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Show that expected number of years till first observation is exceeded for first time is infinite given infinite observations
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Let $X_1,X_2...$ be jointly independently random variables with common pdf given by $f(x_i)$. I am trying to show that $E[N]$ is infinite, where $N$ is the number of years till $X_1$ is exceeded for the first time, and $X_i$ represents the annual rainfall in the i-th year.
First I obtained the distribution function for N. Since $F(X_1)$ is the probability of a particular observation being less than $X_1$, it can be shown that: $$p(N=n)=F(X_1)^n-1(1-F(X_1)).$$
Next, I can express the expected value of $N$ as: $$E[N]=sum_1^infty nF(X_1)^n-1(1-F(X_1))=sum_1^infty nF(X_1)^n-1 - sum_1^infty nF(X_1)^n. $$
At this point, I'm not sure how to proceed. Any advice?
probability-theory probability-distributions
asked Jul 22 at 18:45
DavidS
858
 |Â
show 2 more comments
up vote
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Let $X_1,X_2...$ be jointly independently random variables with common pdf given by $f(x_i)$. I am trying to show that $E[N]$ is infinite, where $N$ is the number of years till $X_1$ is exceeded for the first time, and $X_i$ represents the annual rainfall in the i-th year.
First I obtained the distribution function for N. Since $F(X_1)$ is the probability of a particular observation being less than $X_1$, it can be shown that: $$p(N=n)=F(X_1)^n-1(1-F(X_1)).$$
Next, I can express the expected value of $N$ as: $$E[N]=sum_1^infty nF(X_1)^n-1(1-F(X_1))=sum_1^infty nF(X_1)^n-1 - sum_1^infty nF(X_1)^n. $$
At this point, I'm not sure how to proceed. Any advice?
probability-theory probability-distributions
asked Jul 22 at 18:45
DavidS
858
1
$p(N=n)=F(X_1)^n-1(1-F(X_1))$ makes no sense -- the left-hand side is a constant and the right-hand side is a random variable.
– joriki
Jul 22 at 18:59
@DF: No, these events are correlated, since they all involve the same $X_1$.
– joriki
Jul 22 at 19:00
@DF: They are. That doesn't mean that the events $X_igt X_j$ and $X_igt X_k$ are independent.
– joriki
Jul 22 at 19:02
$X_1, X_2...$ share a common pdf but are jointly independent according to the question.
– DavidS
Jul 22 at 19:03
1
@DavidS: This might be $P(N=nmid X_1)$ (I haven't checked that), but it can't be $P(N=n)$; an equation that depends on $X_1$ on one side and not on the other makes no sense.
– joriki
Jul 22 at 19:10
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_1,X_2...$ be jointly independently random variables with common pdf given by $f(x_i)$. I am trying to show that $E[N]$ is infinite, where $N$ is the number of years till $X_1$ is exceeded for the first time, and $X_i$ represents the annual rainfall in the i-th year.
First I obtained the distribution function for N. Since $F(X_1)$ is the probability of a particular observation being less than $X_1$, it can be shown that: $$p(N=n)=F(X_1)^n-1(1-F(X_1)).$$
Next, I can express the expected value of $N$ as: $$E[N]=sum_1^infty nF(X_1)^n-1(1-F(X_1))=sum_1^infty nF(X_1)^n-1 - sum_1^infty nF(X_1)^n. $$
At this point, I'm not sure how to proceed. Any advice?
probability-theory probability-distributions
asked Jul 22 at 18:45
DavidS
858
Let $X_1,X_2...$ be jointly independently random variables with common pdf given by $f(x_i)$. I am trying to show that $E[N]$ is infinite, where $N$ is the number of years till $X_1$ is exceeded for the first time, and $X_i$ represents the annual rainfall in the i-th year.
First I obtained the distribution function for N. Since $F(X_1)$ is the probability of a particular observation being less than $X_1$, it can be shown that: $$p(N=n)=F(X_1)^n-1(1-F(X_1)).$$
Next, I can express the expected value of $N$ as: $$E[N]=sum_1^infty nF(X_1)^n-1(1-F(X_1))=sum_1^infty nF(X_1)^n-1 - sum_1^infty nF(X_1)^n. $$
At this point, I'm not sure how to proceed. Any advice?
probability-theory probability-distributions
asked Jul 22 at 18:45
DavidS
858
edited Jul 22 at 19:05
asked Jul 22 at 18:45
DavidS
858
asked Jul 22 at 18:45
DavidS
858
asked Jul 22 at 18:45
DavidS
858
858
1
$p(N=n)=F(X_1)^n-1(1-F(X_1))$ makes no sense -- the left-hand side is a constant and the right-hand side is a random variable.
– joriki
Jul 22 at 18:59
@DF: No, these events are correlated, since they all involve the same $X_1$.
– joriki
Jul 22 at 19:00
@DF: They are. That doesn't mean that the events $X_igt X_j$ and $X_igt X_k$ are independent.
– joriki
Jul 22 at 19:02
$X_1, X_2...$ share a common pdf but are jointly independent according to the question.
– DavidS
Jul 22 at 19:03
1
@DavidS: This might be $P(N=nmid X_1)$ (I haven't checked that), but it can't be $P(N=n)$; an equation that depends on $X_1$ on one side and not on the other makes no sense.
– joriki
Jul 22 at 19:10
 |Â
show 2 more comments
1
$p(N=n)=F(X_1)^n-1(1-F(X_1))$ makes no sense -- the left-hand side is a constant and the right-hand side is a random variable.
– joriki
Jul 22 at 18:59
@DF: No, these events are correlated, since they all involve the same $X_1$.
– joriki
Jul 22 at 19:00
@DF: They are. That doesn't mean that the events $X_igt X_j$ and $X_igt X_k$ are independent.
– joriki
Jul 22 at 19:02
$X_1, X_2...$ share a common pdf but are jointly independent according to the question.
– DavidS
Jul 22 at 19:03
1
@DavidS: This might be $P(N=nmid X_1)$ (I haven't checked that), but it can't be $P(N=n)$; an equation that depends on $X_1$ on one side and not on the other makes no sense.
– joriki
Jul 22 at 19:10
1
1
$p(N=n)=F(X_1)^n-1(1-F(X_1))$ makes no sense -- the left-hand side is a constant and the right-hand side is a random variable.
– joriki
Jul 22 at 18:59
$p(N=n)=F(X_1)^n-1(1-F(X_1))$ makes no sense -- the left-hand side is a constant and the right-hand side is a random variable.
– joriki
Jul 22 at 18:59
@DF: No, these events are correlated, since they all involve the same $X_1$.
– joriki
Jul 22 at 19:00
@DF: No, these events are correlated, since they all involve the same $X_1$.
– joriki
Jul 22 at 19:00
@DF: They are. That doesn't mean that the events $X_igt X_j$ and $X_igt X_k$ are independent.
– joriki
Jul 22 at 19:02
@DF: They are. That doesn't mean that the events $X_igt X_j$ and $X_igt X_k$ are independent.
– joriki
Jul 22 at 19:02
$X_1, X_2...$ share a common pdf but are jointly independent according to the question.
– DavidS
Jul 22 at 19:03
$X_1, X_2...$ share a common pdf but are jointly independent according to the question.
– DavidS
Jul 22 at 19:03
1
1
@DavidS: This might be $P(N=nmid X_1)$ (I haven't checked that), but it can't be $P(N=n)$; an equation that depends on $X_1$ on one side and not on the other makes no sense.
– joriki
Jul 22 at 19:10
@DavidS: This might be $P(N=nmid X_1)$ (I haven't checked that), but it can't be $P(N=n)$; an equation that depends on $X_1$ on one side and not on the other makes no sense.
– joriki
Jul 22 at 19:10
 |Â
show 2 more comments
1 Answer
1
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1
down vote
accepted
The probability that $Nge n$ is the probability that $X_1$ is the greatest of the first $n$ values, which is $frac1n$. Thus
$$
mathsf E[N]=sum_n=1^inftymathsf P(Nge n)=sum_n=1^inftyfrac1n=lim_ktoinftyH_k=infty;.
$$
answered Jul 22 at 19:07
joriki
164k10180328
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
The probability that $Nge n$ is the probability that $X_1$ is the greatest of the first $n$ values, which is $frac1n$. Thus
$$
mathsf E[N]=sum_n=1^inftymathsf P(Nge n)=sum_n=1^inftyfrac1n=lim_ktoinftyH_k=infty;.
$$
answered Jul 22 at 19:07
joriki
164k10180328
add a comment |Â
up vote
1
down vote
accepted
The probability that $Nge n$ is the probability that $X_1$ is the greatest of the first $n$ values, which is $frac1n$. Thus
$$
mathsf E[N]=sum_n=1^inftymathsf P(Nge n)=sum_n=1^inftyfrac1n=lim_ktoinftyH_k=infty;.
$$
answered Jul 22 at 19:07
joriki
164k10180328
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The probability that $Nge n$ is the probability that $X_1$ is the greatest of the first $n$ values, which is $frac1n$. Thus
$$
mathsf E[N]=sum_n=1^inftymathsf P(Nge n)=sum_n=1^inftyfrac1n=lim_ktoinftyH_k=infty;.
$$
answered Jul 22 at 19:07
joriki
164k10180328
The probability that $Nge n$ is the probability that $X_1$ is the greatest of the first $n$ values, which is $frac1n$. Thus
$$
mathsf E[N]=sum_n=1^inftymathsf P(Nge n)=sum_n=1^inftyfrac1n=lim_ktoinftyH_k=infty;.
$$
answered Jul 22 at 19:07
joriki
164k10180328
edited Jul 22 at 19:14
answered Jul 22 at 19:07
joriki
164k10180328
answered Jul 22 at 19:07
joriki
164k10180328
answered Jul 22 at 19:07
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
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1
$p(N=n)=F(X_1)^n-1(1-F(X_1))$ makes no sense -- the left-hand side is a constant and the right-hand side is a random variable.
– joriki
Jul 22 at 18:59
@DF: No, these events are correlated, since they all involve the same $X_1$.
– joriki
Jul 22 at 19:00
@DF: They are. That doesn't mean that the events $X_igt X_j$ and $X_igt X_k$ are independent.
– joriki
Jul 22 at 19:02
$X_1, X_2...$ share a common pdf but are jointly independent according to the question.
– DavidS
Jul 22 at 19:03
1
@DavidS: This might be $P(N=nmid X_1)$ (I haven't checked that), but it can't be $P(N=n)$; an equation that depends on $X_1$ on one side and not on the other makes no sense.
– joriki
Jul 22 at 19:10