$T_1, … , T_n$ time of life of instruments with geometric distribution of parameter p independent












0














$T_1, ... , T_n$ time of life of instruments with geometric distribution of parameter p independents.
I define S as the first n instants of failure and U as the last n instant of failure.
I want to find the law of S and the density of U.



This is what I did so far:



To find the law of S I start by calculating:
$$mathbb{P}(S = k) = mathbb{P}(S geq k) - mathbb{P}(S geq k + 1) $$
$$mathbb{P}(S = k) = mathbb{P}(S < k+1) - mathbb{P}(S < k)$$
But here I am stuck on what I should do because I still do not know the distribution.



I do the same reasoning with U as follows:
$$ mathbb{U}(S = k) = mathbb{U}(U leq k) - mathbb{U}(S leq k + 1) $$
But I still get nowhere.



Any suggestions?



EDIT:



Following the suggestions I managed to say that:



$$mathbb{P}(S=k) = [mathbb{P}(T_1 geq k)]^n = [1 -mathbb{P}(T_1 < k)]^n = $$
$$ = [1 - [1-(1-p)^k]^n = [(1-p)^k]^n $$



Now the exercise asks me to identify random variable, saying it is a geometric distribution with parameter $1-(1-p)^n$. However I am struggling to see how this is true.



Moreover, how should I study U? Is it the maximum instead of the minimum? I would love to see a hint of solution using my procedure as well.










share|cite|improve this question




















  • 1




    If I understand correct then $S=min(T_1,dots,T_n)$ so that $P(Sgeq k)=P(T_1geq k,dots,T_ngeq k)=P(T_1geq k)^n$. Does that help?
    – drhab
    Nov 2 '18 at 9:36










  • Can you follow it up? I am not sure where this is going.
    – qcc101
    Nov 2 '18 at 11:07










  • @qcc101 you have an expression for $P(T_1geq k)$ because you know $T_1 sim Geo(p)$ thus substitute it into what drhab found to get an expression for $P(S geq k)$...
    – LoveTooNap29
    Nov 3 '18 at 19:15










  • @LoveTooNap29 From this I found that: $[mathbb{P}(T_1 geq k)]^n = [(1-p)^k]^n$. Is this distribution equivalent to a geometric distribution of parameter $1-(1-p)^n$? Anyway, I would love to solve this exercise also with my procedure if that is possible.
    – qcc101
    Dec 4 '18 at 15:20


















0














$T_1, ... , T_n$ time of life of instruments with geometric distribution of parameter p independents.
I define S as the first n instants of failure and U as the last n instant of failure.
I want to find the law of S and the density of U.



This is what I did so far:



To find the law of S I start by calculating:
$$mathbb{P}(S = k) = mathbb{P}(S geq k) - mathbb{P}(S geq k + 1) $$
$$mathbb{P}(S = k) = mathbb{P}(S < k+1) - mathbb{P}(S < k)$$
But here I am stuck on what I should do because I still do not know the distribution.



I do the same reasoning with U as follows:
$$ mathbb{U}(S = k) = mathbb{U}(U leq k) - mathbb{U}(S leq k + 1) $$
But I still get nowhere.



Any suggestions?



EDIT:



Following the suggestions I managed to say that:



$$mathbb{P}(S=k) = [mathbb{P}(T_1 geq k)]^n = [1 -mathbb{P}(T_1 < k)]^n = $$
$$ = [1 - [1-(1-p)^k]^n = [(1-p)^k]^n $$



Now the exercise asks me to identify random variable, saying it is a geometric distribution with parameter $1-(1-p)^n$. However I am struggling to see how this is true.



Moreover, how should I study U? Is it the maximum instead of the minimum? I would love to see a hint of solution using my procedure as well.










share|cite|improve this question




















  • 1




    If I understand correct then $S=min(T_1,dots,T_n)$ so that $P(Sgeq k)=P(T_1geq k,dots,T_ngeq k)=P(T_1geq k)^n$. Does that help?
    – drhab
    Nov 2 '18 at 9:36










  • Can you follow it up? I am not sure where this is going.
    – qcc101
    Nov 2 '18 at 11:07










  • @qcc101 you have an expression for $P(T_1geq k)$ because you know $T_1 sim Geo(p)$ thus substitute it into what drhab found to get an expression for $P(S geq k)$...
    – LoveTooNap29
    Nov 3 '18 at 19:15










  • @LoveTooNap29 From this I found that: $[mathbb{P}(T_1 geq k)]^n = [(1-p)^k]^n$. Is this distribution equivalent to a geometric distribution of parameter $1-(1-p)^n$? Anyway, I would love to solve this exercise also with my procedure if that is possible.
    – qcc101
    Dec 4 '18 at 15:20
















0












0








0







$T_1, ... , T_n$ time of life of instruments with geometric distribution of parameter p independents.
I define S as the first n instants of failure and U as the last n instant of failure.
I want to find the law of S and the density of U.



This is what I did so far:



To find the law of S I start by calculating:
$$mathbb{P}(S = k) = mathbb{P}(S geq k) - mathbb{P}(S geq k + 1) $$
$$mathbb{P}(S = k) = mathbb{P}(S < k+1) - mathbb{P}(S < k)$$
But here I am stuck on what I should do because I still do not know the distribution.



I do the same reasoning with U as follows:
$$ mathbb{U}(S = k) = mathbb{U}(U leq k) - mathbb{U}(S leq k + 1) $$
But I still get nowhere.



Any suggestions?



EDIT:



Following the suggestions I managed to say that:



$$mathbb{P}(S=k) = [mathbb{P}(T_1 geq k)]^n = [1 -mathbb{P}(T_1 < k)]^n = $$
$$ = [1 - [1-(1-p)^k]^n = [(1-p)^k]^n $$



Now the exercise asks me to identify random variable, saying it is a geometric distribution with parameter $1-(1-p)^n$. However I am struggling to see how this is true.



Moreover, how should I study U? Is it the maximum instead of the minimum? I would love to see a hint of solution using my procedure as well.










share|cite|improve this question















$T_1, ... , T_n$ time of life of instruments with geometric distribution of parameter p independents.
I define S as the first n instants of failure and U as the last n instant of failure.
I want to find the law of S and the density of U.



This is what I did so far:



To find the law of S I start by calculating:
$$mathbb{P}(S = k) = mathbb{P}(S geq k) - mathbb{P}(S geq k + 1) $$
$$mathbb{P}(S = k) = mathbb{P}(S < k+1) - mathbb{P}(S < k)$$
But here I am stuck on what I should do because I still do not know the distribution.



I do the same reasoning with U as follows:
$$ mathbb{U}(S = k) = mathbb{U}(U leq k) - mathbb{U}(S leq k + 1) $$
But I still get nowhere.



Any suggestions?



EDIT:



Following the suggestions I managed to say that:



$$mathbb{P}(S=k) = [mathbb{P}(T_1 geq k)]^n = [1 -mathbb{P}(T_1 < k)]^n = $$
$$ = [1 - [1-(1-p)^k]^n = [(1-p)^k]^n $$



Now the exercise asks me to identify random variable, saying it is a geometric distribution with parameter $1-(1-p)^n$. However I am struggling to see how this is true.



Moreover, how should I study U? Is it the maximum instead of the minimum? I would love to see a hint of solution using my procedure as well.







probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 15:38

























asked Nov 2 '18 at 8:59









qcc101

477113




477113








  • 1




    If I understand correct then $S=min(T_1,dots,T_n)$ so that $P(Sgeq k)=P(T_1geq k,dots,T_ngeq k)=P(T_1geq k)^n$. Does that help?
    – drhab
    Nov 2 '18 at 9:36










  • Can you follow it up? I am not sure where this is going.
    – qcc101
    Nov 2 '18 at 11:07










  • @qcc101 you have an expression for $P(T_1geq k)$ because you know $T_1 sim Geo(p)$ thus substitute it into what drhab found to get an expression for $P(S geq k)$...
    – LoveTooNap29
    Nov 3 '18 at 19:15










  • @LoveTooNap29 From this I found that: $[mathbb{P}(T_1 geq k)]^n = [(1-p)^k]^n$. Is this distribution equivalent to a geometric distribution of parameter $1-(1-p)^n$? Anyway, I would love to solve this exercise also with my procedure if that is possible.
    – qcc101
    Dec 4 '18 at 15:20
















  • 1




    If I understand correct then $S=min(T_1,dots,T_n)$ so that $P(Sgeq k)=P(T_1geq k,dots,T_ngeq k)=P(T_1geq k)^n$. Does that help?
    – drhab
    Nov 2 '18 at 9:36










  • Can you follow it up? I am not sure where this is going.
    – qcc101
    Nov 2 '18 at 11:07










  • @qcc101 you have an expression for $P(T_1geq k)$ because you know $T_1 sim Geo(p)$ thus substitute it into what drhab found to get an expression for $P(S geq k)$...
    – LoveTooNap29
    Nov 3 '18 at 19:15










  • @LoveTooNap29 From this I found that: $[mathbb{P}(T_1 geq k)]^n = [(1-p)^k]^n$. Is this distribution equivalent to a geometric distribution of parameter $1-(1-p)^n$? Anyway, I would love to solve this exercise also with my procedure if that is possible.
    – qcc101
    Dec 4 '18 at 15:20










1




1




If I understand correct then $S=min(T_1,dots,T_n)$ so that $P(Sgeq k)=P(T_1geq k,dots,T_ngeq k)=P(T_1geq k)^n$. Does that help?
– drhab
Nov 2 '18 at 9:36




If I understand correct then $S=min(T_1,dots,T_n)$ so that $P(Sgeq k)=P(T_1geq k,dots,T_ngeq k)=P(T_1geq k)^n$. Does that help?
– drhab
Nov 2 '18 at 9:36












Can you follow it up? I am not sure where this is going.
– qcc101
Nov 2 '18 at 11:07




Can you follow it up? I am not sure where this is going.
– qcc101
Nov 2 '18 at 11:07












@qcc101 you have an expression for $P(T_1geq k)$ because you know $T_1 sim Geo(p)$ thus substitute it into what drhab found to get an expression for $P(S geq k)$...
– LoveTooNap29
Nov 3 '18 at 19:15




@qcc101 you have an expression for $P(T_1geq k)$ because you know $T_1 sim Geo(p)$ thus substitute it into what drhab found to get an expression for $P(S geq k)$...
– LoveTooNap29
Nov 3 '18 at 19:15












@LoveTooNap29 From this I found that: $[mathbb{P}(T_1 geq k)]^n = [(1-p)^k]^n$. Is this distribution equivalent to a geometric distribution of parameter $1-(1-p)^n$? Anyway, I would love to solve this exercise also with my procedure if that is possible.
– qcc101
Dec 4 '18 at 15:20






@LoveTooNap29 From this I found that: $[mathbb{P}(T_1 geq k)]^n = [(1-p)^k]^n$. Is this distribution equivalent to a geometric distribution of parameter $1-(1-p)^n$? Anyway, I would love to solve this exercise also with my procedure if that is possible.
– qcc101
Dec 4 '18 at 15:20












1 Answer
1






active

oldest

votes


















1





+50









First of all defining $S$ as the "first n instants [sic] of failure" makes little sense, as you have only been given $n$ lifetimes, so for there to be $n$ instances of failure would require all $n$ lifetimes to terminate. This confusion is likely why Drhab commented as such, suggesting instead to look at $S=min{T_1,dotsc,T_n}$ and provided the general formula for all minimums of IIDRVs: $P(S> k)=P(T_1> k)^n$. This is interpreted as the first instance of failure out of the $n$ lifetimes, i.e. the minimum lifetime.



Then, with $P(T_1leq k)=1-(1-p)^k$ (assuming CDF corresponding to when the $T_i$ take values in $mathbb{N}$), we first have $P(T_1>k)=(1-p)^k=:q^k$ where $q=1-p$ for ease of notation. Then substituting, we have $P(S>k)=(q^k)^n=(q^n)^k$, so that indeed $P(Sleq k)=1-(q^n)^k$ is the CDF of a Geometric RV with new parameter $tilde{p}=1-q^n=1-(1-p)^n$.



How can we tackle $U$? Well, let's see how that minimum formula was derived in the first place and hope that informs us on a similar manner to derive a formula for the maximum of IIDRVs.



If the minimum of $n$ variables is greater than a number $k$, then all of those variables must be greater than $k$, i.e the following set equalities hold
$${min{T_1,dotsc, T_n} >k}={T_1 >k,dotsc, T_n>k}.$$
That's just from the definition of minimum. Thus, taking probabilities and writing $S=min{S_1,dotsc, S_n}$, we have
$$P(S >k)=P(T_1>k, dotsc, T_n>k),$$
and finally the IID assumption greatly simplifies the RHS further, to
$$P(S>k)=P(T_1>K)P(T_2>k)cdot dotsc cdot P(T_n >k),$$
first using independence, then using the identical distributions, we finally get
$$P(S>k)=P(T_1>k)^n.$$



Can you now, after seeing this argument for minimum, derive the analogous formula for maximums? Please comment for further clarification.



EDIT
If $U=max{T_1,dots, T_n}$, then we know that $P(Uleq k)=P(T_1 leq k)^n=(1-q^{k})^n$ by definition of maximum and since the $T_i$ are IID. To find $mathbb{E}(U)$ we use the alternative formula for expectations for discrete non-negative RVs:
$$E(U)=sum_{k=1}^infty P(Ugeq k).$$
Since $P(Ugeq k)=1-P(Uleq k-1)=1-(1-q^{k-1})^n,$ we must compute the series,
$$sum_{k=1}^infty 1-(1-q^{k-1})^n.$$
In principle this series can always be computed by using the binomial theorem to expand $(1-q^{k-1})^n$ and then use theory of geometric series to compute the resulting series' in closed form. For $n=2$, we have
$$mathbb{E}(U)=frac{3-2p}{p(2-p)}approx 2.667$$
when $p=0.5$, for example.






share|cite|improve this answer























  • Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
    – qcc101
    Dec 5 '18 at 13:38












  • @qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
    – LoveTooNap29
    Dec 5 '18 at 18:47











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1 Answer
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oldest

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+50









First of all defining $S$ as the "first n instants [sic] of failure" makes little sense, as you have only been given $n$ lifetimes, so for there to be $n$ instances of failure would require all $n$ lifetimes to terminate. This confusion is likely why Drhab commented as such, suggesting instead to look at $S=min{T_1,dotsc,T_n}$ and provided the general formula for all minimums of IIDRVs: $P(S> k)=P(T_1> k)^n$. This is interpreted as the first instance of failure out of the $n$ lifetimes, i.e. the minimum lifetime.



Then, with $P(T_1leq k)=1-(1-p)^k$ (assuming CDF corresponding to when the $T_i$ take values in $mathbb{N}$), we first have $P(T_1>k)=(1-p)^k=:q^k$ where $q=1-p$ for ease of notation. Then substituting, we have $P(S>k)=(q^k)^n=(q^n)^k$, so that indeed $P(Sleq k)=1-(q^n)^k$ is the CDF of a Geometric RV with new parameter $tilde{p}=1-q^n=1-(1-p)^n$.



How can we tackle $U$? Well, let's see how that minimum formula was derived in the first place and hope that informs us on a similar manner to derive a formula for the maximum of IIDRVs.



If the minimum of $n$ variables is greater than a number $k$, then all of those variables must be greater than $k$, i.e the following set equalities hold
$${min{T_1,dotsc, T_n} >k}={T_1 >k,dotsc, T_n>k}.$$
That's just from the definition of minimum. Thus, taking probabilities and writing $S=min{S_1,dotsc, S_n}$, we have
$$P(S >k)=P(T_1>k, dotsc, T_n>k),$$
and finally the IID assumption greatly simplifies the RHS further, to
$$P(S>k)=P(T_1>K)P(T_2>k)cdot dotsc cdot P(T_n >k),$$
first using independence, then using the identical distributions, we finally get
$$P(S>k)=P(T_1>k)^n.$$



Can you now, after seeing this argument for minimum, derive the analogous formula for maximums? Please comment for further clarification.



EDIT
If $U=max{T_1,dots, T_n}$, then we know that $P(Uleq k)=P(T_1 leq k)^n=(1-q^{k})^n$ by definition of maximum and since the $T_i$ are IID. To find $mathbb{E}(U)$ we use the alternative formula for expectations for discrete non-negative RVs:
$$E(U)=sum_{k=1}^infty P(Ugeq k).$$
Since $P(Ugeq k)=1-P(Uleq k-1)=1-(1-q^{k-1})^n,$ we must compute the series,
$$sum_{k=1}^infty 1-(1-q^{k-1})^n.$$
In principle this series can always be computed by using the binomial theorem to expand $(1-q^{k-1})^n$ and then use theory of geometric series to compute the resulting series' in closed form. For $n=2$, we have
$$mathbb{E}(U)=frac{3-2p}{p(2-p)}approx 2.667$$
when $p=0.5$, for example.






share|cite|improve this answer























  • Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
    – qcc101
    Dec 5 '18 at 13:38












  • @qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
    – LoveTooNap29
    Dec 5 '18 at 18:47
















1





+50









First of all defining $S$ as the "first n instants [sic] of failure" makes little sense, as you have only been given $n$ lifetimes, so for there to be $n$ instances of failure would require all $n$ lifetimes to terminate. This confusion is likely why Drhab commented as such, suggesting instead to look at $S=min{T_1,dotsc,T_n}$ and provided the general formula for all minimums of IIDRVs: $P(S> k)=P(T_1> k)^n$. This is interpreted as the first instance of failure out of the $n$ lifetimes, i.e. the minimum lifetime.



Then, with $P(T_1leq k)=1-(1-p)^k$ (assuming CDF corresponding to when the $T_i$ take values in $mathbb{N}$), we first have $P(T_1>k)=(1-p)^k=:q^k$ where $q=1-p$ for ease of notation. Then substituting, we have $P(S>k)=(q^k)^n=(q^n)^k$, so that indeed $P(Sleq k)=1-(q^n)^k$ is the CDF of a Geometric RV with new parameter $tilde{p}=1-q^n=1-(1-p)^n$.



How can we tackle $U$? Well, let's see how that minimum formula was derived in the first place and hope that informs us on a similar manner to derive a formula for the maximum of IIDRVs.



If the minimum of $n$ variables is greater than a number $k$, then all of those variables must be greater than $k$, i.e the following set equalities hold
$${min{T_1,dotsc, T_n} >k}={T_1 >k,dotsc, T_n>k}.$$
That's just from the definition of minimum. Thus, taking probabilities and writing $S=min{S_1,dotsc, S_n}$, we have
$$P(S >k)=P(T_1>k, dotsc, T_n>k),$$
and finally the IID assumption greatly simplifies the RHS further, to
$$P(S>k)=P(T_1>K)P(T_2>k)cdot dotsc cdot P(T_n >k),$$
first using independence, then using the identical distributions, we finally get
$$P(S>k)=P(T_1>k)^n.$$



Can you now, after seeing this argument for minimum, derive the analogous formula for maximums? Please comment for further clarification.



EDIT
If $U=max{T_1,dots, T_n}$, then we know that $P(Uleq k)=P(T_1 leq k)^n=(1-q^{k})^n$ by definition of maximum and since the $T_i$ are IID. To find $mathbb{E}(U)$ we use the alternative formula for expectations for discrete non-negative RVs:
$$E(U)=sum_{k=1}^infty P(Ugeq k).$$
Since $P(Ugeq k)=1-P(Uleq k-1)=1-(1-q^{k-1})^n,$ we must compute the series,
$$sum_{k=1}^infty 1-(1-q^{k-1})^n.$$
In principle this series can always be computed by using the binomial theorem to expand $(1-q^{k-1})^n$ and then use theory of geometric series to compute the resulting series' in closed form. For $n=2$, we have
$$mathbb{E}(U)=frac{3-2p}{p(2-p)}approx 2.667$$
when $p=0.5$, for example.






share|cite|improve this answer























  • Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
    – qcc101
    Dec 5 '18 at 13:38












  • @qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
    – LoveTooNap29
    Dec 5 '18 at 18:47














1





+50







1





+50



1




+50




First of all defining $S$ as the "first n instants [sic] of failure" makes little sense, as you have only been given $n$ lifetimes, so for there to be $n$ instances of failure would require all $n$ lifetimes to terminate. This confusion is likely why Drhab commented as such, suggesting instead to look at $S=min{T_1,dotsc,T_n}$ and provided the general formula for all minimums of IIDRVs: $P(S> k)=P(T_1> k)^n$. This is interpreted as the first instance of failure out of the $n$ lifetimes, i.e. the minimum lifetime.



Then, with $P(T_1leq k)=1-(1-p)^k$ (assuming CDF corresponding to when the $T_i$ take values in $mathbb{N}$), we first have $P(T_1>k)=(1-p)^k=:q^k$ where $q=1-p$ for ease of notation. Then substituting, we have $P(S>k)=(q^k)^n=(q^n)^k$, so that indeed $P(Sleq k)=1-(q^n)^k$ is the CDF of a Geometric RV with new parameter $tilde{p}=1-q^n=1-(1-p)^n$.



How can we tackle $U$? Well, let's see how that minimum formula was derived in the first place and hope that informs us on a similar manner to derive a formula for the maximum of IIDRVs.



If the minimum of $n$ variables is greater than a number $k$, then all of those variables must be greater than $k$, i.e the following set equalities hold
$${min{T_1,dotsc, T_n} >k}={T_1 >k,dotsc, T_n>k}.$$
That's just from the definition of minimum. Thus, taking probabilities and writing $S=min{S_1,dotsc, S_n}$, we have
$$P(S >k)=P(T_1>k, dotsc, T_n>k),$$
and finally the IID assumption greatly simplifies the RHS further, to
$$P(S>k)=P(T_1>K)P(T_2>k)cdot dotsc cdot P(T_n >k),$$
first using independence, then using the identical distributions, we finally get
$$P(S>k)=P(T_1>k)^n.$$



Can you now, after seeing this argument for minimum, derive the analogous formula for maximums? Please comment for further clarification.



EDIT
If $U=max{T_1,dots, T_n}$, then we know that $P(Uleq k)=P(T_1 leq k)^n=(1-q^{k})^n$ by definition of maximum and since the $T_i$ are IID. To find $mathbb{E}(U)$ we use the alternative formula for expectations for discrete non-negative RVs:
$$E(U)=sum_{k=1}^infty P(Ugeq k).$$
Since $P(Ugeq k)=1-P(Uleq k-1)=1-(1-q^{k-1})^n,$ we must compute the series,
$$sum_{k=1}^infty 1-(1-q^{k-1})^n.$$
In principle this series can always be computed by using the binomial theorem to expand $(1-q^{k-1})^n$ and then use theory of geometric series to compute the resulting series' in closed form. For $n=2$, we have
$$mathbb{E}(U)=frac{3-2p}{p(2-p)}approx 2.667$$
when $p=0.5$, for example.






share|cite|improve this answer














First of all defining $S$ as the "first n instants [sic] of failure" makes little sense, as you have only been given $n$ lifetimes, so for there to be $n$ instances of failure would require all $n$ lifetimes to terminate. This confusion is likely why Drhab commented as such, suggesting instead to look at $S=min{T_1,dotsc,T_n}$ and provided the general formula for all minimums of IIDRVs: $P(S> k)=P(T_1> k)^n$. This is interpreted as the first instance of failure out of the $n$ lifetimes, i.e. the minimum lifetime.



Then, with $P(T_1leq k)=1-(1-p)^k$ (assuming CDF corresponding to when the $T_i$ take values in $mathbb{N}$), we first have $P(T_1>k)=(1-p)^k=:q^k$ where $q=1-p$ for ease of notation. Then substituting, we have $P(S>k)=(q^k)^n=(q^n)^k$, so that indeed $P(Sleq k)=1-(q^n)^k$ is the CDF of a Geometric RV with new parameter $tilde{p}=1-q^n=1-(1-p)^n$.



How can we tackle $U$? Well, let's see how that minimum formula was derived in the first place and hope that informs us on a similar manner to derive a formula for the maximum of IIDRVs.



If the minimum of $n$ variables is greater than a number $k$, then all of those variables must be greater than $k$, i.e the following set equalities hold
$${min{T_1,dotsc, T_n} >k}={T_1 >k,dotsc, T_n>k}.$$
That's just from the definition of minimum. Thus, taking probabilities and writing $S=min{S_1,dotsc, S_n}$, we have
$$P(S >k)=P(T_1>k, dotsc, T_n>k),$$
and finally the IID assumption greatly simplifies the RHS further, to
$$P(S>k)=P(T_1>K)P(T_2>k)cdot dotsc cdot P(T_n >k),$$
first using independence, then using the identical distributions, we finally get
$$P(S>k)=P(T_1>k)^n.$$



Can you now, after seeing this argument for minimum, derive the analogous formula for maximums? Please comment for further clarification.



EDIT
If $U=max{T_1,dots, T_n}$, then we know that $P(Uleq k)=P(T_1 leq k)^n=(1-q^{k})^n$ by definition of maximum and since the $T_i$ are IID. To find $mathbb{E}(U)$ we use the alternative formula for expectations for discrete non-negative RVs:
$$E(U)=sum_{k=1}^infty P(Ugeq k).$$
Since $P(Ugeq k)=1-P(Uleq k-1)=1-(1-q^{k-1})^n,$ we must compute the series,
$$sum_{k=1}^infty 1-(1-q^{k-1})^n.$$
In principle this series can always be computed by using the binomial theorem to expand $(1-q^{k-1})^n$ and then use theory of geometric series to compute the resulting series' in closed form. For $n=2$, we have
$$mathbb{E}(U)=frac{3-2p}{p(2-p)}approx 2.667$$
when $p=0.5$, for example.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 18:41

























answered Dec 5 '18 at 0:33









LoveTooNap29

1,0041613




1,0041613












  • Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
    – qcc101
    Dec 5 '18 at 13:38












  • @qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
    – LoveTooNap29
    Dec 5 '18 at 18:47


















  • Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
    – qcc101
    Dec 5 '18 at 13:38












  • @qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
    – LoveTooNap29
    Dec 5 '18 at 18:47
















Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
– qcc101
Dec 5 '18 at 13:38






Super clear. I managed to solve the one with U as well, finding that $f_U(k) = [1-(1-p)^k]^n - [1-(1-p)^{k-1}]^n$ which is indeed the right solution. Let's say I want to find $mathbb{E}[U]$ as well. By linearity, I write: $$ mathbb{E}[U] = mathbb{E}[U+S] - mathbb{E}[S] $$ The expected value of S is easy because it is geometrically distributed. How should I find the expected value of the sum of U and S? (Is this a good procedure to simplify the problem?) Edit: with n = 2
– qcc101
Dec 5 '18 at 13:38














@qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
– LoveTooNap29
Dec 5 '18 at 18:47




@qcc101 Thanks. I have added an edit demonstrating how to compute $mathbb{E}(U)$ with the analytic answer for $n=2$, as an example and its numerical value for when $p=0.5$. Your idea would require knowing the PDF or CDF of $U+S$ which is not immediately known. Better to compute the expected value by definition or become familiar with the alternative formula I cite—extremely useful in the case of non-negative RVs and perusing the Wikipedia link, you will notice an analogous version for continuous non-negative RVs, as well. Hope this helps.
– LoveTooNap29
Dec 5 '18 at 18:47


















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