Mixing
Some problems in conditional probability
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A system process like $x_1 to x_2 to x_3 to ldots to x_n, (x_n =0text{ or }1)$ and there are abservations of the process$y_1, y_2 , y_3 , ldots y_n$. Assuming that the system has the following relation. $P(x_{n+1}=c_{n+1}|x_1=c_1,x_2=c_2.ldots)=begin{cases}1-lambda ,,&(c_{n+1}=c_n)\lambda,,&(c_{n+1}ne c_n)end{cases} tag1$Also, due to the noise, the observation of each $x$ has the relation $P(y_n=d|x_n=d')=begin{cases}1-rho &(d=d')\rho &(dne d')end{cases} tag2$ If $P(x_1=1)=p$, what is $P(x_2=1|y_1=1)$ and $P(x_2=1|y_1=1,y_2=1)$.
I am trying to solve it, but there is something I am not sure.
For $P(x_2=1|y_1=1)$, I think there is no relation between the event $y_1=1$ and $x_2=1$ (they are independent), so
$P(x_2=1|y_1=1)=P(x_2=1)=P(x_1=1)P(x_2=1|x_1=1)=p(1-lambda)$. However, I not sure if it is right or not.
In case of $P(x_2=1|y_1=1,y_2=1)$.
$P(x_2=1|y_1=1,y_2=1)\=dfrac{P(x_2=1,y_1=1,y_2=1)}{P(y_1=1,y_2=1)}\=dfrac{P(y_1=1)P(x_2=1,y_2=1)}{P(y_1=1)P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(x_2=0text{ or }1,y_2=1)}\=dfrac{P(x_2=1)}{P(x_2=0text{ or }1)}\=(1-p)lambda+p(1-lambda) $
probability
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
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$begingroup$
A system process like $x_1 to x_2 to x_3 to ldots to x_n, (x_n =0text{ or }1)$ and there are abservations of the process$y_1, y_2 , y_3 , ldots y_n$. Assuming that the system has the following relation. $P(x_{n+1}=c_{n+1}|x_1=c_1,x_2=c_2.ldots)=begin{cases}1-lambda ,,&(c_{n+1}=c_n)\lambda,,&(c_{n+1}ne c_n)end{cases} tag1$Also, due to the noise, the observation of each $x$ has the relation $P(y_n=d|x_n=d')=begin{cases}1-rho &(d=d')\rho &(dne d')end{cases} tag2$ If $P(x_1=1)=p$, what is $P(x_2=1|y_1=1)$ and $P(x_2=1|y_1=1,y_2=1)$.
I am trying to solve it, but there is something I am not sure.
For $P(x_2=1|y_1=1)$, I think there is no relation between the event $y_1=1$ and $x_2=1$ (they are independent), so
$P(x_2=1|y_1=1)=P(x_2=1)=P(x_1=1)P(x_2=1|x_1=1)=p(1-lambda)$. However, I not sure if it is right or not.
In case of $P(x_2=1|y_1=1,y_2=1)$.
$P(x_2=1|y_1=1,y_2=1)\=dfrac{P(x_2=1,y_1=1,y_2=1)}{P(y_1=1,y_2=1)}\=dfrac{P(y_1=1)P(x_2=1,y_2=1)}{P(y_1=1)P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(x_2=0text{ or }1,y_2=1)}\=dfrac{P(x_2=1)}{P(x_2=0text{ or }1)}\=(1-p)lambda+p(1-lambda) $
probability
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
$endgroup$
add a comment |
$begingroup$
A system process like $x_1 to x_2 to x_3 to ldots to x_n, (x_n =0text{ or }1)$ and there are abservations of the process$y_1, y_2 , y_3 , ldots y_n$. Assuming that the system has the following relation. $P(x_{n+1}=c_{n+1}|x_1=c_1,x_2=c_2.ldots)=begin{cases}1-lambda ,,&(c_{n+1}=c_n)\lambda,,&(c_{n+1}ne c_n)end{cases} tag1$Also, due to the noise, the observation of each $x$ has the relation $P(y_n=d|x_n=d')=begin{cases}1-rho &(d=d')\rho &(dne d')end{cases} tag2$ If $P(x_1=1)=p$, what is $P(x_2=1|y_1=1)$ and $P(x_2=1|y_1=1,y_2=1)$.
I am trying to solve it, but there is something I am not sure.
For $P(x_2=1|y_1=1)$, I think there is no relation between the event $y_1=1$ and $x_2=1$ (they are independent), so
$P(x_2=1|y_1=1)=P(x_2=1)=P(x_1=1)P(x_2=1|x_1=1)=p(1-lambda)$. However, I not sure if it is right or not.
In case of $P(x_2=1|y_1=1,y_2=1)$.
$P(x_2=1|y_1=1,y_2=1)\=dfrac{P(x_2=1,y_1=1,y_2=1)}{P(y_1=1,y_2=1)}\=dfrac{P(y_1=1)P(x_2=1,y_2=1)}{P(y_1=1)P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(x_2=0text{ or }1,y_2=1)}\=dfrac{P(x_2=1)}{P(x_2=0text{ or }1)}\=(1-p)lambda+p(1-lambda) $
probability
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
$endgroup$
A system process like $x_1 to x_2 to x_3 to ldots to x_n, (x_n =0text{ or }1)$ and there are abservations of the process$y_1, y_2 , y_3 , ldots y_n$. Assuming that the system has the following relation. $P(x_{n+1}=c_{n+1}|x_1=c_1,x_2=c_2.ldots)=begin{cases}1-lambda ,,&(c_{n+1}=c_n)\lambda,,&(c_{n+1}ne c_n)end{cases} tag1$Also, due to the noise, the observation of each $x$ has the relation $P(y_n=d|x_n=d')=begin{cases}1-rho &(d=d')\rho &(dne d')end{cases} tag2$ If $P(x_1=1)=p$, what is $P(x_2=1|y_1=1)$ and $P(x_2=1|y_1=1,y_2=1)$.
I am trying to solve it, but there is something I am not sure.
For $P(x_2=1|y_1=1)$, I think there is no relation between the event $y_1=1$ and $x_2=1$ (they are independent), so
$P(x_2=1|y_1=1)=P(x_2=1)=P(x_1=1)P(x_2=1|x_1=1)=p(1-lambda)$. However, I not sure if it is right or not.
In case of $P(x_2=1|y_1=1,y_2=1)$.
$P(x_2=1|y_1=1,y_2=1)\=dfrac{P(x_2=1,y_1=1,y_2=1)}{P(y_1=1,y_2=1)}\=dfrac{P(y_1=1)P(x_2=1,y_2=1)}{P(y_1=1)P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(y_2=1)}\=dfrac{P(x_2=1,y_2=1)}{P(x_2=0text{ or }1,y_2=1)}\=dfrac{P(x_2=1)}{P(x_2=0text{ or }1)}\=(1-p)lambda+p(1-lambda) $
probability
probability
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
asked Jan 23 at 5:07
Weihao HuangWeihao Huang
112
112
add a comment |
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