Proving $mathbb{P}(sum_{n=1}^{infty} S_n = infty) = 1$ where $S_i sim exp(lambda_i)$ and $sup_n lambda_n <...
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I was working through J.R. Norris Markov Chains and got stuck at exercise 2.3.1. The question is the following:
Let $S_1, S_2, ldots$ be independent exponential random variables of parameters $lambda_1, lambda_2, ldots$ respectively. Show that $lambda_1 S_1$ is exponential with parameter $1$.
Use the strong law of large numbers to show, first in the special case $lambda_n = 1$ for all $n$, and then subject only to the condition $sup_n lambda_n < infty$, that
$$
mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$$
Is the condition $sup_n lambda_n < infty$.
I was able to prove that $lambda_i S_i sim exp(1)$, but I was not able to prove that $mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$, even for the special case $lambda_n = 1 , forall n$.
Intuitively I have concluded the following: Since $mathbb{E}(S_n) = frac{1}{lambda_n}$ and $sup_n lambda_n < infty$, we can conclude that
$$
mathbb{E}(sum_{n=1}^{infty} S_n) = sum_{n=1}^{infty} mathbb{E}(S_n) = sum_{n=1}^{infty} frac{1}{lambda_n} = infty,
$$
since $sup_n lambda_n < infty$.
Is there any way to build on this intuition? How do I exactly use the strong law of large numbers to solve the exercise?
Thanks in advance!
probability random-variables exponential-distribution
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I was working through J.R. Norris Markov Chains and got stuck at exercise 2.3.1. The question is the following:
Let $S_1, S_2, ldots$ be independent exponential random variables of parameters $lambda_1, lambda_2, ldots$ respectively. Show that $lambda_1 S_1$ is exponential with parameter $1$.
Use the strong law of large numbers to show, first in the special case $lambda_n = 1$ for all $n$, and then subject only to the condition $sup_n lambda_n < infty$, that
$$
mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$$
Is the condition $sup_n lambda_n < infty$.
I was able to prove that $lambda_i S_i sim exp(1)$, but I was not able to prove that $mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$, even for the special case $lambda_n = 1 , forall n$.
Intuitively I have concluded the following: Since $mathbb{E}(S_n) = frac{1}{lambda_n}$ and $sup_n lambda_n < infty$, we can conclude that
$$
mathbb{E}(sum_{n=1}^{infty} S_n) = sum_{n=1}^{infty} mathbb{E}(S_n) = sum_{n=1}^{infty} frac{1}{lambda_n} = infty,
$$
since $sup_n lambda_n < infty$.
Is there any way to build on this intuition? How do I exactly use the strong law of large numbers to solve the exercise?
Thanks in advance!
probability random-variables exponential-distribution
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was working through J.R. Norris Markov Chains and got stuck at exercise 2.3.1. The question is the following:
Let $S_1, S_2, ldots$ be independent exponential random variables of parameters $lambda_1, lambda_2, ldots$ respectively. Show that $lambda_1 S_1$ is exponential with parameter $1$.
Use the strong law of large numbers to show, first in the special case $lambda_n = 1$ for all $n$, and then subject only to the condition $sup_n lambda_n < infty$, that
$$
mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$$
Is the condition $sup_n lambda_n < infty$.
I was able to prove that $lambda_i S_i sim exp(1)$, but I was not able to prove that $mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$, even for the special case $lambda_n = 1 , forall n$.
Intuitively I have concluded the following: Since $mathbb{E}(S_n) = frac{1}{lambda_n}$ and $sup_n lambda_n < infty$, we can conclude that
$$
mathbb{E}(sum_{n=1}^{infty} S_n) = sum_{n=1}^{infty} mathbb{E}(S_n) = sum_{n=1}^{infty} frac{1}{lambda_n} = infty,
$$
since $sup_n lambda_n < infty$.
Is there any way to build on this intuition? How do I exactly use the strong law of large numbers to solve the exercise?
Thanks in advance!
probability random-variables exponential-distribution
I was working through J.R. Norris Markov Chains and got stuck at exercise 2.3.1. The question is the following:
Let $S_1, S_2, ldots$ be independent exponential random variables of parameters $lambda_1, lambda_2, ldots$ respectively. Show that $lambda_1 S_1$ is exponential with parameter $1$.
Use the strong law of large numbers to show, first in the special case $lambda_n = 1$ for all $n$, and then subject only to the condition $sup_n lambda_n < infty$, that
$$
mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$$
Is the condition $sup_n lambda_n < infty$.
I was able to prove that $lambda_i S_i sim exp(1)$, but I was not able to prove that $mathbb{P}left(sum_{n=1}^{infty} S_n = inftyright) = 1
$, even for the special case $lambda_n = 1 , forall n$.
Intuitively I have concluded the following: Since $mathbb{E}(S_n) = frac{1}{lambda_n}$ and $sup_n lambda_n < infty$, we can conclude that
$$
mathbb{E}(sum_{n=1}^{infty} S_n) = sum_{n=1}^{infty} mathbb{E}(S_n) = sum_{n=1}^{infty} frac{1}{lambda_n} = infty,
$$
since $sup_n lambda_n < infty$.
Is there any way to build on this intuition? How do I exactly use the strong law of large numbers to solve the exercise?
Thanks in advance!
probability random-variables exponential-distribution
probability random-variables exponential-distribution
edited Nov 14 at 18:55
user159517
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4,338929
asked Nov 14 at 18:33
Sliem el Ela
11511
11511
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First let $lambda_n = 1$. Then, by the strong law of large numbers, $$frac{1}{N} sum_{n=1}^{N} S_n to 1$$ almost surely. In particular, the sequence of partial sums $(sum_{n=1}^{N} S_n)_{Ngeq 1}$ is almost surely unbounded (otherwise the limit would be $0$ on a set of nonzero probability).
Now assume $Lambda:=sup_n lambda_n < infty$. By what we've just shown and because $lambda_i S_i sim exp(1)$, we know that $$mathbb{P}left(sum_{n=1}^{infty}lambda_n S_n = inftyright) = 1.$$
But if $sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, then also $frac{1}{Lambda}sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, and
$$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First let $lambda_n = 1$. Then, by the strong law of large numbers, $$frac{1}{N} sum_{n=1}^{N} S_n to 1$$ almost surely. In particular, the sequence of partial sums $(sum_{n=1}^{N} S_n)_{Ngeq 1}$ is almost surely unbounded (otherwise the limit would be $0$ on a set of nonzero probability).
Now assume $Lambda:=sup_n lambda_n < infty$. By what we've just shown and because $lambda_i S_i sim exp(1)$, we know that $$mathbb{P}left(sum_{n=1}^{infty}lambda_n S_n = inftyright) = 1.$$
But if $sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, then also $frac{1}{Lambda}sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, and
$$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$
add a comment |
up vote
1
down vote
accepted
First let $lambda_n = 1$. Then, by the strong law of large numbers, $$frac{1}{N} sum_{n=1}^{N} S_n to 1$$ almost surely. In particular, the sequence of partial sums $(sum_{n=1}^{N} S_n)_{Ngeq 1}$ is almost surely unbounded (otherwise the limit would be $0$ on a set of nonzero probability).
Now assume $Lambda:=sup_n lambda_n < infty$. By what we've just shown and because $lambda_i S_i sim exp(1)$, we know that $$mathbb{P}left(sum_{n=1}^{infty}lambda_n S_n = inftyright) = 1.$$
But if $sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, then also $frac{1}{Lambda}sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, and
$$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First let $lambda_n = 1$. Then, by the strong law of large numbers, $$frac{1}{N} sum_{n=1}^{N} S_n to 1$$ almost surely. In particular, the sequence of partial sums $(sum_{n=1}^{N} S_n)_{Ngeq 1}$ is almost surely unbounded (otherwise the limit would be $0$ on a set of nonzero probability).
Now assume $Lambda:=sup_n lambda_n < infty$. By what we've just shown and because $lambda_i S_i sim exp(1)$, we know that $$mathbb{P}left(sum_{n=1}^{infty}lambda_n S_n = inftyright) = 1.$$
But if $sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, then also $frac{1}{Lambda}sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, and
$$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$
First let $lambda_n = 1$. Then, by the strong law of large numbers, $$frac{1}{N} sum_{n=1}^{N} S_n to 1$$ almost surely. In particular, the sequence of partial sums $(sum_{n=1}^{N} S_n)_{Ngeq 1}$ is almost surely unbounded (otherwise the limit would be $0$ on a set of nonzero probability).
Now assume $Lambda:=sup_n lambda_n < infty$. By what we've just shown and because $lambda_i S_i sim exp(1)$, we know that $$mathbb{P}left(sum_{n=1}^{infty}lambda_n S_n = inftyright) = 1.$$
But if $sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, then also $frac{1}{Lambda}sum_{n=1}^{infty}lambda_n S_n (omega) = infty$, and
$$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$
answered Nov 14 at 18:54
user159517
4,338929
4,338929
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