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Lifespan of a machine with two breaking parts.
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Let us consider a machine with two parts X and Y. The time before part X goes out of service has an Exponential distribution with $lambda = 1$ and the time before part Y goes out of service has an Exponential distribution with $lambda = 0.05$ (its life expectation is 20 time units). When the part X breaks it is replaced in 1 time unit and the whole machine is not working during this time (hence the part Y is not ageing while the part X is being replaced). When part Y goes out of service, the whole machine is considered dead. What is the expectation of the random variable T, which stands for the time before death of the whole machine?
I have tried to consider conditioning on the lifespan of the part Y (I denoted this random variable T_Y) and on the number of times the part X breaks before the part Y goes out of service (N).
$$ E(T) = sum_k=0^inftyint_0^inftyE(T|T_Y, N)P(N=k|T_Y)P(T_Y=t_y)dt_y $$
But I am unsure of the formula. Also I would say that $E(T|T_Y, N) = t_y + k$ but I don't see why exactly. And what would be $ P(N=k|T_y)$?
probability probability-distributions
asked Aug 6 at 15:03
Vwann
224
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Let us consider a machine with two parts X and Y. The time before part X goes out of service has an Exponential distribution with $lambda = 1$ and the time before part Y goes out of service has an Exponential distribution with $lambda = 0.05$ (its life expectation is 20 time units). When the part X breaks it is replaced in 1 time unit and the whole machine is not working during this time (hence the part Y is not ageing while the part X is being replaced). When part Y goes out of service, the whole machine is considered dead. What is the expectation of the random variable T, which stands for the time before death of the whole machine?
I have tried to consider conditioning on the lifespan of the part Y (I denoted this random variable T_Y) and on the number of times the part X breaks before the part Y goes out of service (N).
$$ E(T) = sum_k=0^inftyint_0^inftyE(T|T_Y, N)P(N=k|T_Y)P(T_Y=t_y)dt_y $$
But I am unsure of the formula. Also I would say that $E(T|T_Y, N) = t_y + k$ but I don't see why exactly. And what would be $ P(N=k|T_y)$?
probability probability-distributions
asked Aug 6 at 15:03
Vwann
224
Just checking that I understand the question. $E(T)$ is the expected time to failure of part $Y$ plus the expected number of times that part $X$ fails before part $Y$ fails. Is that correct?
– saulspatz
Aug 6 at 15:13
Yes, I think so.
– Vwann
Aug 6 at 15:22
add a comment |Â
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0
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up vote
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down vote
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Let us consider a machine with two parts X and Y. The time before part X goes out of service has an Exponential distribution with $lambda = 1$ and the time before part Y goes out of service has an Exponential distribution with $lambda = 0.05$ (its life expectation is 20 time units). When the part X breaks it is replaced in 1 time unit and the whole machine is not working during this time (hence the part Y is not ageing while the part X is being replaced). When part Y goes out of service, the whole machine is considered dead. What is the expectation of the random variable T, which stands for the time before death of the whole machine?
I have tried to consider conditioning on the lifespan of the part Y (I denoted this random variable T_Y) and on the number of times the part X breaks before the part Y goes out of service (N).
$$ E(T) = sum_k=0^inftyint_0^inftyE(T|T_Y, N)P(N=k|T_Y)P(T_Y=t_y)dt_y $$
But I am unsure of the formula. Also I would say that $E(T|T_Y, N) = t_y + k$ but I don't see why exactly. And what would be $ P(N=k|T_y)$?
probability probability-distributions
asked Aug 6 at 15:03
Vwann
224
Let us consider a machine with two parts X and Y. The time before part X goes out of service has an Exponential distribution with $lambda = 1$ and the time before part Y goes out of service has an Exponential distribution with $lambda = 0.05$ (its life expectation is 20 time units). When the part X breaks it is replaced in 1 time unit and the whole machine is not working during this time (hence the part Y is not ageing while the part X is being replaced). When part Y goes out of service, the whole machine is considered dead. What is the expectation of the random variable T, which stands for the time before death of the whole machine?
I have tried to consider conditioning on the lifespan of the part Y (I denoted this random variable T_Y) and on the number of times the part X breaks before the part Y goes out of service (N).
$$ E(T) = sum_k=0^inftyint_0^inftyE(T|T_Y, N)P(N=k|T_Y)P(T_Y=t_y)dt_y $$
But I am unsure of the formula. Also I would say that $E(T|T_Y, N) = t_y + k$ but I don't see why exactly. And what would be $ P(N=k|T_y)$?
probability probability-distributions
asked Aug 6 at 15:03
Vwann
224
edited Aug 6 at 15:23
asked Aug 6 at 15:03
Vwann
224
asked Aug 6 at 15:03
Vwann
224
asked Aug 6 at 15:03
Vwann
224
224
Just checking that I understand the question. $E(T)$ is the expected time to failure of part $Y$ plus the expected number of times that part $X$ fails before part $Y$ fails. Is that correct?
– saulspatz
Aug 6 at 15:13
Yes, I think so.
– Vwann
Aug 6 at 15:22
add a comment |Â
Just checking that I understand the question. $E(T)$ is the expected time to failure of part $Y$ plus the expected number of times that part $X$ fails before part $Y$ fails. Is that correct?
– saulspatz
Aug 6 at 15:13
Yes, I think so.
– Vwann
Aug 6 at 15:22
Just checking that I understand the question. $E(T)$ is the expected time to failure of part $Y$ plus the expected number of times that part $X$ fails before part $Y$ fails. Is that correct?
– saulspatz
Aug 6 at 15:13
Just checking that I understand the question. $E(T)$ is the expected time to failure of part $Y$ plus the expected number of times that part $X$ fails before part $Y$ fails. Is that correct?
– saulspatz
Aug 6 at 15:13
Yes, I think so.
– Vwann
Aug 6 at 15:22
Yes, I think so.
– Vwann
Aug 6 at 15:22
add a comment |Â
1 Answer
1
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up vote
1
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accepted
Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p=.05over1.05$$ and we know that the expected number of trials until success is $$1over p=21$$
That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$
just as we would have guessed.
answered Aug 6 at 16:49
saulspatz
10.7k21323
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
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
Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p=.05over1.05$$ and we know that the expected number of trials until success is $$1over p=21$$
That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$
just as we would have guessed.
answered Aug 6 at 16:49
saulspatz
10.7k21323
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
add a comment |Â
up vote
1
down vote
accepted
Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p=.05over1.05$$ and we know that the expected number of trials until success is $$1over p=21$$
That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$
just as we would have guessed.
answered Aug 6 at 16:49
saulspatz
10.7k21323
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p=.05over1.05$$ and we know that the expected number of trials until success is $$1over p=21$$
That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$
just as we would have guessed.
answered Aug 6 at 16:49
saulspatz
10.7k21323
Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p=.05over1.05$$ and we know that the expected number of trials until success is $$1over p=21$$
That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$
just as we would have guessed.
answered Aug 6 at 16:49
saulspatz
10.7k21323
answered Aug 6 at 16:49
saulspatz
10.7k21323
answered Aug 6 at 16:49
saulspatz
10.7k21323
answered Aug 6 at 16:49
saulspatz
10.7k21323
10.7k21323
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
add a comment |Â
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$?
– Vwann
Aug 6 at 20:22
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(lambda^2) = 20*21 +20^2$?
– Vwann
Aug 6 at 20:27
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
"The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe.
– saulspatz
Aug 6 at 20:52
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work.
– saulspatz
Aug 6 at 20:58
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
You are right, I can not compute the variance as a sum of variances.
– Vwann
Aug 6 at 21:14
add a comment |Â
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Just checking that I understand the question. $E(T)$ is the expected time to failure of part $Y$ plus the expected number of times that part $X$ fails before part $Y$ fails. Is that correct?
– saulspatz
Aug 6 at 15:13
Yes, I think so.
– Vwann
Aug 6 at 15:22