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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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?







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












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?







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?







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?









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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
















active

oldest

votes











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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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%2f2865236%2fbinary-symmetric-channel-calculating-problemthe-answer-of-two-method-should-be%23new-answer', 'question_page');

);

Post as a guest



































active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865236%2fbinary-symmetric-channel-calculating-problemthe-answer-of-two-method-should-be%23new-answer', 'question_page');

);

Post as a guest













































































Comments

Popular posts from this blog

Color the edges and diagonals of a regular polygon

Relationship between determinant of matrix and determinant of adjoint?

What is the equation of a 3D cone with generalised tilt?