Calculating necessary assumptions on simple Poisson process
I'm reading a textbook on probability theory, and the author has recently introduced the Poisson process. We're counting occurrences in an interval of time; this quantity is a random variable for each time interval. One of the requirements is that this counting has no double points: at any point in time, the counter goes off at most once.
To formalize these requirements, we let $mathcal I = {(a,b] : a, b in [0, infty), a leq b}$, and let $N_I$ be the number of counts in the interval $I in mathcal I$, defining $N_t = N_{(0,t]}$ for $t > 0$. So the Poisson process satisfies:
$N_{I cup J} = N_I + N_J$ for disjoint $I, J$;
$mathbb P_{N_I} = mathbb P_{N_J}$ for $I, J$ of the same length;
$(N_J , J in mathcal J)$ is an independent family of random variables for $mathcal J subset mathcal I$ with $I cap J = emptyset$ $forall I, J in mathcal J$;
$mathbb E[N_I] < infty$ for all $I in mathcal I$;
$limsup_{epsilon to 0} epsilon^{-1}mathbb P[N_epsilon geq 2] = 0$.
This last requirement ensures we have no double points. Proving we have no double points from 5 is a mostly straightforward calculation, but the author makes the following claim:
For any $n in mathbb N$ and $epsilon > 0$, we have $$mathbb P[N_{2^n} geq 2] geq leftlfloor 2^{-n}/epsilon rightrfloor mathbb P[N_{epsilon} geq 2] - leftlfloor 2^{-n}/epsilonrightrfloor^2mathbb P[N_{epsilon} geq 2]^2.$$
I'm trying to figure out where the author is getting this estimate from, but I'm having trouble. I have a feeling there's something very simple I'm overlooking. Any suggestions?
probability probability-theory poisson-distribution poisson-process
asked Nov 28 '18 at 15:49
D Ford
549213
add a comment |
I'm reading a textbook on probability theory, and the author has recently introduced the Poisson process. We're counting occurrences in an interval of time; this quantity is a random variable for each time interval. One of the requirements is that this counting has no double points: at any point in time, the counter goes off at most once.
To formalize these requirements, we let $mathcal I = {(a,b] : a, b in [0, infty), a leq b}$, and let $N_I$ be the number of counts in the interval $I in mathcal I$, defining $N_t = N_{(0,t]}$ for $t > 0$. So the Poisson process satisfies:
$N_{I cup J} = N_I + N_J$ for disjoint $I, J$;
$mathbb P_{N_I} = mathbb P_{N_J}$ for $I, J$ of the same length;
$(N_J , J in mathcal J)$ is an independent family of random variables for $mathcal J subset mathcal I$ with $I cap J = emptyset$ $forall I, J in mathcal J$;
$mathbb E[N_I] < infty$ for all $I in mathcal I$;
$limsup_{epsilon to 0} epsilon^{-1}mathbb P[N_epsilon geq 2] = 0$.
This last requirement ensures we have no double points. Proving we have no double points from 5 is a mostly straightforward calculation, but the author makes the following claim:
For any $n in mathbb N$ and $epsilon > 0$, we have $$mathbb P[N_{2^n} geq 2] geq leftlfloor 2^{-n}/epsilon rightrfloor mathbb P[N_{epsilon} geq 2] - leftlfloor 2^{-n}/epsilonrightrfloor^2mathbb P[N_{epsilon} geq 2]^2.$$
I'm trying to figure out where the author is getting this estimate from, but I'm having trouble. I have a feeling there's something very simple I'm overlooking. Any suggestions?
probability probability-theory poisson-distribution poisson-process
asked Nov 28 '18 at 15:49
D Ford
549213
add a comment |
I'm reading a textbook on probability theory, and the author has recently introduced the Poisson process. We're counting occurrences in an interval of time; this quantity is a random variable for each time interval. One of the requirements is that this counting has no double points: at any point in time, the counter goes off at most once.
To formalize these requirements, we let $mathcal I = {(a,b] : a, b in [0, infty), a leq b}$, and let $N_I$ be the number of counts in the interval $I in mathcal I$, defining $N_t = N_{(0,t]}$ for $t > 0$. So the Poisson process satisfies:
$N_{I cup J} = N_I + N_J$ for disjoint $I, J$;
$mathbb P_{N_I} = mathbb P_{N_J}$ for $I, J$ of the same length;
$(N_J , J in mathcal J)$ is an independent family of random variables for $mathcal J subset mathcal I$ with $I cap J = emptyset$ $forall I, J in mathcal J$;
$mathbb E[N_I] < infty$ for all $I in mathcal I$;
$limsup_{epsilon to 0} epsilon^{-1}mathbb P[N_epsilon geq 2] = 0$.
This last requirement ensures we have no double points. Proving we have no double points from 5 is a mostly straightforward calculation, but the author makes the following claim:
For any $n in mathbb N$ and $epsilon > 0$, we have $$mathbb P[N_{2^n} geq 2] geq leftlfloor 2^{-n}/epsilon rightrfloor mathbb P[N_{epsilon} geq 2] - leftlfloor 2^{-n}/epsilonrightrfloor^2mathbb P[N_{epsilon} geq 2]^2.$$
I'm trying to figure out where the author is getting this estimate from, but I'm having trouble. I have a feeling there's something very simple I'm overlooking. Any suggestions?
probability probability-theory poisson-distribution poisson-process
asked Nov 28 '18 at 15:49
D Ford
549213
I'm reading a textbook on probability theory, and the author has recently introduced the Poisson process. We're counting occurrences in an interval of time; this quantity is a random variable for each time interval. One of the requirements is that this counting has no double points: at any point in time, the counter goes off at most once.
To formalize these requirements, we let $mathcal I = {(a,b] : a, b in [0, infty), a leq b}$, and let $N_I$ be the number of counts in the interval $I in mathcal I$, defining $N_t = N_{(0,t]}$ for $t > 0$. So the Poisson process satisfies:
$N_{I cup J} = N_I + N_J$ for disjoint $I, J$;
$mathbb P_{N_I} = mathbb P_{N_J}$ for $I, J$ of the same length;
$(N_J , J in mathcal J)$ is an independent family of random variables for $mathcal J subset mathcal I$ with $I cap J = emptyset$ $forall I, J in mathcal J$;
$mathbb E[N_I] < infty$ for all $I in mathcal I$;
$limsup_{epsilon to 0} epsilon^{-1}mathbb P[N_epsilon geq 2] = 0$.
This last requirement ensures we have no double points. Proving we have no double points from 5 is a mostly straightforward calculation, but the author makes the following claim:
For any $n in mathbb N$ and $epsilon > 0$, we have $$mathbb P[N_{2^n} geq 2] geq leftlfloor 2^{-n}/epsilon rightrfloor mathbb P[N_{epsilon} geq 2] - leftlfloor 2^{-n}/epsilonrightrfloor^2mathbb P[N_{epsilon} geq 2]^2.$$
I'm trying to figure out where the author is getting this estimate from, but I'm having trouble. I have a feeling there's something very simple I'm overlooking. Any suggestions?
probability probability-theory poisson-distribution poisson-process
probability probability-theory poisson-distribution poisson-process
asked Nov 28 '18 at 15:49
D Ford
549213
asked Nov 28 '18 at 15:49
D Ford
549213
asked Nov 28 '18 at 15:49
D Ford
549213
asked Nov 28 '18 at 15:49
D Ford
549213
asked Nov 28 '18 at 15:49
D Ford
549213
549213
add a comment |
add a comment |
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