# Probability error - binary message

Please, I need some support (not solution) and input to know if the way is right to go on.

Consider the communication of binary messages in a transmission medium. Any message sent is selected from two possible symbols, $0$ or $1$. Each symbol occurs with equal probability. It is also known that any numerical value sent on that channel is subjected to distortion. If a value $x$ is transmitted, the $y$ value is received at its destination, described by $y = x + n$, where n represents a random variable that additive noise is independent of $x$. The noise has a normal distribution with parameters $σ^2 = 4$ and $\mu = 0$.

1. Suppose the transmitter encodes the symbol $0$ with the value $x = -2$ and $1$ with the symbol value $x = 2$. At the destination, the received message is decoded according to the following rules:
2. not yet ....

• If $y ≥ 0$, one concludes that the symbol $1$ was sent.
• If $y <0$, conclude the symbol $0$ was sent.

Q: Determine the probability of error for this schema encoding / decoding.

So, I know that....

Bit error probability: $P(x=0|y=1) ~\&~ P(x=1|y=0)$. The probabilities of transmitting each signal $(0,1)$ are equal (i.e., $1/2$).

$y_t = x_t + n_t$ where $x_t$ is a free signal of noise.

Assuming that the noise has a Gaussian distribution.

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Welcome to the site, @Rykardu. I have edited your question using the markup supported by CV to make it easier to read. Please make sure it still says what you want. Note that #2 (previously "(b)") doesn't make sense to me; & I don't know what "$e$" means in your bit error probability, is it "="? I also added the 'homework' tag, which indicates you want hints, not a full solution. – gung Nov 4 '12 at 13:38
Dear Gung, Tks for your note. Sorry, letter (b) is another question. (If "Y" .....) are information from the original question and not from de letter (b) that will be answer after (a). Bit error probability .... is the beginning of my reasoning "e" = "and" about tag that´s ok. tks – rykardu Nov 7 '12 at 18:19
Thanks for letting me know, @rykardu. I tweaked the question. Let me know if it doesn't say what you want to be asking. – gung Nov 7 '12 at 18:26

Let $E$ be the event that you make an error in reading the message.

$$P\left\{E\right\}=P\left\{\left(E\cap\left(x=-2\right)\right)\cup\left(E\cap\left(x=2\right)\right)\right\}$$

The events $\left(E\cap\left(x=-2\right)\right)$ and $\left(E\cap\left(x=2\right)\right)$ are mutually exclusive because $x$ may only take on a single value. Now we have that

$$P\left\{E\right\}=P\left\{E\cap\left(x=-2\right)\right\}+P\left\{E\cap\left(x=2\right)\right\}$$

Conditional probability tells us that $$P\left\{E\right\}=P\left\{E\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{E\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$

You should be able to rewrite the last probability statement in terms of the observed values $y$ rather than the event $E$, and by extension the probability statement may be written in terms of only $x$ and $n$. That should lead to a simple answer.

When $x=-2$, an error is made if $y\geq0$. When $x=2$, an error is if $y<0$. The probability statement becomes $$P\left\{E\right\}=P\left\{y\geq0\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{y<0\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$

We know that $y=x+n$, so change the $y$'s accordingly.

$$P\left\{E\right\}=P\left\{x+n\geq0\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{x+n<0\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$

When $x=-2$, $x+n\geq0$ is exactly the same as $n\geq2$. When $x=2$, $x+n<0$ is exactly the same as $n<-2$.

$$P\left\{E\right\}=P\left\{n\geq2\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{n<-2\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$

We also know that $x$ an $n$ are independent, so the conditional probabilities become marginals.

$$P\left\{E\right\}=P\left\{n\geq2\right\}P\left\{x=-2\right\}+P\left\{n<-2\right\}P\left\{x=2\right\}$$

You know that $n\sim N(0,4)$ and $x$ is equally likely to be either message.

\begin{align} P\left\{E\right\}&=P\left\{n\geq2\right\}P\left\{x=-2\right\}+P\left\{n<-2\right\}P\left\{x=2\right\}\\ &=P\left\{\frac{n-0}{2}\geq\frac{2-0}{2}\right\}P\left\{x=-2\right\}+P\left\{\frac{n-0}{2}<\frac{-2-0}{2}\right\}P\left\{x=2\right\}\\ &=\left(1-\Phi(1)\right)\times\frac{1}{2}+\Phi(-1)\times\frac{1}{2}\\ \end{align}

Using the $68-95-99.7$ rule, the probability is $\approx0.16/2+0.16/2=0.16$.

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Dear Max, tks for y help. I understand your information but I could not see how to rewrite the probability. – rykardu Nov 10 '12 at 10:57
I edited my answer. – Max Nov 10 '12 at 14:27
In fact, it is worth noting that in the right side of the last displayed equation $\Phi(-1) = 1 - \Phi(1)$ and so $P\{E\} = 1 - \Phi(1)$ even if the messages are not equally likely, as long as the maximum-likelihood decision rule (item 2 in the OP's question) is being used. Bayesians would use a different threshold than $0$ when the messages are not equally likely and achieve a smaller average error probability, but for the ML decision rule with symmetric noise density, the error probability is just the one-sided tail probability. – Dilip Sarwate Nov 10 '12 at 16:07
@DilipSarwate That's a very good point. – Max Nov 10 '12 at 16:10
Each symbol occurs with equal probability. The next question (2) for this original problema request is: "To reduce the probability of error, the following modifications are made...." I will edit the full question next time. Tks Dilip. – rykardu Nov 11 '12 at 14:49