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!










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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!










    share|cite|improve this question


























      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!










      share|cite|improve this question















      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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      edited Nov 14 at 18:55









      user159517

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      asked Nov 14 at 18:33









      Sliem el Ela

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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



          $$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$






          share|cite|improve this answer





















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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



            $$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$






            share|cite|improve this answer

























              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



              $$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$






              share|cite|improve this answer























                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



                $$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$






                share|cite|improve this answer












                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



                $$infty = sum_{n=1}^{infty}frac{lambda_n}{Lambda} S_n leq sum_{n=1}^{infty} S_n .$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 14 at 18:54









                user159517

                4,338929




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