We consider a noisy communication link in which the message is encoded into binary digits (0,1) (bits) before being transmitted. We will denote the length of the encoded message by n. Since the channel is noisy, the bits can get flipped; a 0 to 1 or a 1 to 0. We will assume that each bit is flipped independently with probability p. In order to be able to detect that the received message is in error, a simple method is to add a parity bit at the transmitter. There can two types of parity - even parity and odd parity. If even (odd) parity is used , the parity bit is set such that total number of 1s in the encoded message is even (odd). For the sake of this problem, we will only consider even parity. Here are a few examples 1) message: 0001 + parity bit: 1 --> encoded message: 00011 2) message: 0101 + parity bit: 0 --> encoded message: 01010 3) message: 0111 + parity bit: 1 --> encoded message: 01111 At the receiver, the parity bit is computed from the message bits. If the received parity bit does not match the computed parity bit, then the message is flagged to be in error. For example, if the received message 1 is 10011, then it is in error. 1. Suppose n = 7 and p = 0.1, what is the probability that the received message has errors which go undetected? 2. For general n and p, write down an expression (as a sum) for the probability that the received message has errors which go undetected.
We consider a noisy communication link in which the message is encoded into binary digits (0,1) (bits)
before being transmitted. We will denote the length of the encoded message by n. Since the channel is
noisy, the bits can get flipped; a 0 to 1 or a 1 to 0. We will assume that each bit is flipped independently with
probability p. In order to be able to detect that the received message is in error, a simple method is to add a
parity bit at the transmitter. There can two types of parity - even parity and odd parity. If even (odd) parity
is used , the parity bit is set such that total number of 1s in the encoded message is even (odd). For the sake
of this problem, we will only consider even parity. Here are a few examples
1) message: 0001 + parity bit: 1 --> encoded message: 00011
2) message: 0101 + parity bit: 0 --> encoded message: 01010
3) message: 0111 + parity bit: 1 --> encoded message: 01111
At the receiver, the parity bit is computed from the message bits. If the received parity bit does not match
the computed parity bit, then the message is flagged to be in error. For example, if the received message 1 is
10011, then it is in error.
1. Suppose n = 7 and p = 0.1, what is the probability that the received message has errors which go
undetected?
2. For general n and p, write down an expression (as a sum) for the probability that the received message
has errors which go undetected.
ANSWER 1:
- The probability that the received message has errors which go undetected can be calculated by finding the probability of having an even number of errors in the transmitted message. In this case, with n = 7 bits and p = 0.1, the probability of having an even number of errors in the message can be calculated as the sum of the probabilities of having 0, 2, 4, 6 errors, since the parity bit requires an even number of errors to go undetected. The probability of having i errors can be calculated as (7 choose i) * (p^i) * ((1-p)^(7-i)), where (7 choose i) is the binomial coefficient. So the probability that the received message has errors which go undetected is:
(7 choose 0) * (p^0) * ((1-p)^7) + (7 choose 2) * (p^2) * ((1-p)^5) + (7 choose 4) * (p^4) * ((1-p)^3) + (7 choose 6) * (p^6) * ((1-p)^1)
To calculate the probability that the received message has errors that go undetected, we need to consider the possible number of bit flips. If 0 bits are flipped, the received message will match the transmitted message, so the probability of no errors is (1-p)^n. If 1 bit is flipped, the received message will have a different parity bit from the transmitted message, so it will be flagged as an error. There are n possible bit positions for the flip, so the probability of 1 undetected error is np(1-p)^(n-1). If 2 or more bits are flipped, the received message will have a different number of ones from the transmitted message, so it will be flagged as an error. The probability of 2 undetected errors is n(n-1)/2 * p^2 * (1-p)^(n-2). We can add up the probabilities for each number of undetected errors to get the total probability of undetected errors:
P_undetected = (1-p)^n + np(1-p)^(n-1) + n(n-1)/2 * p^2 * (1-p)^(n-2) + ...
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