2-components machine
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Can you help me just making me understand the point?
I have a 2-components machine. Both components have independent lifetimes, distributed as exponential r.v. with expected value $mu$ = 5. To make a job they simultaneously work and, if components break, they will be fixed simultaneously (there are moments in which just a component is working).
Considering X(t) the number of functioning componet at time t (t>0), compute:
lim (t--> ∞) of P[X(t+s) = 1] for every s in [0,5]
I don't understand why it should be different from computing the limit of P[X(t) = 1], which is a previous point.
Can you help me understanding this?
probability stochastic-processes
asked Jan 3 at 16:30
CronoPoundCronoPound
11
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add a comment |
$begingroup$
Can you help me just making me understand the point?
I have a 2-components machine. Both components have independent lifetimes, distributed as exponential r.v. with expected value $mu$ = 5. To make a job they simultaneously work and, if components break, they will be fixed simultaneously (there are moments in which just a component is working).
Considering X(t) the number of functioning componet at time t (t>0), compute:
lim (t--> ∞) of P[X(t+s) = 1] for every s in [0,5]
I don't understand why it should be different from computing the limit of P[X(t) = 1], which is a previous point.
Can you help me understanding this?
probability stochastic-processes
asked Jan 3 at 16:30
CronoPoundCronoPound
11
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Don't we have to know how long it takes to fix a component?
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– saulspatz
Jan 3 at 17:34
$begingroup$
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
$endgroup$
– CronoPound
Jan 4 at 8:07
add a comment |
$begingroup$
Can you help me just making me understand the point?
I have a 2-components machine. Both components have independent lifetimes, distributed as exponential r.v. with expected value $mu$ = 5. To make a job they simultaneously work and, if components break, they will be fixed simultaneously (there are moments in which just a component is working).
Considering X(t) the number of functioning componet at time t (t>0), compute:
lim (t--> ∞) of P[X(t+s) = 1] for every s in [0,5]
I don't understand why it should be different from computing the limit of P[X(t) = 1], which is a previous point.
Can you help me understanding this?
probability stochastic-processes
asked Jan 3 at 16:30
CronoPoundCronoPound
11
$endgroup$
Can you help me just making me understand the point?
I have a 2-components machine. Both components have independent lifetimes, distributed as exponential r.v. with expected value $mu$ = 5. To make a job they simultaneously work and, if components break, they will be fixed simultaneously (there are moments in which just a component is working).
Considering X(t) the number of functioning componet at time t (t>0), compute:
lim (t--> ∞) of P[X(t+s) = 1] for every s in [0,5]
I don't understand why it should be different from computing the limit of P[X(t) = 1], which is a previous point.
Can you help me understanding this?
probability stochastic-processes
probability stochastic-processes
asked Jan 3 at 16:30
CronoPoundCronoPound
11
asked Jan 3 at 16:30
CronoPoundCronoPound
11
asked Jan 3 at 16:30
CronoPoundCronoPound
11
asked Jan 3 at 16:30
CronoPoundCronoPound
11
asked Jan 3 at 16:30
CronoPoundCronoPound
11
11
$begingroup$
Don't we have to know how long it takes to fix a component?
$endgroup$
– saulspatz
Jan 3 at 17:34
$begingroup$
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
$endgroup$
– CronoPound
Jan 4 at 8:07
add a comment |
$begingroup$
Don't we have to know how long it takes to fix a component?
$endgroup$
– saulspatz
Jan 3 at 17:34
$begingroup$
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
$endgroup$
– CronoPound
Jan 4 at 8:07
$begingroup$
Don't we have to know how long it takes to fix a component?
$endgroup$
– saulspatz
Jan 3 at 17:34
$begingroup$
Don't we have to know how long it takes to fix a component?
$endgroup$
– saulspatz
Jan 3 at 17:34
$begingroup$
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
$endgroup$
– CronoPound
Jan 4 at 8:07
$begingroup$
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
$endgroup$
– CronoPound
Jan 4 at 8:07
add a comment |
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$begingroup$
Don't we have to know how long it takes to fix a component?
$endgroup$
– saulspatz
Jan 3 at 17:34
$begingroup$
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
$endgroup$
– CronoPound
Jan 4 at 8:07