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













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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




















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