Binary Symmetric Channel calculating problem(the answer of two method should be the same,however,its not )

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This is a Binary Symmetric Channel ,now Let $P(X)=(p,1-p)$ where $0 le p le 1$,and $X=Y=(0,1)$



$P(X|Y)=$
beginbmatrix
1-epsilon & epsilon\
epsilon & 1-epsilon
endbmatrix
and $0le epsilon le 1 , P_e=Pr[X neq Y]$,



$ X neq Y $ means $X=0,Y=1$ or $X=1,Y=0$,and we have two methods to calculate it.One is $P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]$ , the other is $P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]$ . Theoretically,their answer should be the same.However,my answer is not the same,so i think i must have a mistake in somewhere,hoping someone can tell me



$P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]=epsilon+epsilon=2epsilon$



$P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]=1-[(1-epsilon)+(1-epsilon)]=1-2+2epsilon=-1+2epsilon$



Where am i wrong?




probability




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  • First mistake: $Pr[X=0,Y=1]neepsilon$, actually, $$Pr[X=0,Y=1]=Pr[X=0]Pr[Y=1mid X=0]$$ with probably $Pr[X=0]=1-p$ and $\Pr[Y=1mid X=0]=epsilon$ (except that you mangled the notations big time).
    – Did
    Jul 28 at 13:10











  • If you add the entries of the matrix then you get $2$ so they cannot correspond with probabilities of mutually exclusive events. Most probably the entries are conditional probabilities.
    – Vera
    Jul 28 at 13:12















up vote
0
down vote

favorite












This is a Binary Symmetric Channel ,now Let $P(X)=(p,1-p)$ where $0 le p le 1$,and $X=Y=(0,1)$



$P(X|Y)=$
beginbmatrix
1-epsilon & epsilon\
epsilon & 1-epsilon
endbmatrix
and $0le epsilon le 1 , P_e=Pr[X neq Y]$,



$ X neq Y $ means $X=0,Y=1$ or $X=1,Y=0$,and we have two methods to calculate it.One is $P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]$ , the other is $P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]$ . Theoretically,their answer should be the same.However,my answer is not the same,so i think i must have a mistake in somewhere,hoping someone can tell me



$P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]=epsilon+epsilon=2epsilon$



$P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]=1-[(1-epsilon)+(1-epsilon)]=1-2+2epsilon=-1+2epsilon$



Where am i wrong?




probability




share|cite|improve this question



















  • First mistake: $Pr[X=0,Y=1]neepsilon$, actually, $$Pr[X=0,Y=1]=Pr[X=0]Pr[Y=1mid X=0]$$ with probably $Pr[X=0]=1-p$ and $\Pr[Y=1mid X=0]=epsilon$ (except that you mangled the notations big time).
    – Did
    Jul 28 at 13:10











  • If you add the entries of the matrix then you get $2$ so they cannot correspond with probabilities of mutually exclusive events. Most probably the entries are conditional probabilities.
    – Vera
    Jul 28 at 13:12













up vote
0
down vote

favorite









up vote
0
down vote

favorite











This is a Binary Symmetric Channel ,now Let $P(X)=(p,1-p)$ where $0 le p le 1$,and $X=Y=(0,1)$



$P(X|Y)=$
beginbmatrix
1-epsilon & epsilon\
epsilon & 1-epsilon
endbmatrix
and $0le epsilon le 1 , P_e=Pr[X neq Y]$,



$ X neq Y $ means $X=0,Y=1$ or $X=1,Y=0$,and we have two methods to calculate it.One is $P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]$ , the other is $P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]$ . Theoretically,their answer should be the same.However,my answer is not the same,so i think i must have a mistake in somewhere,hoping someone can tell me



$P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]=epsilon+epsilon=2epsilon$



$P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]=1-[(1-epsilon)+(1-epsilon)]=1-2+2epsilon=-1+2epsilon$



Where am i wrong?




probability




share|cite|improve this question











This is a Binary Symmetric Channel ,now Let $P(X)=(p,1-p)$ where $0 le p le 1$,and $X=Y=(0,1)$



$P(X|Y)=$
beginbmatrix
1-epsilon & epsilon\
epsilon & 1-epsilon
endbmatrix
and $0le epsilon le 1 , P_e=Pr[X neq Y]$,



$ X neq Y $ means $X=0,Y=1$ or $X=1,Y=0$,and we have two methods to calculate it.One is $P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]$ , the other is $P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]$ . Theoretically,their answer should be the same.However,my answer is not the same,so i think i must have a mistake in somewhere,hoping someone can tell me



$P_e=Pr[X=0,Y=1]+Pr[X=1,Y=0]=epsilon+epsilon=2epsilon$



$P_e=1-[Pr[X=1,Y=1]+Pr[X=0,Y=0]]=1-[(1-epsilon)+(1-epsilon)]=1-2+2epsilon=-1+2epsilon$



Where am i wrong?





probability





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asked Jul 28 at 13:05









Shine Sun

1258




1258











  • First mistake: $Pr[X=0,Y=1]neepsilon$, actually, $$Pr[X=0,Y=1]=Pr[X=0]Pr[Y=1mid X=0]$$ with probably $Pr[X=0]=1-p$ and $\Pr[Y=1mid X=0]=epsilon$ (except that you mangled the notations big time).
    – Did
    Jul 28 at 13:10











  • If you add the entries of the matrix then you get $2$ so they cannot correspond with probabilities of mutually exclusive events. Most probably the entries are conditional probabilities.
    – Vera
    Jul 28 at 13:12

















  • First mistake: $Pr[X=0,Y=1]neepsilon$, actually, $$Pr[X=0,Y=1]=Pr[X=0]Pr[Y=1mid X=0]$$ with probably $Pr[X=0]=1-p$ and $\Pr[Y=1mid X=0]=epsilon$ (except that you mangled the notations big time).
    – Did
    Jul 28 at 13:10











  • If you add the entries of the matrix then you get $2$ so they cannot correspond with probabilities of mutually exclusive events. Most probably the entries are conditional probabilities.
    – Vera
    Jul 28 at 13:12
















First mistake: $Pr[X=0,Y=1]neepsilon$, actually, $$Pr[X=0,Y=1]=Pr[X=0]Pr[Y=1mid X=0]$$ with probably $Pr[X=0]=1-p$ and $\Pr[Y=1mid X=0]=epsilon$ (except that you mangled the notations big time).
– Did
Jul 28 at 13:10





First mistake: $Pr[X=0,Y=1]neepsilon$, actually, $$Pr[X=0,Y=1]=Pr[X=0]Pr[Y=1mid X=0]$$ with probably $Pr[X=0]=1-p$ and $\Pr[Y=1mid X=0]=epsilon$ (except that you mangled the notations big time).
– Did
Jul 28 at 13:10













If you add the entries of the matrix then you get $2$ so they cannot correspond with probabilities of mutually exclusive events. Most probably the entries are conditional probabilities.
– Vera
Jul 28 at 13:12





If you add the entries of the matrix then you get $2$ so they cannot correspond with probabilities of mutually exclusive events. Most probably the entries are conditional probabilities.
– Vera
Jul 28 at 13:12
















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