Updated belief with Bayes' rule
$begingroup$
Let ${x_0,x_1}$ be to states. Suppose the reward is $1$ in state $x_1$, and in state $x_0$ it is $1$ with probability $r$ and $0$ with probability $1-r$, with $rin(0,1)$.
Bob has a belief $pin [0,1]$ over the states $x_0$ and $x_1$, i.e., $p$ is the probability that the true state is $x_1$ according Bob.
Suppose the true state is $x_0$. If $pin[0,1)$ and Bob observes a reward of $0$, he updates his belief according to Bayes' rule and is new belief is $0$.
Suppose now that $p=1$, that is Bob is sure that the true state is $x_1$ when it is actually $x_0$, and suppose the Bob observes a reward of $0$.
Bayes' rule in this case would give "$frac{0}{0}$".
My question is, Bob being Bayesian, will his new belief be $0$ or $1$, and why?
bayesian
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
$endgroup$
add a comment |
$begingroup$
Let ${x_0,x_1}$ be to states. Suppose the reward is $1$ in state $x_1$, and in state $x_0$ it is $1$ with probability $r$ and $0$ with probability $1-r$, with $rin(0,1)$.
Bob has a belief $pin [0,1]$ over the states $x_0$ and $x_1$, i.e., $p$ is the probability that the true state is $x_1$ according Bob.
Suppose the true state is $x_0$. If $pin[0,1)$ and Bob observes a reward of $0$, he updates his belief according to Bayes' rule and is new belief is $0$.
Suppose now that $p=1$, that is Bob is sure that the true state is $x_1$ when it is actually $x_0$, and suppose the Bob observes a reward of $0$.
Bayes' rule in this case would give "$frac{0}{0}$".
My question is, Bob being Bayesian, will his new belief be $0$ or $1$, and why?
bayesian
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
$endgroup$
1
$begingroup$
What does it mean for a state to be true?
$endgroup$
– user593746
Dec 18 '18 at 20:29
$begingroup$
@Zvi It could be for example raining/not raining. Bob (at home) would believe that it is raining when actually it is not.
$endgroup$
– S. Pel
Dec 19 '18 at 10:41
add a comment |
$begingroup$
Let ${x_0,x_1}$ be to states. Suppose the reward is $1$ in state $x_1$, and in state $x_0$ it is $1$ with probability $r$ and $0$ with probability $1-r$, with $rin(0,1)$.
Bob has a belief $pin [0,1]$ over the states $x_0$ and $x_1$, i.e., $p$ is the probability that the true state is $x_1$ according Bob.
Suppose the true state is $x_0$. If $pin[0,1)$ and Bob observes a reward of $0$, he updates his belief according to Bayes' rule and is new belief is $0$.
Suppose now that $p=1$, that is Bob is sure that the true state is $x_1$ when it is actually $x_0$, and suppose the Bob observes a reward of $0$.
Bayes' rule in this case would give "$frac{0}{0}$".
My question is, Bob being Bayesian, will his new belief be $0$ or $1$, and why?
bayesian
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
$endgroup$
Let ${x_0,x_1}$ be to states. Suppose the reward is $1$ in state $x_1$, and in state $x_0$ it is $1$ with probability $r$ and $0$ with probability $1-r$, with $rin(0,1)$.
Bob has a belief $pin [0,1]$ over the states $x_0$ and $x_1$, i.e., $p$ is the probability that the true state is $x_1$ according Bob.
Suppose the true state is $x_0$. If $pin[0,1)$ and Bob observes a reward of $0$, he updates his belief according to Bayes' rule and is new belief is $0$.
Suppose now that $p=1$, that is Bob is sure that the true state is $x_1$ when it is actually $x_0$, and suppose the Bob observes a reward of $0$.
Bayes' rule in this case would give "$frac{0}{0}$".
My question is, Bob being Bayesian, will his new belief be $0$ or $1$, and why?
bayesian
bayesian
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
asked Dec 18 '18 at 15:47
S. PelS. Pel
12
12
1
$begingroup$
What does it mean for a state to be true?
$endgroup$
– user593746
Dec 18 '18 at 20:29
$begingroup$
@Zvi It could be for example raining/not raining. Bob (at home) would believe that it is raining when actually it is not.
$endgroup$
– S. Pel
Dec 19 '18 at 10:41
add a comment |
1
$begingroup$
What does it mean for a state to be true?
$endgroup$
– user593746
Dec 18 '18 at 20:29
$begingroup$
@Zvi It could be for example raining/not raining. Bob (at home) would believe that it is raining when actually it is not.
$endgroup$
– S. Pel
Dec 19 '18 at 10:41
1
1
$begingroup$
What does it mean for a state to be true?
$endgroup$
– user593746
Dec 18 '18 at 20:29
$begingroup$
What does it mean for a state to be true?
$endgroup$
– user593746
Dec 18 '18 at 20:29
$begingroup$
@Zvi It could be for example raining/not raining. Bob (at home) would believe that it is raining when actually it is not.
$endgroup$
– S. Pel
Dec 19 '18 at 10:41
$begingroup$
@Zvi It could be for example raining/not raining. Bob (at home) would believe that it is raining when actually it is not.
$endgroup$
– S. Pel
Dec 19 '18 at 10:41
add a comment |
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1
$begingroup$
What does it mean for a state to be true?
$endgroup$
– user593746
Dec 18 '18 at 20:29
$begingroup$
@Zvi It could be for example raining/not raining. Bob (at home) would believe that it is raining when actually it is not.
$endgroup$
– S. Pel
Dec 19 '18 at 10:41