What is actually being asked here? (Measure theoretic probability)
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I am reading a probability text that asks the following question.
If $X_1, X_2, dots$ are independent $mathrm{Ber}(p)$ random variables, where $0 < p < 1$, then define $T : min{k : X_k = 1}$, and give a complete, measure theoretic proof of the fact that $T sim mathrm{Geo}(p)$.
I guess I don't really understand what this means. First of all, what even is the underlying probability space here? It seems that $T$ is taking integer values and $X$ is taking $0, 1$ values.
probability-theory measure-theory
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
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add a comment |
$begingroup$
I am reading a probability text that asks the following question.
If $X_1, X_2, dots$ are independent $mathrm{Ber}(p)$ random variables, where $0 < p < 1$, then define $T : min{k : X_k = 1}$, and give a complete, measure theoretic proof of the fact that $T sim mathrm{Geo}(p)$.
I guess I don't really understand what this means. First of all, what even is the underlying probability space here? It seems that $T$ is taking integer values and $X$ is taking $0, 1$ values.
probability-theory measure-theory
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
$endgroup$
$begingroup$
Forget about measure theory for the moment and think about the event T <= k from a set-theoretic point of view.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 23:00
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This is saying that T is indicating the first success. The geometric distribution models probability to the first success.
$endgroup$
– Sean Roberson
Dec 8 '18 at 23:13
add a comment |
$begingroup$
I am reading a probability text that asks the following question.
If $X_1, X_2, dots$ are independent $mathrm{Ber}(p)$ random variables, where $0 < p < 1$, then define $T : min{k : X_k = 1}$, and give a complete, measure theoretic proof of the fact that $T sim mathrm{Geo}(p)$.
I guess I don't really understand what this means. First of all, what even is the underlying probability space here? It seems that $T$ is taking integer values and $X$ is taking $0, 1$ values.
probability-theory measure-theory
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
$endgroup$
I am reading a probability text that asks the following question.
If $X_1, X_2, dots$ are independent $mathrm{Ber}(p)$ random variables, where $0 < p < 1$, then define $T : min{k : X_k = 1}$, and give a complete, measure theoretic proof of the fact that $T sim mathrm{Geo}(p)$.
I guess I don't really understand what this means. First of all, what even is the underlying probability space here? It seems that $T$ is taking integer values and $X$ is taking $0, 1$ values.
probability-theory measure-theory
probability-theory measure-theory
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 22:56
Drew BradyDrew Brady
719315
719315
$begingroup$
Forget about measure theory for the moment and think about the event T <= k from a set-theoretic point of view.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 23:00
$begingroup$
This is saying that T is indicating the first success. The geometric distribution models probability to the first success.
$endgroup$
– Sean Roberson
Dec 8 '18 at 23:13
add a comment |
$begingroup$
Forget about measure theory for the moment and think about the event T <= k from a set-theoretic point of view.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 23:00
$begingroup$
This is saying that T is indicating the first success. The geometric distribution models probability to the first success.
$endgroup$
– Sean Roberson
Dec 8 '18 at 23:13
$begingroup$
Forget about measure theory for the moment and think about the event T <= k from a set-theoretic point of view.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 23:00
$begingroup$
Forget about measure theory for the moment and think about the event T <= k from a set-theoretic point of view.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 23:00
$begingroup$
This is saying that T is indicating the first success. The geometric distribution models probability to the first success.
$endgroup$
– Sean Roberson
Dec 8 '18 at 23:13
$begingroup$
This is saying that T is indicating the first success. The geometric distribution models probability to the first success.
$endgroup$
– Sean Roberson
Dec 8 '18 at 23:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It can be a little confusing at first that in probability, we often don't mention the probability space $Omega$ on which random variables are defined. This is because $Omega$ often plays a secondary role in what we are interested in, and the idea is that any "suitable" $Omega$ would do. So here, you are supposed to just assume that $X_1,X_2,...$ are an infinite sequence of random variables all defined on the same probability space $Omega$. You then want to prove that the function $T: Omega to mathbb{N}$ defined in terms of the $X_i$ is also a measurable function with the desired distribution.
(I should say that the probability space $Omega$ also comes with a probability measure $mathbb{P}$ and a sigma algebra $mathcal{F}$.)
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
It can be a little confusing at first that in probability, we often don't mention the probability space $Omega$ on which random variables are defined. This is because $Omega$ often plays a secondary role in what we are interested in, and the idea is that any "suitable" $Omega$ would do. So here, you are supposed to just assume that $X_1,X_2,...$ are an infinite sequence of random variables all defined on the same probability space $Omega$. You then want to prove that the function $T: Omega to mathbb{N}$ defined in terms of the $X_i$ is also a measurable function with the desired distribution.
(I should say that the probability space $Omega$ also comes with a probability measure $mathbb{P}$ and a sigma algebra $mathcal{F}$.)
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
$endgroup$
add a comment |
$begingroup$
It can be a little confusing at first that in probability, we often don't mention the probability space $Omega$ on which random variables are defined. This is because $Omega$ often plays a secondary role in what we are interested in, and the idea is that any "suitable" $Omega$ would do. So here, you are supposed to just assume that $X_1,X_2,...$ are an infinite sequence of random variables all defined on the same probability space $Omega$. You then want to prove that the function $T: Omega to mathbb{N}$ defined in terms of the $X_i$ is also a measurable function with the desired distribution.
(I should say that the probability space $Omega$ also comes with a probability measure $mathbb{P}$ and a sigma algebra $mathcal{F}$.)
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
$endgroup$
add a comment |
$begingroup$
It can be a little confusing at first that in probability, we often don't mention the probability space $Omega$ on which random variables are defined. This is because $Omega$ often plays a secondary role in what we are interested in, and the idea is that any "suitable" $Omega$ would do. So here, you are supposed to just assume that $X_1,X_2,...$ are an infinite sequence of random variables all defined on the same probability space $Omega$. You then want to prove that the function $T: Omega to mathbb{N}$ defined in terms of the $X_i$ is also a measurable function with the desired distribution.
(I should say that the probability space $Omega$ also comes with a probability measure $mathbb{P}$ and a sigma algebra $mathcal{F}$.)
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
$endgroup$
It can be a little confusing at first that in probability, we often don't mention the probability space $Omega$ on which random variables are defined. This is because $Omega$ often plays a secondary role in what we are interested in, and the idea is that any "suitable" $Omega$ would do. So here, you are supposed to just assume that $X_1,X_2,...$ are an infinite sequence of random variables all defined on the same probability space $Omega$. You then want to prove that the function $T: Omega to mathbb{N}$ defined in terms of the $X_i$ is also a measurable function with the desired distribution.
(I should say that the probability space $Omega$ also comes with a probability measure $mathbb{P}$ and a sigma algebra $mathcal{F}$.)
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
edited Dec 8 '18 at 23:08
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
answered Dec 8 '18 at 23:02
zoidbergzoidberg
1,070113
1,070113
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$begingroup$
Forget about measure theory for the moment and think about the event T <= k from a set-theoretic point of view.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 23:00
$begingroup$
This is saying that T is indicating the first success. The geometric distribution models probability to the first success.
$endgroup$
– Sean Roberson
Dec 8 '18 at 23:13