q-ary symmetric channel error probability
$begingroup$
Why is it valid to assume that $0< p < (q-1)/q$, where $p$ is the probability of symbol error. When $(q-1) / q< p <1$, why are we able to flip it? We are not in binary channel, so we are not able to flip all 1 to 0, all 0 to 1, how can we do that?
My thought is to randomly pick any other symbol at an equal probability
abstract-algebra coding-theory
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
asked Jan 15 at 15:59
JoeJoe
223
$endgroup$
add a comment |
$begingroup$
Why is it valid to assume that $0< p < (q-1)/q$, where $p$ is the probability of symbol error. When $(q-1) / q< p <1$, why are we able to flip it? We are not in binary channel, so we are not able to flip all 1 to 0, all 0 to 1, how can we do that?
My thought is to randomly pick any other symbol at an equal probability
abstract-algebra coding-theory
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
asked Jan 15 at 15:59
JoeJoe
223
$endgroup$
1
$begingroup$
Perhaps you should add some more context (details of the channel, and where you've read that assumption)
$endgroup$
– leonbloy
Jan 15 at 20:52
$begingroup$
Since $q$ is an integer that has value at least $2$ and $p$ is a nonzero probability, the inequalities $$0 < p < q-1/q$$ are trivially true. What are you really trying to ask?
$endgroup$
– Dilip Sarwate
Jan 17 at 15:02
$begingroup$
@DilipSarwate The most likely explanation for that strange thing you observed was that when an earlier editor TeXified the post, they didn't think about this, and didn't add the parentheses I just added. When written in ASCII it is a bit ambiguos whether the parens were intended. My educated guess is that they were intended :-)
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:04
$begingroup$
Even so, the question is a bit unclear. I'm seconding leonbloy's request.
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:05
add a comment |
$begingroup$
Why is it valid to assume that $0< p < (q-1)/q$, where $p$ is the probability of symbol error. When $(q-1) / q< p <1$, why are we able to flip it? We are not in binary channel, so we are not able to flip all 1 to 0, all 0 to 1, how can we do that?
My thought is to randomly pick any other symbol at an equal probability
abstract-algebra coding-theory
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
asked Jan 15 at 15:59
JoeJoe
223
$endgroup$
Why is it valid to assume that $0< p < (q-1)/q$, where $p$ is the probability of symbol error. When $(q-1) / q< p <1$, why are we able to flip it? We are not in binary channel, so we are not able to flip all 1 to 0, all 0 to 1, how can we do that?
My thought is to randomly pick any other symbol at an equal probability
abstract-algebra coding-theory
abstract-algebra coding-theory
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
asked Jan 15 at 15:59
JoeJoe
223
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
asked Jan 15 at 15:59
JoeJoe
223
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
edited Jan 17 at 23:02
Jyrki Lahtonen
109k13169372
109k13169372
asked Jan 15 at 15:59
JoeJoe
223
asked Jan 15 at 15:59
JoeJoe
223
asked Jan 15 at 15:59
JoeJoe
223
223
1
$begingroup$
Perhaps you should add some more context (details of the channel, and where you've read that assumption)
$endgroup$
– leonbloy
Jan 15 at 20:52
$begingroup$
Since $q$ is an integer that has value at least $2$ and $p$ is a nonzero probability, the inequalities $$0 < p < q-1/q$$ are trivially true. What are you really trying to ask?
$endgroup$
– Dilip Sarwate
Jan 17 at 15:02
$begingroup$
@DilipSarwate The most likely explanation for that strange thing you observed was that when an earlier editor TeXified the post, they didn't think about this, and didn't add the parentheses I just added. When written in ASCII it is a bit ambiguos whether the parens were intended. My educated guess is that they were intended :-)
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:04
$begingroup$
Even so, the question is a bit unclear. I'm seconding leonbloy's request.
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:05
add a comment |
1
$begingroup$
Perhaps you should add some more context (details of the channel, and where you've read that assumption)
$endgroup$
– leonbloy
Jan 15 at 20:52
$begingroup$
Since $q$ is an integer that has value at least $2$ and $p$ is a nonzero probability, the inequalities $$0 < p < q-1/q$$ are trivially true. What are you really trying to ask?
$endgroup$
– Dilip Sarwate
Jan 17 at 15:02
$begingroup$
@DilipSarwate The most likely explanation for that strange thing you observed was that when an earlier editor TeXified the post, they didn't think about this, and didn't add the parentheses I just added. When written in ASCII it is a bit ambiguos whether the parens were intended. My educated guess is that they were intended :-)
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:04
$begingroup$
Even so, the question is a bit unclear. I'm seconding leonbloy's request.
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:05
1
1
$begingroup$
Perhaps you should add some more context (details of the channel, and where you've read that assumption)
$endgroup$
– leonbloy
Jan 15 at 20:52
$begingroup$
Perhaps you should add some more context (details of the channel, and where you've read that assumption)
$endgroup$
– leonbloy
Jan 15 at 20:52
$begingroup$
Since $q$ is an integer that has value at least $2$ and $p$ is a nonzero probability, the inequalities $$0 < p < q-1/q$$ are trivially true. What are you really trying to ask?
$endgroup$
– Dilip Sarwate
Jan 17 at 15:02
$begingroup$
Since $q$ is an integer that has value at least $2$ and $p$ is a nonzero probability, the inequalities $$0 < p < q-1/q$$ are trivially true. What are you really trying to ask?
$endgroup$
– Dilip Sarwate
Jan 17 at 15:02
$begingroup$
@DilipSarwate The most likely explanation for that strange thing you observed was that when an earlier editor TeXified the post, they didn't think about this, and didn't add the parentheses I just added. When written in ASCII it is a bit ambiguos whether the parens were intended. My educated guess is that they were intended :-)
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:04
$begingroup$
@DilipSarwate The most likely explanation for that strange thing you observed was that when an earlier editor TeXified the post, they didn't think about this, and didn't add the parentheses I just added. When written in ASCII it is a bit ambiguos whether the parens were intended. My educated guess is that they were intended :-)
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:04
$begingroup$
Even so, the question is a bit unclear. I'm seconding leonbloy's request.
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:05
$begingroup$
Even so, the question is a bit unclear. I'm seconding leonbloy's request.
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In a $q$-ary channel with input and output alphabet ${0,1, ldots, q-1}$, consider the transition probabilities $p_{i,j}$ that tell us the probability that a transmitted $i$ is received as a $j$. Note that for each $i$, $sum_{j=0}^{q-1} p_{i,j} = 1$ since something must be received when $i$ is transmitted. In a $q$-ary symmetric channel, all the $p_{i,j}$ have the same value $hat{p}$ for all $ineq j$ while all the $p_{i,i}$ have the same value $alpha$. Thus,
$$hat{p}(q-1) + alpha = 1.$$ The OP defines the probability of symbol error as $p = hat{p}(q-1)$ -- it is the probability that an incorrect symbol is received -- and so we have that $p+alpha = 1$.
Now, of the $q$ numbers $p_{i,0}, p_{i,0}, ldots, p_{i,q-1}$ whose sum is $1$, at least one, hopefully $p_{i,i} = alpha$, must be larger than $frac 1q$ because if none of the $p_{I,j}$ exceed $frac 1q$, then they must all will equal $frac 1q$ and the channel will be useless: the output will be independent of the input. So, since $p+alpha = 1$ and $alpha > frac 1q$, we see that $$p + frac 1q < 1 implies p < 1 - frac 1q = frac{q-1}{q}.$$
So why does this argument not work when $q = 2$?? well for the binary symmetric channel, if $p > frac 12 = frac{q-1}{q}bigvert_{q=2}$, as mathematicians we can just cuss the engineer for mis-labeling the channel outputs and tell him to flip the output bits, thus converting the binary symmetric channel with crossover probability $p > frac 12$ into a binary symmetric channel with crossover probability $1-p < frac 12$. Unfortunately such "flipping" does not work when $q>2$
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
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add a comment |
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$begingroup$
In a $q$-ary channel with input and output alphabet ${0,1, ldots, q-1}$, consider the transition probabilities $p_{i,j}$ that tell us the probability that a transmitted $i$ is received as a $j$. Note that for each $i$, $sum_{j=0}^{q-1} p_{i,j} = 1$ since something must be received when $i$ is transmitted. In a $q$-ary symmetric channel, all the $p_{i,j}$ have the same value $hat{p}$ for all $ineq j$ while all the $p_{i,i}$ have the same value $alpha$. Thus,
$$hat{p}(q-1) + alpha = 1.$$ The OP defines the probability of symbol error as $p = hat{p}(q-1)$ -- it is the probability that an incorrect symbol is received -- and so we have that $p+alpha = 1$.
Now, of the $q$ numbers $p_{i,0}, p_{i,0}, ldots, p_{i,q-1}$ whose sum is $1$, at least one, hopefully $p_{i,i} = alpha$, must be larger than $frac 1q$ because if none of the $p_{I,j}$ exceed $frac 1q$, then they must all will equal $frac 1q$ and the channel will be useless: the output will be independent of the input. So, since $p+alpha = 1$ and $alpha > frac 1q$, we see that $$p + frac 1q < 1 implies p < 1 - frac 1q = frac{q-1}{q}.$$
So why does this argument not work when $q = 2$?? well for the binary symmetric channel, if $p > frac 12 = frac{q-1}{q}bigvert_{q=2}$, as mathematicians we can just cuss the engineer for mis-labeling the channel outputs and tell him to flip the output bits, thus converting the binary symmetric channel with crossover probability $p > frac 12$ into a binary symmetric channel with crossover probability $1-p < frac 12$. Unfortunately such "flipping" does not work when $q>2$
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
$endgroup$
add a comment |
$begingroup$
In a $q$-ary channel with input and output alphabet ${0,1, ldots, q-1}$, consider the transition probabilities $p_{i,j}$ that tell us the probability that a transmitted $i$ is received as a $j$. Note that for each $i$, $sum_{j=0}^{q-1} p_{i,j} = 1$ since something must be received when $i$ is transmitted. In a $q$-ary symmetric channel, all the $p_{i,j}$ have the same value $hat{p}$ for all $ineq j$ while all the $p_{i,i}$ have the same value $alpha$. Thus,
$$hat{p}(q-1) + alpha = 1.$$ The OP defines the probability of symbol error as $p = hat{p}(q-1)$ -- it is the probability that an incorrect symbol is received -- and so we have that $p+alpha = 1$.
Now, of the $q$ numbers $p_{i,0}, p_{i,0}, ldots, p_{i,q-1}$ whose sum is $1$, at least one, hopefully $p_{i,i} = alpha$, must be larger than $frac 1q$ because if none of the $p_{I,j}$ exceed $frac 1q$, then they must all will equal $frac 1q$ and the channel will be useless: the output will be independent of the input. So, since $p+alpha = 1$ and $alpha > frac 1q$, we see that $$p + frac 1q < 1 implies p < 1 - frac 1q = frac{q-1}{q}.$$
So why does this argument not work when $q = 2$?? well for the binary symmetric channel, if $p > frac 12 = frac{q-1}{q}bigvert_{q=2}$, as mathematicians we can just cuss the engineer for mis-labeling the channel outputs and tell him to flip the output bits, thus converting the binary symmetric channel with crossover probability $p > frac 12$ into a binary symmetric channel with crossover probability $1-p < frac 12$. Unfortunately such "flipping" does not work when $q>2$
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
$endgroup$
add a comment |
$begingroup$
In a $q$-ary channel with input and output alphabet ${0,1, ldots, q-1}$, consider the transition probabilities $p_{i,j}$ that tell us the probability that a transmitted $i$ is received as a $j$. Note that for each $i$, $sum_{j=0}^{q-1} p_{i,j} = 1$ since something must be received when $i$ is transmitted. In a $q$-ary symmetric channel, all the $p_{i,j}$ have the same value $hat{p}$ for all $ineq j$ while all the $p_{i,i}$ have the same value $alpha$. Thus,
$$hat{p}(q-1) + alpha = 1.$$ The OP defines the probability of symbol error as $p = hat{p}(q-1)$ -- it is the probability that an incorrect symbol is received -- and so we have that $p+alpha = 1$.
Now, of the $q$ numbers $p_{i,0}, p_{i,0}, ldots, p_{i,q-1}$ whose sum is $1$, at least one, hopefully $p_{i,i} = alpha$, must be larger than $frac 1q$ because if none of the $p_{I,j}$ exceed $frac 1q$, then they must all will equal $frac 1q$ and the channel will be useless: the output will be independent of the input. So, since $p+alpha = 1$ and $alpha > frac 1q$, we see that $$p + frac 1q < 1 implies p < 1 - frac 1q = frac{q-1}{q}.$$
So why does this argument not work when $q = 2$?? well for the binary symmetric channel, if $p > frac 12 = frac{q-1}{q}bigvert_{q=2}$, as mathematicians we can just cuss the engineer for mis-labeling the channel outputs and tell him to flip the output bits, thus converting the binary symmetric channel with crossover probability $p > frac 12$ into a binary symmetric channel with crossover probability $1-p < frac 12$. Unfortunately such "flipping" does not work when $q>2$
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
$endgroup$
In a $q$-ary channel with input and output alphabet ${0,1, ldots, q-1}$, consider the transition probabilities $p_{i,j}$ that tell us the probability that a transmitted $i$ is received as a $j$. Note that for each $i$, $sum_{j=0}^{q-1} p_{i,j} = 1$ since something must be received when $i$ is transmitted. In a $q$-ary symmetric channel, all the $p_{i,j}$ have the same value $hat{p}$ for all $ineq j$ while all the $p_{i,i}$ have the same value $alpha$. Thus,
$$hat{p}(q-1) + alpha = 1.$$ The OP defines the probability of symbol error as $p = hat{p}(q-1)$ -- it is the probability that an incorrect symbol is received -- and so we have that $p+alpha = 1$.
Now, of the $q$ numbers $p_{i,0}, p_{i,0}, ldots, p_{i,q-1}$ whose sum is $1$, at least one, hopefully $p_{i,i} = alpha$, must be larger than $frac 1q$ because if none of the $p_{I,j}$ exceed $frac 1q$, then they must all will equal $frac 1q$ and the channel will be useless: the output will be independent of the input. So, since $p+alpha = 1$ and $alpha > frac 1q$, we see that $$p + frac 1q < 1 implies p < 1 - frac 1q = frac{q-1}{q}.$$
So why does this argument not work when $q = 2$?? well for the binary symmetric channel, if $p > frac 12 = frac{q-1}{q}bigvert_{q=2}$, as mathematicians we can just cuss the engineer for mis-labeling the channel outputs and tell him to flip the output bits, thus converting the binary symmetric channel with crossover probability $p > frac 12$ into a binary symmetric channel with crossover probability $1-p < frac 12$. Unfortunately such "flipping" does not work when $q>2$
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
answered Jan 18 at 4:31
Dilip SarwateDilip Sarwate
19.1k13076
19.1k13076
add a comment |
add a comment |
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$begingroup$
Perhaps you should add some more context (details of the channel, and where you've read that assumption)
$endgroup$
– leonbloy
Jan 15 at 20:52
$begingroup$
Since $q$ is an integer that has value at least $2$ and $p$ is a nonzero probability, the inequalities $$0 < p < q-1/q$$ are trivially true. What are you really trying to ask?
$endgroup$
– Dilip Sarwate
Jan 17 at 15:02
$begingroup$
@DilipSarwate The most likely explanation for that strange thing you observed was that when an earlier editor TeXified the post, they didn't think about this, and didn't add the parentheses I just added. When written in ASCII it is a bit ambiguos whether the parens were intended. My educated guess is that they were intended :-)
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:04
$begingroup$
Even so, the question is a bit unclear. I'm seconding leonbloy's request.
$endgroup$
– Jyrki Lahtonen
Jan 17 at 23:05