If $bar X_nleq X_n$ for every $n$ then the associated counting processes are such that $bar N(t) geq N(t)$...












-3












$begingroup$



${X_n, ngeq 1}$ is a renewal process. Let, there exists an $alpha>0$ such that $P(X_ngeq alpha)>0$. Now, define another related renewal process, ${bar X_n, ngeq 1}$ by,
begin{equation}
bar X_n=
begin{cases}
0, & text{if} X_n <alpha \
alpha, & text{otherwise}
end{cases}
end{equation}
For, $nge 1$, let $S_n = X_1+dots + X_n$ and $bar S_n = bar X_1+dots + bar X_n$. Also, let, $N(t) = max {nmid S_n leq t}$ and $bar N(t) = max {nmid bar S_n leq t}$



Show that, $bar N(t) geq N(t)$.




What I did: It is easy to see that $bar X_n leq X_n$, so $bar S_nleq S_n$. We also know that ${N(t)geq n}Leftrightarrow{S_nleq t}$.






probability-theory stochastic-processes







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  • 1




    $begingroup$
    Well, $bar S_nleq S_n$ hence ${N(t)geq n}={S_nleq t}subseteq{bar S_nleq t}={bar N(t)geq n}$. And if ${N(t)geq n}subseteq{bar N(t)geq n}$ for every $n$, then $bar N(t)ge N(t)$ almost surely.
    $endgroup$
    – Did
    Dec 17 '18 at 9:15


















-3












$begingroup$



${X_n, ngeq 1}$ is a renewal process. Let, there exists an $alpha>0$ such that $P(X_ngeq alpha)>0$. Now, define another related renewal process, ${bar X_n, ngeq 1}$ by,
begin{equation}
bar X_n=
begin{cases}
0, & text{if} X_n <alpha \
alpha, & text{otherwise}
end{cases}
end{equation}
For, $nge 1$, let $S_n = X_1+dots + X_n$ and $bar S_n = bar X_1+dots + bar X_n$. Also, let, $N(t) = max {nmid S_n leq t}$ and $bar N(t) = max {nmid bar S_n leq t}$



Show that, $bar N(t) geq N(t)$.




What I did: It is easy to see that $bar X_n leq X_n$, so $bar S_nleq S_n$. We also know that ${N(t)geq n}Leftrightarrow{S_nleq t}$.






probability-theory stochastic-processes







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, $bar S_nleq S_n$ hence ${N(t)geq n}={S_nleq t}subseteq{bar S_nleq t}={bar N(t)geq n}$. And if ${N(t)geq n}subseteq{bar N(t)geq n}$ for every $n$, then $bar N(t)ge N(t)$ almost surely.
    $endgroup$
    – Did
    Dec 17 '18 at 9:15
















-3












-3








-3





$begingroup$



${X_n, ngeq 1}$ is a renewal process. Let, there exists an $alpha>0$ such that $P(X_ngeq alpha)>0$. Now, define another related renewal process, ${bar X_n, ngeq 1}$ by,
begin{equation}
bar X_n=
begin{cases}
0, & text{if} X_n <alpha \
alpha, & text{otherwise}
end{cases}
end{equation}
For, $nge 1$, let $S_n = X_1+dots + X_n$ and $bar S_n = bar X_1+dots + bar X_n$. Also, let, $N(t) = max {nmid S_n leq t}$ and $bar N(t) = max {nmid bar S_n leq t}$



Show that, $bar N(t) geq N(t)$.




What I did: It is easy to see that $bar X_n leq X_n$, so $bar S_nleq S_n$. We also know that ${N(t)geq n}Leftrightarrow{S_nleq t}$.






probability-theory stochastic-processes







share|cite|improve this question











$endgroup$





${X_n, ngeq 1}$ is a renewal process. Let, there exists an $alpha>0$ such that $P(X_ngeq alpha)>0$. Now, define another related renewal process, ${bar X_n, ngeq 1}$ by,
begin{equation}
bar X_n=
begin{cases}
0, & text{if} X_n <alpha \
alpha, & text{otherwise}
end{cases}
end{equation}
For, $nge 1$, let $S_n = X_1+dots + X_n$ and $bar S_n = bar X_1+dots + bar X_n$. Also, let, $N(t) = max {nmid S_n leq t}$ and $bar N(t) = max {nmid bar S_n leq t}$



Show that, $bar N(t) geq N(t)$.




What I did: It is easy to see that $bar X_n leq X_n$, so $bar S_nleq S_n$. We also know that ${N(t)geq n}Leftrightarrow{S_nleq t}$.






probability-theory stochastic-processes




probability-theory stochastic-processes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 9:17









Did

248k23225463




248k23225463










asked Dec 17 '18 at 8:11









Stat_prob_001Stat_prob_001

321113




321113








  • 1




    $begingroup$
    Well, $bar S_nleq S_n$ hence ${N(t)geq n}={S_nleq t}subseteq{bar S_nleq t}={bar N(t)geq n}$. And if ${N(t)geq n}subseteq{bar N(t)geq n}$ for every $n$, then $bar N(t)ge N(t)$ almost surely.
    $endgroup$
    – Did
    Dec 17 '18 at 9:15
















  • 1




    $begingroup$
    Well, $bar S_nleq S_n$ hence ${N(t)geq n}={S_nleq t}subseteq{bar S_nleq t}={bar N(t)geq n}$. And if ${N(t)geq n}subseteq{bar N(t)geq n}$ for every $n$, then $bar N(t)ge N(t)$ almost surely.
    $endgroup$
    – Did
    Dec 17 '18 at 9:15










1




1




$begingroup$
Well, $bar S_nleq S_n$ hence ${N(t)geq n}={S_nleq t}subseteq{bar S_nleq t}={bar N(t)geq n}$. And if ${N(t)geq n}subseteq{bar N(t)geq n}$ for every $n$, then $bar N(t)ge N(t)$ almost surely.
$endgroup$
– Did
Dec 17 '18 at 9:15






$begingroup$
Well, $bar S_nleq S_n$ hence ${N(t)geq n}={S_nleq t}subseteq{bar S_nleq t}={bar N(t)geq n}$. And if ${N(t)geq n}subseteq{bar N(t)geq n}$ for every $n$, then $bar N(t)ge N(t)$ almost surely.
$endgroup$
– Did
Dec 17 '18 at 9:15












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$begingroup$

The braces ${$ and $}$ in your last statement should not be there. Then we get:$$N(t)geq niff S_nleq t$$as it should.



Combining this with $overline S_nleq S_n$ we find:$$N(t)geq nimplies overline S_nleq t$$or - on base of $overline N(t)geq niff overline S_nleq t$ - equivalently:$$N(t)geq nimplies overline N(t)geq ntag1$$



This for every $n$, so it implies that: $$overline N(t)geq N(t)tag2$$




Formally pick out an arbitrary $omegainOmega$ an let it be that $n=N(t)(omega)$. Then $N(t)(omega)geq n$ so according to $(1)$ we have $overline N(t)(omega)geq n=N(t)(omega)$. This proves that $overline N(t)(omega)geq N(t)(omega)$ for every $omegainOmega$ or shortly that $(2)$ is valid.






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    $begingroup$

    The braces ${$ and $}$ in your last statement should not be there. Then we get:$$N(t)geq niff S_nleq t$$as it should.



    Combining this with $overline S_nleq S_n$ we find:$$N(t)geq nimplies overline S_nleq t$$or - on base of $overline N(t)geq niff overline S_nleq t$ - equivalently:$$N(t)geq nimplies overline N(t)geq ntag1$$



    This for every $n$, so it implies that: $$overline N(t)geq N(t)tag2$$




    Formally pick out an arbitrary $omegainOmega$ an let it be that $n=N(t)(omega)$. Then $N(t)(omega)geq n$ so according to $(1)$ we have $overline N(t)(omega)geq n=N(t)(omega)$. This proves that $overline N(t)(omega)geq N(t)(omega)$ for every $omegainOmega$ or shortly that $(2)$ is valid.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The braces ${$ and $}$ in your last statement should not be there. Then we get:$$N(t)geq niff S_nleq t$$as it should.



      Combining this with $overline S_nleq S_n$ we find:$$N(t)geq nimplies overline S_nleq t$$or - on base of $overline N(t)geq niff overline S_nleq t$ - equivalently:$$N(t)geq nimplies overline N(t)geq ntag1$$



      This for every $n$, so it implies that: $$overline N(t)geq N(t)tag2$$




      Formally pick out an arbitrary $omegainOmega$ an let it be that $n=N(t)(omega)$. Then $N(t)(omega)geq n$ so according to $(1)$ we have $overline N(t)(omega)geq n=N(t)(omega)$. This proves that $overline N(t)(omega)geq N(t)(omega)$ for every $omegainOmega$ or shortly that $(2)$ is valid.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The braces ${$ and $}$ in your last statement should not be there. Then we get:$$N(t)geq niff S_nleq t$$as it should.



        Combining this with $overline S_nleq S_n$ we find:$$N(t)geq nimplies overline S_nleq t$$or - on base of $overline N(t)geq niff overline S_nleq t$ - equivalently:$$N(t)geq nimplies overline N(t)geq ntag1$$



        This for every $n$, so it implies that: $$overline N(t)geq N(t)tag2$$




        Formally pick out an arbitrary $omegainOmega$ an let it be that $n=N(t)(omega)$. Then $N(t)(omega)geq n$ so according to $(1)$ we have $overline N(t)(omega)geq n=N(t)(omega)$. This proves that $overline N(t)(omega)geq N(t)(omega)$ for every $omegainOmega$ or shortly that $(2)$ is valid.






        share|cite|improve this answer











        $endgroup$



        The braces ${$ and $}$ in your last statement should not be there. Then we get:$$N(t)geq niff S_nleq t$$as it should.



        Combining this with $overline S_nleq S_n$ we find:$$N(t)geq nimplies overline S_nleq t$$or - on base of $overline N(t)geq niff overline S_nleq t$ - equivalently:$$N(t)geq nimplies overline N(t)geq ntag1$$



        This for every $n$, so it implies that: $$overline N(t)geq N(t)tag2$$




        Formally pick out an arbitrary $omegainOmega$ an let it be that $n=N(t)(omega)$. Then $N(t)(omega)geq n$ so according to $(1)$ we have $overline N(t)(omega)geq n=N(t)(omega)$. This proves that $overline N(t)(omega)geq N(t)(omega)$ for every $omegainOmega$ or shortly that $(2)$ is valid.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 10:40

























        answered Dec 17 '18 at 9:23









        drhabdrhab

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