2-components machine
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 2 days ago
CronoPound
11
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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 2 days ago
CronoPound
11
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Don't we have to know how long it takes to fix a component?
– saulspatz
2 days ago
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
– CronoPound
yesterday
add a comment |
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 2 days ago
CronoPound
11
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
CronoPound
11
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
CronoPound
11
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
CronoPound
11
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
CronoPound
11
asked 2 days ago
CronoPound
11
11
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CronoPound is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Don't we have to know how long it takes to fix a component?
– saulspatz
2 days ago
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
– CronoPound
yesterday
add a comment |
Don't we have to know how long it takes to fix a component?
– saulspatz
2 days ago
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
– CronoPound
yesterday
Don't we have to know how long it takes to fix a component?
– saulspatz
2 days ago
Don't we have to know how long it takes to fix a component?
– saulspatz
2 days ago
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
– CronoPound
yesterday
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
– CronoPound
yesterday
add a comment |
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Don't we have to know how long it takes to fix a component?
– saulspatz
2 days ago
@saulspatz components are fixed simultaneously and immediatly, we can ignore fixing time
– CronoPound
yesterday