Mixing
Gärtner-Ellis theorem on Markov chains
Clash Royale CLAN TAG#URR8PPP
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Let $Z_n in mathcalX$ be a sequence of independent random variables where $mathcalX$ is a topological vector space and let $mu_n$ the probability measures associated with $Z_n$. Suppose that $Z_n$ that satisfies the conditions for the abstract Gärtner-Ellis Theorem. So the rate function $I(cdot)$ associated with $Z_n$ is the Fenchel–Legendre transform of $M(lambda) =lim_n rightarrow infty frac1n log E[exp langle nlambda, Z_n rangle ]$.
Now consider a second random variable $Y_n$ that is a Markov chain ($Y_n$ is sampled from $Y_n-1$) and the probability measure associated with $Y_n$ is $mu_n$.
My question is:
Is $I(cdot)$ the rate function associated with $Y_n$?
If answer is yes to the above question, let me provide a counter-example of the equivalence of the rate function of $Z_n$ and $Y_n$.
Consider two random variables on $0,1$
let $g_n$ be 1 with probability $1/2$, else 0
let $h_1$ = $g_1$ and then for $n>1$, let $h_n=1-h_n-1$ with probability
$(1/2)^n$ and else $h_n=h_n-1$.
Note that for all $n$, $h_n$ and $g_n$ have the same probability distribution,
$Prob(g_n = 1) = 1/2$ (by definition).
$$
Prob(h_n = 1) = Prob(h_n-1=1)cdot(1/2)^n + Prob(h_n-1=0)cdot(1-(1/2)^n) = (1/2) cdot (1/2)^n+(1/2) cdot (1-(1/2)^n) = (1/2)cdot((1/2)^n + 1-(1/2)^n) = 1/2
$$
Hence $Prob(g_n=1) = Prob(h_n=1)$.
At the limit both random variables have the same distribution,
but the convergence of the sequences is rather different. ($h_n$
converges, $g_n$ doesn't)
My other question is: Does this example invalidate the positive answer to first question?
probability-theory large-deviation-theory
asked Jul 27 at 6:01
jaogye
405413
 |Â
show 4 more comments
up vote
0
down vote
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Let $Z_n in mathcalX$ be a sequence of independent random variables where $mathcalX$ is a topological vector space and let $mu_n$ the probability measures associated with $Z_n$. Suppose that $Z_n$ that satisfies the conditions for the abstract Gärtner-Ellis Theorem. So the rate function $I(cdot)$ associated with $Z_n$ is the Fenchel–Legendre transform of $M(lambda) =lim_n rightarrow infty frac1n log E[exp langle nlambda, Z_n rangle ]$.
Now consider a second random variable $Y_n$ that is a Markov chain ($Y_n$ is sampled from $Y_n-1$) and the probability measure associated with $Y_n$ is $mu_n$.
My question is:
Is $I(cdot)$ the rate function associated with $Y_n$?
If answer is yes to the above question, let me provide a counter-example of the equivalence of the rate function of $Z_n$ and $Y_n$.
Consider two random variables on $0,1$
let $g_n$ be 1 with probability $1/2$, else 0
let $h_1$ = $g_1$ and then for $n>1$, let $h_n=1-h_n-1$ with probability
$(1/2)^n$ and else $h_n=h_n-1$.
Note that for all $n$, $h_n$ and $g_n$ have the same probability distribution,
$Prob(g_n = 1) = 1/2$ (by definition).
$$
Prob(h_n = 1) = Prob(h_n-1=1)cdot(1/2)^n + Prob(h_n-1=0)cdot(1-(1/2)^n) = (1/2) cdot (1/2)^n+(1/2) cdot (1-(1/2)^n) = (1/2)cdot((1/2)^n + 1-(1/2)^n) = 1/2
$$
Hence $Prob(g_n=1) = Prob(h_n=1)$.
At the limit both random variables have the same distribution,
but the convergence of the sequences is rather different. ($h_n$
converges, $g_n$ doesn't)
My other question is: Does this example invalidate the positive answer to first question?
probability-theory large-deviation-theory
asked Jul 27 at 6:01
jaogye
405413
You really need to rephrase the paragraph "Now consider a second random variable $Y_n$ sampled from $Y_n-1$ such that the probability measures assoociate with $Y_n$ is $mu_n$." At the moment, one cannot know what you mean by this.
– Did
Jul 27 at 6:28
Dear @Did, Is now the question more clear?.
– jaogye
Jul 27 at 6:39
It seems you are assuming that the distribution of $Z_n$ is $mu_n$ for every $n$, that $(Z_n)$ satisfies a LDP with rate $I$, and that $(Y_n)$ is a Markov chain with marginal distributions $mu_n$, and that you are asking whether, then, $(Y_n)$ also satisies a LDP with rate $I$. But this is trivially so since the LDP for $(Z_n)$ describes probabilities $P(Z_nin B)$, likewise a LDP for $(Y_n)$ would describe probabilities $P(Y_nin B)$, and you are assuming that for every $n$ and $B$, $P(Z_nin B)=P(Y_nin B)$...
– Did
Jul 27 at 7:56
Dear @Did yes, the answer could be trivial for an expert. However this equivalence can produces an example that at least for me it is counter-intuitive. I will add it to the question
– jaogye
Jul 27 at 9:19
In the example you added, $(h_n)$ and $g_n)$ both satisfy the same LDPs and fail the same LDPs. No counterexample at all.
– Did
Jul 27 at 9:49
 |Â
show 4 more comments
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0
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up vote
0
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Let $Z_n in mathcalX$ be a sequence of independent random variables where $mathcalX$ is a topological vector space and let $mu_n$ the probability measures associated with $Z_n$. Suppose that $Z_n$ that satisfies the conditions for the abstract Gärtner-Ellis Theorem. So the rate function $I(cdot)$ associated with $Z_n$ is the Fenchel–Legendre transform of $M(lambda) =lim_n rightarrow infty frac1n log E[exp langle nlambda, Z_n rangle ]$.
Now consider a second random variable $Y_n$ that is a Markov chain ($Y_n$ is sampled from $Y_n-1$) and the probability measure associated with $Y_n$ is $mu_n$.
My question is:
Is $I(cdot)$ the rate function associated with $Y_n$?
If answer is yes to the above question, let me provide a counter-example of the equivalence of the rate function of $Z_n$ and $Y_n$.
Consider two random variables on $0,1$
let $g_n$ be 1 with probability $1/2$, else 0
let $h_1$ = $g_1$ and then for $n>1$, let $h_n=1-h_n-1$ with probability
$(1/2)^n$ and else $h_n=h_n-1$.
Note that for all $n$, $h_n$ and $g_n$ have the same probability distribution,
$Prob(g_n = 1) = 1/2$ (by definition).
$$
Prob(h_n = 1) = Prob(h_n-1=1)cdot(1/2)^n + Prob(h_n-1=0)cdot(1-(1/2)^n) = (1/2) cdot (1/2)^n+(1/2) cdot (1-(1/2)^n) = (1/2)cdot((1/2)^n + 1-(1/2)^n) = 1/2
$$
Hence $Prob(g_n=1) = Prob(h_n=1)$.
At the limit both random variables have the same distribution,
but the convergence of the sequences is rather different. ($h_n$
converges, $g_n$ doesn't)
My other question is: Does this example invalidate the positive answer to first question?
probability-theory large-deviation-theory
asked Jul 27 at 6:01
jaogye
405413
Let $Z_n in mathcalX$ be a sequence of independent random variables where $mathcalX$ is a topological vector space and let $mu_n$ the probability measures associated with $Z_n$. Suppose that $Z_n$ that satisfies the conditions for the abstract Gärtner-Ellis Theorem. So the rate function $I(cdot)$ associated with $Z_n$ is the Fenchel–Legendre transform of $M(lambda) =lim_n rightarrow infty frac1n log E[exp langle nlambda, Z_n rangle ]$.
Now consider a second random variable $Y_n$ that is a Markov chain ($Y_n$ is sampled from $Y_n-1$) and the probability measure associated with $Y_n$ is $mu_n$.
My question is:
Is $I(cdot)$ the rate function associated with $Y_n$?
If answer is yes to the above question, let me provide a counter-example of the equivalence of the rate function of $Z_n$ and $Y_n$.
Consider two random variables on $0,1$
let $g_n$ be 1 with probability $1/2$, else 0
let $h_1$ = $g_1$ and then for $n>1$, let $h_n=1-h_n-1$ with probability
$(1/2)^n$ and else $h_n=h_n-1$.
Note that for all $n$, $h_n$ and $g_n$ have the same probability distribution,
$Prob(g_n = 1) = 1/2$ (by definition).
$$
Prob(h_n = 1) = Prob(h_n-1=1)cdot(1/2)^n + Prob(h_n-1=0)cdot(1-(1/2)^n) = (1/2) cdot (1/2)^n+(1/2) cdot (1-(1/2)^n) = (1/2)cdot((1/2)^n + 1-(1/2)^n) = 1/2
$$
Hence $Prob(g_n=1) = Prob(h_n=1)$.
At the limit both random variables have the same distribution,
but the convergence of the sequences is rather different. ($h_n$
converges, $g_n$ doesn't)
My other question is: Does this example invalidate the positive answer to first question?
probability-theory large-deviation-theory
asked Jul 27 at 6:01
jaogye
405413
edited Jul 27 at 9:33
asked Jul 27 at 6:01
jaogye
405413
asked Jul 27 at 6:01
jaogye
405413
asked Jul 27 at 6:01
jaogye
405413
405413
You really need to rephrase the paragraph "Now consider a second random variable $Y_n$ sampled from $Y_n-1$ such that the probability measures assoociate with $Y_n$ is $mu_n$." At the moment, one cannot know what you mean by this.
– Did
Jul 27 at 6:28
Dear @Did, Is now the question more clear?.
– jaogye
Jul 27 at 6:39
It seems you are assuming that the distribution of $Z_n$ is $mu_n$ for every $n$, that $(Z_n)$ satisfies a LDP with rate $I$, and that $(Y_n)$ is a Markov chain with marginal distributions $mu_n$, and that you are asking whether, then, $(Y_n)$ also satisies a LDP with rate $I$. But this is trivially so since the LDP for $(Z_n)$ describes probabilities $P(Z_nin B)$, likewise a LDP for $(Y_n)$ would describe probabilities $P(Y_nin B)$, and you are assuming that for every $n$ and $B$, $P(Z_nin B)=P(Y_nin B)$...
– Did
Jul 27 at 7:56
Dear @Did yes, the answer could be trivial for an expert. However this equivalence can produces an example that at least for me it is counter-intuitive. I will add it to the question
– jaogye
Jul 27 at 9:19
In the example you added, $(h_n)$ and $g_n)$ both satisfy the same LDPs and fail the same LDPs. No counterexample at all.
– Did
Jul 27 at 9:49
 |Â
show 4 more comments
You really need to rephrase the paragraph "Now consider a second random variable $Y_n$ sampled from $Y_n-1$ such that the probability measures assoociate with $Y_n$ is $mu_n$." At the moment, one cannot know what you mean by this.
– Did
Jul 27 at 6:28
Dear @Did, Is now the question more clear?.
– jaogye
Jul 27 at 6:39
It seems you are assuming that the distribution of $Z_n$ is $mu_n$ for every $n$, that $(Z_n)$ satisfies a LDP with rate $I$, and that $(Y_n)$ is a Markov chain with marginal distributions $mu_n$, and that you are asking whether, then, $(Y_n)$ also satisies a LDP with rate $I$. But this is trivially so since the LDP for $(Z_n)$ describes probabilities $P(Z_nin B)$, likewise a LDP for $(Y_n)$ would describe probabilities $P(Y_nin B)$, and you are assuming that for every $n$ and $B$, $P(Z_nin B)=P(Y_nin B)$...
– Did
Jul 27 at 7:56
Dear @Did yes, the answer could be trivial for an expert. However this equivalence can produces an example that at least for me it is counter-intuitive. I will add it to the question
– jaogye
Jul 27 at 9:19
In the example you added, $(h_n)$ and $g_n)$ both satisfy the same LDPs and fail the same LDPs. No counterexample at all.
– Did
Jul 27 at 9:49
You really need to rephrase the paragraph "Now consider a second random variable $Y_n$ sampled from $Y_n-1$ such that the probability measures assoociate with $Y_n$ is $mu_n$." At the moment, one cannot know what you mean by this.
– Did
Jul 27 at 6:28
You really need to rephrase the paragraph "Now consider a second random variable $Y_n$ sampled from $Y_n-1$ such that the probability measures assoociate with $Y_n$ is $mu_n$." At the moment, one cannot know what you mean by this.
– Did
Jul 27 at 6:28
Dear @Did, Is now the question more clear?.
– jaogye
Jul 27 at 6:39
Dear @Did, Is now the question more clear?.
– jaogye
Jul 27 at 6:39
It seems you are assuming that the distribution of $Z_n$ is $mu_n$ for every $n$, that $(Z_n)$ satisfies a LDP with rate $I$, and that $(Y_n)$ is a Markov chain with marginal distributions $mu_n$, and that you are asking whether, then, $(Y_n)$ also satisies a LDP with rate $I$. But this is trivially so since the LDP for $(Z_n)$ describes probabilities $P(Z_nin B)$, likewise a LDP for $(Y_n)$ would describe probabilities $P(Y_nin B)$, and you are assuming that for every $n$ and $B$, $P(Z_nin B)=P(Y_nin B)$...
– Did
Jul 27 at 7:56
It seems you are assuming that the distribution of $Z_n$ is $mu_n$ for every $n$, that $(Z_n)$ satisfies a LDP with rate $I$, and that $(Y_n)$ is a Markov chain with marginal distributions $mu_n$, and that you are asking whether, then, $(Y_n)$ also satisies a LDP with rate $I$. But this is trivially so since the LDP for $(Z_n)$ describes probabilities $P(Z_nin B)$, likewise a LDP for $(Y_n)$ would describe probabilities $P(Y_nin B)$, and you are assuming that for every $n$ and $B$, $P(Z_nin B)=P(Y_nin B)$...
– Did
Jul 27 at 7:56
Dear @Did yes, the answer could be trivial for an expert. However this equivalence can produces an example that at least for me it is counter-intuitive. I will add it to the question
– jaogye
Jul 27 at 9:19
Dear @Did yes, the answer could be trivial for an expert. However this equivalence can produces an example that at least for me it is counter-intuitive. I will add it to the question
– jaogye
Jul 27 at 9:19
In the example you added, $(h_n)$ and $g_n)$ both satisfy the same LDPs and fail the same LDPs. No counterexample at all.
– Did
Jul 27 at 9:49
In the example you added, $(h_n)$ and $g_n)$ both satisfy the same LDPs and fail the same LDPs. No counterexample at all.
– Did
Jul 27 at 9:49
 |Â
show 4 more comments
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You really need to rephrase the paragraph "Now consider a second random variable $Y_n$ sampled from $Y_n-1$ such that the probability measures assoociate with $Y_n$ is $mu_n$." At the moment, one cannot know what you mean by this.
– Did
Jul 27 at 6:28
Dear @Did, Is now the question more clear?.
– jaogye
Jul 27 at 6:39
It seems you are assuming that the distribution of $Z_n$ is $mu_n$ for every $n$, that $(Z_n)$ satisfies a LDP with rate $I$, and that $(Y_n)$ is a Markov chain with marginal distributions $mu_n$, and that you are asking whether, then, $(Y_n)$ also satisies a LDP with rate $I$. But this is trivially so since the LDP for $(Z_n)$ describes probabilities $P(Z_nin B)$, likewise a LDP for $(Y_n)$ would describe probabilities $P(Y_nin B)$, and you are assuming that for every $n$ and $B$, $P(Z_nin B)=P(Y_nin B)$...
– Did
Jul 27 at 7:56
Dear @Did yes, the answer could be trivial for an expert. However this equivalence can produces an example that at least for me it is counter-intuitive. I will add it to the question
– jaogye
Jul 27 at 9:19
In the example you added, $(h_n)$ and $g_n)$ both satisfy the same LDPs and fail the same LDPs. No counterexample at all.
– Did
Jul 27 at 9:49