N exponentially distributed components with mean τ
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I am stuck with the following problem. Let the mean life time of a component be exponentially distributed with mean τ. If we switch on all of them at the same time, how is the waiting time until the first failure distributed? What is its mean? How is the waiting time until the last failure distributed?
Now, I know that the exponential distribution is special in the sense that it does not matter how much time has past, the porbability of failure in the next moment is always the same. However, I don't know if this is asked here and, if yes, how I would go into the problem.
Thank you all for your answers already.
EDIT: I believe I got the first part of the problem:
Let $X_1,X_2 cdots X_N$ ~ $Exp(lambda)$, then the cumulative distribution function
is given by
begin{align}
F_{X_i}(x) = P(X_i leq x)=1-e^{-lambda x_i}
end{align}
So if we define $Y = min{X_1,X_2,cdots X_N}$, we can write the cumulative distribution function
of Y as:
begin{align}
F_Y(y) &= P(Yleq y) =\
&= 1 - P(Ygeq y) =\
&=1 - P(min{X_1,X_2,cdots X_N} geq y) =\
&= 1 - P(X_1 geq y) P(X_2geq y) cdots P(X_Ngeq y) =\
&=1 - e^{-lambda y}e^{-lambda y}cdots e^{-lambda y} = \
&= 1 - e^{-nlambda y}
end{align}
which means that $Y$ ~ Exp ($nlambda$). Still I don't see what the distribution of the maximum should be.
statistics exponential-distribution
asked Nov 21 at 19:50
F. Moe
155
add a comment |
up vote
1
down vote
favorite
I am stuck with the following problem. Let the mean life time of a component be exponentially distributed with mean τ. If we switch on all of them at the same time, how is the waiting time until the first failure distributed? What is its mean? How is the waiting time until the last failure distributed?
Now, I know that the exponential distribution is special in the sense that it does not matter how much time has past, the porbability of failure in the next moment is always the same. However, I don't know if this is asked here and, if yes, how I would go into the problem.
Thank you all for your answers already.
EDIT: I believe I got the first part of the problem:
Let $X_1,X_2 cdots X_N$ ~ $Exp(lambda)$, then the cumulative distribution function
is given by
begin{align}
F_{X_i}(x) = P(X_i leq x)=1-e^{-lambda x_i}
end{align}
So if we define $Y = min{X_1,X_2,cdots X_N}$, we can write the cumulative distribution function
of Y as:
begin{align}
F_Y(y) &= P(Yleq y) =\
&= 1 - P(Ygeq y) =\
&=1 - P(min{X_1,X_2,cdots X_N} geq y) =\
&= 1 - P(X_1 geq y) P(X_2geq y) cdots P(X_Ngeq y) =\
&=1 - e^{-lambda y}e^{-lambda y}cdots e^{-lambda y} = \
&= 1 - e^{-nlambda y}
end{align}
which means that $Y$ ~ Exp ($nlambda$). Still I don't see what the distribution of the maximum should be.
statistics exponential-distribution
asked Nov 21 at 19:50
F. Moe
155
It seems you are asking for the distribution of the minimum of $n$ independent exponential random variables all with rate $tau,$ which has an exponential distribution. Also, for the distribution of the maximum, which does not have an exponential distribution. // These are standard problems, which must have answers here already, if you search for them. // The usual approach is to use the CDF of an exponential distribution to find the CDF of the minimum (or maximum).
– BruceET
Nov 21 at 20:52
Thank you. This should help me for now!
– F. Moe
Nov 21 at 21:18
Okay, I have seen now that the minimum of these N random variables is again exponential (with mean life time n/τ or rate nλ). Also, I have read that the maximum is not exponential, but I have not seen a proof of this. Can someone help?
– F. Moe
Nov 21 at 22:21
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am stuck with the following problem. Let the mean life time of a component be exponentially distributed with mean τ. If we switch on all of them at the same time, how is the waiting time until the first failure distributed? What is its mean? How is the waiting time until the last failure distributed?
Now, I know that the exponential distribution is special in the sense that it does not matter how much time has past, the porbability of failure in the next moment is always the same. However, I don't know if this is asked here and, if yes, how I would go into the problem.
Thank you all for your answers already.
EDIT: I believe I got the first part of the problem:
Let $X_1,X_2 cdots X_N$ ~ $Exp(lambda)$, then the cumulative distribution function
is given by
begin{align}
F_{X_i}(x) = P(X_i leq x)=1-e^{-lambda x_i}
end{align}
So if we define $Y = min{X_1,X_2,cdots X_N}$, we can write the cumulative distribution function
of Y as:
begin{align}
F_Y(y) &= P(Yleq y) =\
&= 1 - P(Ygeq y) =\
&=1 - P(min{X_1,X_2,cdots X_N} geq y) =\
&= 1 - P(X_1 geq y) P(X_2geq y) cdots P(X_Ngeq y) =\
&=1 - e^{-lambda y}e^{-lambda y}cdots e^{-lambda y} = \
&= 1 - e^{-nlambda y}
end{align}
which means that $Y$ ~ Exp ($nlambda$). Still I don't see what the distribution of the maximum should be.
statistics exponential-distribution
asked Nov 21 at 19:50
F. Moe
155
I am stuck with the following problem. Let the mean life time of a component be exponentially distributed with mean τ. If we switch on all of them at the same time, how is the waiting time until the first failure distributed? What is its mean? How is the waiting time until the last failure distributed?
Now, I know that the exponential distribution is special in the sense that it does not matter how much time has past, the porbability of failure in the next moment is always the same. However, I don't know if this is asked here and, if yes, how I would go into the problem.
Thank you all for your answers already.
EDIT: I believe I got the first part of the problem:
Let $X_1,X_2 cdots X_N$ ~ $Exp(lambda)$, then the cumulative distribution function
is given by
begin{align}
F_{X_i}(x) = P(X_i leq x)=1-e^{-lambda x_i}
end{align}
So if we define $Y = min{X_1,X_2,cdots X_N}$, we can write the cumulative distribution function
of Y as:
begin{align}
F_Y(y) &= P(Yleq y) =\
&= 1 - P(Ygeq y) =\
&=1 - P(min{X_1,X_2,cdots X_N} geq y) =\
&= 1 - P(X_1 geq y) P(X_2geq y) cdots P(X_Ngeq y) =\
&=1 - e^{-lambda y}e^{-lambda y}cdots e^{-lambda y} = \
&= 1 - e^{-nlambda y}
end{align}
which means that $Y$ ~ Exp ($nlambda$). Still I don't see what the distribution of the maximum should be.
statistics exponential-distribution
statistics exponential-distribution
asked Nov 21 at 19:50
F. Moe
155
asked Nov 21 at 19:50
F. Moe
155
edited Nov 21 at 22:59
asked Nov 21 at 19:50
F. Moe
155
asked Nov 21 at 19:50
F. Moe
155
asked Nov 21 at 19:50
F. Moe
155
155
It seems you are asking for the distribution of the minimum of $n$ independent exponential random variables all with rate $tau,$ which has an exponential distribution. Also, for the distribution of the maximum, which does not have an exponential distribution. // These are standard problems, which must have answers here already, if you search for them. // The usual approach is to use the CDF of an exponential distribution to find the CDF of the minimum (or maximum).
– BruceET
Nov 21 at 20:52
Thank you. This should help me for now!
– F. Moe
Nov 21 at 21:18
Okay, I have seen now that the minimum of these N random variables is again exponential (with mean life time n/τ or rate nλ). Also, I have read that the maximum is not exponential, but I have not seen a proof of this. Can someone help?
– F. Moe
Nov 21 at 22:21
add a comment |
It seems you are asking for the distribution of the minimum of $n$ independent exponential random variables all with rate $tau,$ which has an exponential distribution. Also, for the distribution of the maximum, which does not have an exponential distribution. // These are standard problems, which must have answers here already, if you search for them. // The usual approach is to use the CDF of an exponential distribution to find the CDF of the minimum (or maximum).
– BruceET
Nov 21 at 20:52
Thank you. This should help me for now!
– F. Moe
Nov 21 at 21:18
Okay, I have seen now that the minimum of these N random variables is again exponential (with mean life time n/τ or rate nλ). Also, I have read that the maximum is not exponential, but I have not seen a proof of this. Can someone help?
– F. Moe
Nov 21 at 22:21
It seems you are asking for the distribution of the minimum of $n$ independent exponential random variables all with rate $tau,$ which has an exponential distribution. Also, for the distribution of the maximum, which does not have an exponential distribution. // These are standard problems, which must have answers here already, if you search for them. // The usual approach is to use the CDF of an exponential distribution to find the CDF of the minimum (or maximum).
– BruceET
Nov 21 at 20:52
It seems you are asking for the distribution of the minimum of $n$ independent exponential random variables all with rate $tau,$ which has an exponential distribution. Also, for the distribution of the maximum, which does not have an exponential distribution. // These are standard problems, which must have answers here already, if you search for them. // The usual approach is to use the CDF of an exponential distribution to find the CDF of the minimum (or maximum).
– BruceET
Nov 21 at 20:52
Thank you. This should help me for now!
– F. Moe
Nov 21 at 21:18
Thank you. This should help me for now!
– F. Moe
Nov 21 at 21:18
Okay, I have seen now that the minimum of these N random variables is again exponential (with mean life time n/τ or rate nλ). Also, I have read that the maximum is not exponential, but I have not seen a proof of this. Can someone help?
– F. Moe
Nov 21 at 22:21
Okay, I have seen now that the minimum of these N random variables is again exponential (with mean life time n/τ or rate nλ). Also, I have read that the maximum is not exponential, but I have not seen a proof of this. Can someone help?
– F. Moe
Nov 21 at 22:21
add a comment |
1 Answer
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1
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As you've already shown, the minimum has an $text{Exp}(Nlambda)$ distribution, with mean $frac{1}{Nlambda}$. The maximum is a little different: $$P(max X_ile x)=P(forall i (X_ile x))=prod_iP(X_ile x)=(1-e^{-lambda x})^N.$$I don't think this has a nice name, although it does imply $exp-lambda max X_i$ has a Beta distribution, as I'll leave you to show.
Although the moments of $max X_i$ aren't easy to obtain, there's an elegant result for its cumulants. Define $H_{n,,m}:=sum_{k=1}^{n}k^{-m}$ so the $n$th Harmonic number $H_{n}=H_{n,,1}$. An $text{Exp}left(lambdaright)$ variable has cgf $-lnleft(1-frac{t}{lambda}right)$, so its cumulants are $kappa_{m}=left(m-1right)!lambda^{-m}$. To get $N$ failures we wait for the initial one, which takes a period $simtext{Exp}(lambda)$; then we wait for the last of the other decays. Induction therefore gives $kappa_{m}=left(m-1right)!H_{N,,m}lambda^{-m}$.
answered Nov 21 at 23:10
J.G.
21k21933
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
add a comment |
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As you've already shown, the minimum has an $text{Exp}(Nlambda)$ distribution, with mean $frac{1}{Nlambda}$. The maximum is a little different: $$P(max X_ile x)=P(forall i (X_ile x))=prod_iP(X_ile x)=(1-e^{-lambda x})^N.$$I don't think this has a nice name, although it does imply $exp-lambda max X_i$ has a Beta distribution, as I'll leave you to show.
Although the moments of $max X_i$ aren't easy to obtain, there's an elegant result for its cumulants. Define $H_{n,,m}:=sum_{k=1}^{n}k^{-m}$ so the $n$th Harmonic number $H_{n}=H_{n,,1}$. An $text{Exp}left(lambdaright)$ variable has cgf $-lnleft(1-frac{t}{lambda}right)$, so its cumulants are $kappa_{m}=left(m-1right)!lambda^{-m}$. To get $N$ failures we wait for the initial one, which takes a period $simtext{Exp}(lambda)$; then we wait for the last of the other decays. Induction therefore gives $kappa_{m}=left(m-1right)!H_{N,,m}lambda^{-m}$.
answered Nov 21 at 23:10
J.G.
21k21933
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
add a comment |
up vote
1
down vote
accepted
As you've already shown, the minimum has an $text{Exp}(Nlambda)$ distribution, with mean $frac{1}{Nlambda}$. The maximum is a little different: $$P(max X_ile x)=P(forall i (X_ile x))=prod_iP(X_ile x)=(1-e^{-lambda x})^N.$$I don't think this has a nice name, although it does imply $exp-lambda max X_i$ has a Beta distribution, as I'll leave you to show.
Although the moments of $max X_i$ aren't easy to obtain, there's an elegant result for its cumulants. Define $H_{n,,m}:=sum_{k=1}^{n}k^{-m}$ so the $n$th Harmonic number $H_{n}=H_{n,,1}$. An $text{Exp}left(lambdaright)$ variable has cgf $-lnleft(1-frac{t}{lambda}right)$, so its cumulants are $kappa_{m}=left(m-1right)!lambda^{-m}$. To get $N$ failures we wait for the initial one, which takes a period $simtext{Exp}(lambda)$; then we wait for the last of the other decays. Induction therefore gives $kappa_{m}=left(m-1right)!H_{N,,m}lambda^{-m}$.
answered Nov 21 at 23:10
J.G.
21k21933
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As you've already shown, the minimum has an $text{Exp}(Nlambda)$ distribution, with mean $frac{1}{Nlambda}$. The maximum is a little different: $$P(max X_ile x)=P(forall i (X_ile x))=prod_iP(X_ile x)=(1-e^{-lambda x})^N.$$I don't think this has a nice name, although it does imply $exp-lambda max X_i$ has a Beta distribution, as I'll leave you to show.
Although the moments of $max X_i$ aren't easy to obtain, there's an elegant result for its cumulants. Define $H_{n,,m}:=sum_{k=1}^{n}k^{-m}$ so the $n$th Harmonic number $H_{n}=H_{n,,1}$. An $text{Exp}left(lambdaright)$ variable has cgf $-lnleft(1-frac{t}{lambda}right)$, so its cumulants are $kappa_{m}=left(m-1right)!lambda^{-m}$. To get $N$ failures we wait for the initial one, which takes a period $simtext{Exp}(lambda)$; then we wait for the last of the other decays. Induction therefore gives $kappa_{m}=left(m-1right)!H_{N,,m}lambda^{-m}$.
answered Nov 21 at 23:10
J.G.
21k21933
As you've already shown, the minimum has an $text{Exp}(Nlambda)$ distribution, with mean $frac{1}{Nlambda}$. The maximum is a little different: $$P(max X_ile x)=P(forall i (X_ile x))=prod_iP(X_ile x)=(1-e^{-lambda x})^N.$$I don't think this has a nice name, although it does imply $exp-lambda max X_i$ has a Beta distribution, as I'll leave you to show.
Although the moments of $max X_i$ aren't easy to obtain, there's an elegant result for its cumulants. Define $H_{n,,m}:=sum_{k=1}^{n}k^{-m}$ so the $n$th Harmonic number $H_{n}=H_{n,,1}$. An $text{Exp}left(lambdaright)$ variable has cgf $-lnleft(1-frac{t}{lambda}right)$, so its cumulants are $kappa_{m}=left(m-1right)!lambda^{-m}$. To get $N$ failures we wait for the initial one, which takes a period $simtext{Exp}(lambda)$; then we wait for the last of the other decays. Induction therefore gives $kappa_{m}=left(m-1right)!H_{N,,m}lambda^{-m}$.
answered Nov 21 at 23:10
J.G.
21k21933
answered Nov 21 at 23:10
J.G.
21k21933
answered Nov 21 at 23:10
J.G.
21k21933
answered Nov 21 at 23:10
J.G.
21k21933
21k21933
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
add a comment |
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
Great. Thank you very much, also for all the additional information. The connection to the Beta distribution seems really interesting and I will look into this!
– F. Moe
Nov 21 at 23:19
add a comment |
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It seems you are asking for the distribution of the minimum of $n$ independent exponential random variables all with rate $tau,$ which has an exponential distribution. Also, for the distribution of the maximum, which does not have an exponential distribution. // These are standard problems, which must have answers here already, if you search for them. // The usual approach is to use the CDF of an exponential distribution to find the CDF of the minimum (or maximum).
– BruceET
Nov 21 at 20:52
Thank you. This should help me for now!
– F. Moe
Nov 21 at 21:18
Okay, I have seen now that the minimum of these N random variables is again exponential (with mean life time n/τ or rate nλ). Also, I have read that the maximum is not exponential, but I have not seen a proof of this. Can someone help?
– F. Moe
Nov 21 at 22:21