# Probability of Error (Noise in Channel)

In this channel a random input $$X$$ is chosen from $$\{-1,0,1\}$$ with equal probabilities and an outputs a value $$Y = X + Z$$. Where $$Z \sim N(0,4)$$ is the noise and is independent from $$X$$ with $$\mu=0$$ and $$\sigma=2$$. With the noise, the channel decides the final output $$\bar Y$$ with the following conditions:

• $$\bar Y = 1$$ if $$Y > \frac{1}2$$.
• $$\bar Y = -1$$ if $$Y < -\frac{1}2$$.
• $$\bar Y = 0$$ if $$-\frac1{2}.

Find the probability of making an error.

Attempt:

What I understand, to calculate the error I need to:

$$P_{err}(X=0)= P(\bar Y=1)P(\bar Y = 1| X = 0) + P(\bar Y=-1)P(\bar Y = -1| X = 0)$$ $$P_{err}(X=1)= P(\bar Y=0)P(\bar Y = 0| X = 1) + P(\bar Y=-1)P(\bar Y = -1| X = 1)$$ $$P_{err}(X=-1)= P(\bar Y=0)P(\bar Y = 0| X = -1) + P(\bar Y=1)P(\bar Y = 1| X = -1)$$

Then get: $$P_{err} = P_{err}(X=0) + P_{err}(X=-1) + P_{err}(X=1)$$

And to calculate the conditional probabilities say $$P(\bar Y = 1 | X = 0)$$ I did:

\begin{aligned} P(\bar Y = 1| X = 0) &= P(Y > \frac{1}2 | X = 0) \\ & = P(X+Z>\frac{1}2|X=0) \\ &= P(Z>\frac{1}2 |X=0) \\ &= P(Z>\frac{1}2) P(X=0) \\ &= \frac{1}3P(Z>\frac{1}2) \end{aligned}

Where $$P(X=x)=P(\bar Y=y) = \frac{1}3$$ since they have equal probabilities. With this I found that $$P_{err}(1) = P_{err}(-1) = 0.04459$$ and $$P_{err}(0) = 0.08918$$

The total error I got was:

$$P_{err}(total) = P_{err}(-1) + P_{err}(0) + P_{err}(1) = 0.17836$$

Is this right?

First I do not understand the probability for $$X$$ to be respectively $$\{-1;0;1\}$$.

For the sake of simplicity let's suppose that these probability are equal, say the rv $$X$$ is Discrete Uniform in the given support.

Thus, the error is the following

$$ERR=\frac{1}{3}\mathbb{P}\Bigg\{Y>-\frac{1}{2}|X=-1\Bigg\}+\frac{1}{3}\mathbb{P}\Bigg\{|Y|\geq\frac{1}{2}|X=0\Bigg\}+\frac{1}{3}\mathbb{P}\Bigg\{Y<\frac{1}{2}|X=1\Bigg\}$$

The probability $$\mathbb{P}[Y \in A|X=x]$$ is very simple because the realization of $$X$$ just modify the location parameter of Z.

$$\frac{1}{3}\mathbb{P}[Y>-0.5|Y\sim N(-1;4)]+ \frac{1}{3}\mathbb{P}[|Y|>0.5|Y\sim N(0;4)]+\frac{1}{3}\mathbb{P}[Y<0.5|Y\sim N(1;4)]$$

$$\frac{1}{3}\Phi(-0.25)+\frac{2}{3}\Phi(-0.25)+\frac{1}{3}\Phi(-0.25)=\frac{4}{3}\cdot 0.4013\approx 0.5351$$

If $$X$$ is not uniform distributed, modify its probabilities accordingly

EDITING AND CORRECTION

1. If you want to write $$\{0;1\}$$ you have to write a backslash \ before the parenthesis (I edited your text)

2. You wrote that the output is

$$\bbox[5px,border:2px solid red] { \overline{Y}=-1 \text{ if } Y<\frac{1}{2} \ }$$

I corrected it in $$\bbox[5px,border:2px solid lightgreen] { \overline{Y}=-1 \text{ if } Y<-\frac{1}{2} \ }$$

Since you wrote did not make sense in the gobal context.

• I noticed that in the third term you excluded P(X=x) is that intended? Oct 30, 2020 at 2:17
• @Fer-de-lance : no, it's a typo. Amended Oct 30, 2020 at 4:21
• In my calculation, Since X and Z are independent then I assumed that P(Y<y|X=x) = P(X+Z<y | X=x) = P(Z+x < y). Is this reasoning correct? Oct 30, 2020 at 4:36
• @Fer-de-lance : the total errer results to me 53.5%. Check my solution Oct 30, 2020 at 5:11
• Thank you, I got it and don't mind my previous comment my assumption was correct. Oct 30, 2020 at 5:11