.
– saulspatz
Jul 17 at 17:46