Markov chain - how to navigate in transition matrix?












1












$begingroup$



Let $X_0,X_1,...$ be a Markov Chain with transition matrix



$$P=begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ p & 1-p & 0
end{pmatrix} $$ for $0<p<1.$ Let $g$ be a function defined by $$
g(x)=left{ begin{array}{rcr}
0, & & text{if} x=&1 \
1, & & text{if} x=&2,3\ end{array} right. $$



Let $Y_n=g(X_n)$, for $ngeq 0$. Show that $Y_0,Y_1,...$ is not a
Markov chain.




My attempt:



So I want to show that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i).
end{align}



does not hold. Substituting in $X_i$ for $Y_i$ i get that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)&=mathbb{P}(g(X_n)=j|g(X_0)=x_0,...,g(x_{n-1})=i)\
&=...?
end{align}



How do I know which states to substitute my $X_i$'s for? I'm pretty sure I should use $P$ to do this but I have no idea how.






markov-chains conditional-probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think you have a typo in your definition of $g(x)$, since both options contain $x=1$..
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:38










  • $begingroup$
    Thanks, editing!
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:39










  • $begingroup$
    You want to show it is not a Markov chain, so you just need to find some $y_0, ldots, y_{n-2}$ and $i$ and $j$ such that the Markov condition fails to hold.
    $endgroup$
    – angryavian
    Dec 24 '18 at 21:45










  • $begingroup$
    $mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i)$ is known as the Markov property and is the defining property of Markov chains. So you want to show that this doesn't hold. In particular, you'll want to find a counterexample
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:45










  • $begingroup$
    @pwerth I forgot to add the text "does not hold" below the Markov property, sorry about that. Oh, so I can simply choose any states that I like such that the proeprty fails to hold?
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:49
















1












$begingroup$



Let $X_0,X_1,...$ be a Markov Chain with transition matrix



$$P=begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ p & 1-p & 0
end{pmatrix} $$ for $0<p<1.$ Let $g$ be a function defined by $$
g(x)=left{ begin{array}{rcr}
0, & & text{if} x=&1 \
1, & & text{if} x=&2,3\ end{array} right. $$



Let $Y_n=g(X_n)$, for $ngeq 0$. Show that $Y_0,Y_1,...$ is not a
Markov chain.




My attempt:



So I want to show that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i).
end{align}



does not hold. Substituting in $X_i$ for $Y_i$ i get that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)&=mathbb{P}(g(X_n)=j|g(X_0)=x_0,...,g(x_{n-1})=i)\
&=...?
end{align}



How do I know which states to substitute my $X_i$'s for? I'm pretty sure I should use $P$ to do this but I have no idea how.






markov-chains conditional-probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think you have a typo in your definition of $g(x)$, since both options contain $x=1$..
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:38










  • $begingroup$
    Thanks, editing!
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:39










  • $begingroup$
    You want to show it is not a Markov chain, so you just need to find some $y_0, ldots, y_{n-2}$ and $i$ and $j$ such that the Markov condition fails to hold.
    $endgroup$
    – angryavian
    Dec 24 '18 at 21:45










  • $begingroup$
    $mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i)$ is known as the Markov property and is the defining property of Markov chains. So you want to show that this doesn't hold. In particular, you'll want to find a counterexample
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:45










  • $begingroup$
    @pwerth I forgot to add the text "does not hold" below the Markov property, sorry about that. Oh, so I can simply choose any states that I like such that the proeprty fails to hold?
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:49














1












1








1





$begingroup$



Let $X_0,X_1,...$ be a Markov Chain with transition matrix



$$P=begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ p & 1-p & 0
end{pmatrix} $$ for $0<p<1.$ Let $g$ be a function defined by $$
g(x)=left{ begin{array}{rcr}
0, & & text{if} x=&1 \
1, & & text{if} x=&2,3\ end{array} right. $$



Let $Y_n=g(X_n)$, for $ngeq 0$. Show that $Y_0,Y_1,...$ is not a
Markov chain.




My attempt:



So I want to show that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i).
end{align}



does not hold. Substituting in $X_i$ for $Y_i$ i get that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)&=mathbb{P}(g(X_n)=j|g(X_0)=x_0,...,g(x_{n-1})=i)\
&=...?
end{align}



How do I know which states to substitute my $X_i$'s for? I'm pretty sure I should use $P$ to do this but I have no idea how.






markov-chains conditional-probability







share|cite|improve this question











$endgroup$





Let $X_0,X_1,...$ be a Markov Chain with transition matrix



$$P=begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ p & 1-p & 0
end{pmatrix} $$ for $0<p<1.$ Let $g$ be a function defined by $$
g(x)=left{ begin{array}{rcr}
0, & & text{if} x=&1 \
1, & & text{if} x=&2,3\ end{array} right. $$



Let $Y_n=g(X_n)$, for $ngeq 0$. Show that $Y_0,Y_1,...$ is not a
Markov chain.




My attempt:



So I want to show that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i).
end{align}



does not hold. Substituting in $X_i$ for $Y_i$ i get that



begin{align}
mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)&=mathbb{P}(g(X_n)=j|g(X_0)=x_0,...,g(x_{n-1})=i)\
&=...?
end{align}



How do I know which states to substitute my $X_i$'s for? I'm pretty sure I should use $P$ to do this but I have no idea how.






markov-chains conditional-probability




markov-chains conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 21:47







Parseval

















asked Dec 24 '18 at 21:37









ParsevalParseval

2,9261719




2,9261719








  • 1




    $begingroup$
    I think you have a typo in your definition of $g(x)$, since both options contain $x=1$..
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:38










  • $begingroup$
    Thanks, editing!
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:39










  • $begingroup$
    You want to show it is not a Markov chain, so you just need to find some $y_0, ldots, y_{n-2}$ and $i$ and $j$ such that the Markov condition fails to hold.
    $endgroup$
    – angryavian
    Dec 24 '18 at 21:45










  • $begingroup$
    $mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i)$ is known as the Markov property and is the defining property of Markov chains. So you want to show that this doesn't hold. In particular, you'll want to find a counterexample
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:45










  • $begingroup$
    @pwerth I forgot to add the text "does not hold" below the Markov property, sorry about that. Oh, so I can simply choose any states that I like such that the proeprty fails to hold?
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:49














  • 1




    $begingroup$
    I think you have a typo in your definition of $g(x)$, since both options contain $x=1$..
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:38










  • $begingroup$
    Thanks, editing!
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:39










  • $begingroup$
    You want to show it is not a Markov chain, so you just need to find some $y_0, ldots, y_{n-2}$ and $i$ and $j$ such that the Markov condition fails to hold.
    $endgroup$
    – angryavian
    Dec 24 '18 at 21:45










  • $begingroup$
    $mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i)$ is known as the Markov property and is the defining property of Markov chains. So you want to show that this doesn't hold. In particular, you'll want to find a counterexample
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:45










  • $begingroup$
    @pwerth I forgot to add the text "does not hold" below the Markov property, sorry about that. Oh, so I can simply choose any states that I like such that the proeprty fails to hold?
    $endgroup$
    – Parseval
    Dec 24 '18 at 21:49








1




1




$begingroup$
I think you have a typo in your definition of $g(x)$, since both options contain $x=1$..
$endgroup$
– pwerth
Dec 24 '18 at 21:38




$begingroup$
I think you have a typo in your definition of $g(x)$, since both options contain $x=1$..
$endgroup$
– pwerth
Dec 24 '18 at 21:38












$begingroup$
Thanks, editing!
$endgroup$
– Parseval
Dec 24 '18 at 21:39




$begingroup$
Thanks, editing!
$endgroup$
– Parseval
Dec 24 '18 at 21:39












$begingroup$
You want to show it is not a Markov chain, so you just need to find some $y_0, ldots, y_{n-2}$ and $i$ and $j$ such that the Markov condition fails to hold.
$endgroup$
– angryavian
Dec 24 '18 at 21:45




$begingroup$
You want to show it is not a Markov chain, so you just need to find some $y_0, ldots, y_{n-2}$ and $i$ and $j$ such that the Markov condition fails to hold.
$endgroup$
– angryavian
Dec 24 '18 at 21:45












$begingroup$
$mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i)$ is known as the Markov property and is the defining property of Markov chains. So you want to show that this doesn't hold. In particular, you'll want to find a counterexample
$endgroup$
– pwerth
Dec 24 '18 at 21:45




$begingroup$
$mathbb{P}(Y_n=j|Y_0=y_0,...,Y_{n-1}=i)=mathbb{P}(Y_n=j|Y_{n-1}=i)$ is known as the Markov property and is the defining property of Markov chains. So you want to show that this doesn't hold. In particular, you'll want to find a counterexample
$endgroup$
– pwerth
Dec 24 '18 at 21:45












$begingroup$
@pwerth I forgot to add the text "does not hold" below the Markov property, sorry about that. Oh, so I can simply choose any states that I like such that the proeprty fails to hold?
$endgroup$
– Parseval
Dec 24 '18 at 21:49




$begingroup$
@pwerth I forgot to add the text "does not hold" below the Markov property, sorry about that. Oh, so I can simply choose any states that I like such that the proeprty fails to hold?
$endgroup$
– Parseval
Dec 24 '18 at 21:49










1 Answer
1






active

oldest

votes


















2












$begingroup$

Consider $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)$. Given that $Y_{0}=0$, we must have $X_{0}=1$, from which the only possibility is $X_{1}=2$ and $X_{2}=3$. Therefore $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0) = 1$.



Now consider $mathbb{P}(Y_{2}=1|Y_{1}=1)$. If $X_{1}=3$, then $Y_{1}=1$ but $mathbb{P}(X_{2}=1)=p$ and if $X_{2}=1$ then $Y_{2}=0$. Therefore this probability is not $1$ (there is a nonzero probability that $Y_{2}$ will equal $0$).



Since $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)neqmathbb{P}(Y_{2}=1|Y_{1}=1)$, the chain is not Markov.



The intuitive reason that $Y_{n}$ is not a Markov chain is because probabilities related to its values depend on knowledge of multiple prior states, whereas the Markov property means that probabilities of values of the chain depend only on the previous state.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great explanation!
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:40










  • $begingroup$
    Thank you! Glad it helped
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:47










  • $begingroup$
    May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:47












  • $begingroup$
    Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:52






  • 1




    $begingroup$
    From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:55













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051658%2fmarkov-chain-how-to-navigate-in-transition-matrix%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Consider $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)$. Given that $Y_{0}=0$, we must have $X_{0}=1$, from which the only possibility is $X_{1}=2$ and $X_{2}=3$. Therefore $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0) = 1$.



Now consider $mathbb{P}(Y_{2}=1|Y_{1}=1)$. If $X_{1}=3$, then $Y_{1}=1$ but $mathbb{P}(X_{2}=1)=p$ and if $X_{2}=1$ then $Y_{2}=0$. Therefore this probability is not $1$ (there is a nonzero probability that $Y_{2}$ will equal $0$).



Since $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)neqmathbb{P}(Y_{2}=1|Y_{1}=1)$, the chain is not Markov.



The intuitive reason that $Y_{n}$ is not a Markov chain is because probabilities related to its values depend on knowledge of multiple prior states, whereas the Markov property means that probabilities of values of the chain depend only on the previous state.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great explanation!
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:40










  • $begingroup$
    Thank you! Glad it helped
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:47










  • $begingroup$
    May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:47












  • $begingroup$
    Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:52






  • 1




    $begingroup$
    From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:55


















2












$begingroup$

Consider $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)$. Given that $Y_{0}=0$, we must have $X_{0}=1$, from which the only possibility is $X_{1}=2$ and $X_{2}=3$. Therefore $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0) = 1$.



Now consider $mathbb{P}(Y_{2}=1|Y_{1}=1)$. If $X_{1}=3$, then $Y_{1}=1$ but $mathbb{P}(X_{2}=1)=p$ and if $X_{2}=1$ then $Y_{2}=0$. Therefore this probability is not $1$ (there is a nonzero probability that $Y_{2}$ will equal $0$).



Since $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)neqmathbb{P}(Y_{2}=1|Y_{1}=1)$, the chain is not Markov.



The intuitive reason that $Y_{n}$ is not a Markov chain is because probabilities related to its values depend on knowledge of multiple prior states, whereas the Markov property means that probabilities of values of the chain depend only on the previous state.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great explanation!
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:40










  • $begingroup$
    Thank you! Glad it helped
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:47










  • $begingroup$
    May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:47












  • $begingroup$
    Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:52






  • 1




    $begingroup$
    From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:55
















2












2








2





$begingroup$

Consider $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)$. Given that $Y_{0}=0$, we must have $X_{0}=1$, from which the only possibility is $X_{1}=2$ and $X_{2}=3$. Therefore $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0) = 1$.



Now consider $mathbb{P}(Y_{2}=1|Y_{1}=1)$. If $X_{1}=3$, then $Y_{1}=1$ but $mathbb{P}(X_{2}=1)=p$ and if $X_{2}=1$ then $Y_{2}=0$. Therefore this probability is not $1$ (there is a nonzero probability that $Y_{2}$ will equal $0$).



Since $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)neqmathbb{P}(Y_{2}=1|Y_{1}=1)$, the chain is not Markov.



The intuitive reason that $Y_{n}$ is not a Markov chain is because probabilities related to its values depend on knowledge of multiple prior states, whereas the Markov property means that probabilities of values of the chain depend only on the previous state.






share|cite|improve this answer











$endgroup$



Consider $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)$. Given that $Y_{0}=0$, we must have $X_{0}=1$, from which the only possibility is $X_{1}=2$ and $X_{2}=3$. Therefore $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0) = 1$.



Now consider $mathbb{P}(Y_{2}=1|Y_{1}=1)$. If $X_{1}=3$, then $Y_{1}=1$ but $mathbb{P}(X_{2}=1)=p$ and if $X_{2}=1$ then $Y_{2}=0$. Therefore this probability is not $1$ (there is a nonzero probability that $Y_{2}$ will equal $0$).



Since $mathbb{P}(Y_{2}=1|Y_{1}=1,Y_{0}=0)neqmathbb{P}(Y_{2}=1|Y_{1}=1)$, the chain is not Markov.



The intuitive reason that $Y_{n}$ is not a Markov chain is because probabilities related to its values depend on knowledge of multiple prior states, whereas the Markov property means that probabilities of values of the chain depend only on the previous state.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 22:01

























answered Dec 24 '18 at 21:53









pwerthpwerth

3,243417




3,243417












  • $begingroup$
    Great explanation!
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:40










  • $begingroup$
    Thank you! Glad it helped
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:47










  • $begingroup$
    May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:47












  • $begingroup$
    Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:52






  • 1




    $begingroup$
    From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:55




















  • $begingroup$
    Great explanation!
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:40










  • $begingroup$
    Thank you! Glad it helped
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:47










  • $begingroup$
    May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
    $endgroup$
    – Parseval
    Dec 24 '18 at 22:47












  • $begingroup$
    Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:52






  • 1




    $begingroup$
    From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
    $endgroup$
    – pwerth
    Dec 24 '18 at 22:55


















$begingroup$
Great explanation!
$endgroup$
– Parseval
Dec 24 '18 at 22:40




$begingroup$
Great explanation!
$endgroup$
– Parseval
Dec 24 '18 at 22:40












$begingroup$
Thank you! Glad it helped
$endgroup$
– pwerth
Dec 24 '18 at 22:47




$begingroup$
Thank you! Glad it helped
$endgroup$
– pwerth
Dec 24 '18 at 22:47












$begingroup$
May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
$endgroup$
– Parseval
Dec 24 '18 at 22:47






$begingroup$
May I ask, why did you only go to $Y_2$ and not $Y_{16}$ or so? Did you just use the lowest possible index just for convenience?
$endgroup$
– Parseval
Dec 24 '18 at 22:47














$begingroup$
Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
$endgroup$
– pwerth
Dec 24 '18 at 22:52




$begingroup$
Yes, I simply used the lowest possible index for clarity. But if you found a counterexample going up to $Y_{16}$, that would still be perfectly valid
$endgroup$
– pwerth
Dec 24 '18 at 22:52




1




1




$begingroup$
From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
$endgroup$
– pwerth
Dec 24 '18 at 22:55






$begingroup$
From your question: "Let $Y_{n} = g(X_{n}), $for $ngeq 0"$
$endgroup$
– pwerth
Dec 24 '18 at 22:55




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051658%2fmarkov-chain-how-to-navigate-in-transition-matrix%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more