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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 $-\frac1{2}<Y < \frac{1}2$.

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?

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1 Answer 1

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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.

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  • $\begingroup$ I noticed that in the third term you excluded P(X=x) is that intended? $\endgroup$ Oct 30, 2020 at 2:17
  • $\begingroup$ @Fer-de-lance : no, it's a typo. Amended $\endgroup$
    – tommik
    Oct 30, 2020 at 4:21
  • $\begingroup$ 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? $\endgroup$ Oct 30, 2020 at 4:36
  • $\begingroup$ @Fer-de-lance : the total errer results to me 53.5%. Check my solution $\endgroup$
    – tommik
    Oct 30, 2020 at 5:11
  • $\begingroup$ Thank you, I got it and don't mind my previous comment my assumption was correct. $\endgroup$ Oct 30, 2020 at 5:11

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