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Given random varaible X with distribution

$$ \begin{cases} \mathbb{P}\left(X=1\right)=\alpha\\ \mathbb{P}\left(X=-1\right)=1-\alpha \end{cases} $$

Where $ \alpha $ is a given parameter($X$ is a binary signal), and noise $Z$ with normal distribution $ Z\sim\mathcal{N}\left(0,\sigma^{2}\right) $, we define the random varaible $$ Y=X+Z $$

(We assume $X$,$Z$ are independent random variables)

Now, in order to estimate $ X $ by the samples $ Y $, we decide that if $ Y \geq \gamma $ then $ X=1 $, and if $ Y< \gamma $ then $ X=-1 $, where $ \gamma $ is some value.

Now, the probability of a mistake is given by:

$$ \mathbb{P}_{\text{error}}\left(\gamma\right)=\left(1-\alpha\right)-\left(1-\alpha\right)F_{Z}\left(\gamma+1\right)+\alpha F_{Z}\left(\gamma-1\right) $$

And I was supposed to find $ \gamma $ such that the error would be minimal. I did the calculations and proved that this probability accepts its minimal value for $$ \gamma=\frac{1}{2}\sigma^{2}\ln\left(\frac{1-\alpha}{\alpha}\right) $$

Now I'm asked to explain why this result makes sense. Putting the calculations aside, Im not sure why would this particular $\gamma $ lead to minimal error. I'd really appreciate an idea or explanation.

Other results I have calculated and may be important for the explanation

The density of $ Y $ : $$ f_{Y}\left(y\right)=\alpha\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{\left(y-1\right)^{2}}{2\sigma^{2}}}+\left(1-\alpha\right)\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{\left(y+1\right)^{2}}{2\sigma^{2}}} $$

Explanation for the probability of the eroor

The cases where we have determined $X$ and got a false result,So the cases are exactly where $ Y\geq\gamma\cap X=-1 $ or $ Y<\gamma\cap X=1 $.

So that

\begin{align} \mathbb{P}_{\text{error}} &=\mathbb{P}\left(Y\geq\gamma\cap X=-1\right)+\mathbb{P}\left(Y<\gamma\cap X=1\right) \\ &=\mathbb{P}\left(X+Z\geq\gamma,X=-1\right)+\mathbb{P}\left(X+Z<\gamma,X=1\right)\\ & =\mathbb{P}\left(Z\geq\gamma+1,X=-1\right)+\mathbb{P}\left(Z<\gamma-1,X=1\right) \\ &\underset{X,Z\text{ ind.}}{=}\mathbb{P}\left(X=-1\right)\mathbb{P}\left(Z\geq\gamma+1\right)+\mathbb{P}\left(X=1\right)\mathbb{P}\left(Z<\gamma-1\right) \\ &=\left(1-\alpha\right)\left(1-F_{Z}\left(\gamma+1\right)\right)+\alpha F_{Z}\left(\gamma-1\right) \\ \end{align}

Claculation of the minimum I'll differntiate with respect to $\gamma $ and then find roots:

$ \frac{d}{d\gamma}\mathbb{P}_{\text{error}}=-\left(1-\alpha\right)f_{Z}\left(\gamma+1\right)+\alpha f_{Z}\left(\gamma-1\right)=0 $

$ \alpha f_{Z}\left(\gamma-1\right)=\left(1-\alpha\right)f_{Z}\left(\gamma+1\right) $

$ \frac{\alpha}{1-\alpha}=\frac{f_{Z}\left(\gamma+1\right)}{f_{Z}\left(\gamma-1\right)}=\frac{e^{-\frac{\left(\gamma+1\right)^{2}}{2\sigma^{2}}}}{e^{-\frac{\left(\gamma-1\right)^{2}}{2\sigma^{2}}}}=e^{-\frac{\left(\gamma+1\right)^{2}}{2\sigma^{2}}+\frac{\left(\gamma-1\right)^{2}}{2\sigma^{2}}}=e^{\frac{-2\gamma-2\gamma}{2\sigma^{2}}}=e^{-\frac{2\gamma}{\sigma^{2}}} $

So

$ \frac{2\gamma}{\sigma^{2}}=\ln(\frac{1-\alpha}{\alpha}) $

and we have

$ \gamma=\frac{1}{2}\sigma^{2}\ln\left(\frac{1-\alpha}{\alpha}\right) $

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    $\begingroup$ can you explain your formula for the error probability? I'm a bit confused how you end up with $\gamma\pm1$ in the argument of the CDF! $\endgroup$ Dec 5 '21 at 13:49
  • $\begingroup$ @MarcusMüller Sure. $\endgroup$
    – FreeZe
    Dec 5 '21 at 14:33
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    $\begingroup$ Tried to align things a bit better, can you check whether I made a mistake when I inserted a = at the beginning of the last line? $\endgroup$ Dec 5 '21 at 15:06
  • $\begingroup$ @MarcusMüller Looks fine. $\endgroup$
    – FreeZe
    Dec 5 '21 at 15:07
  • $\begingroup$ The "why" should drop out of the actual calculation of $\gamma$, yet you do not show that calculation -- please add it in. $\endgroup$
    – TimWescott
    Dec 5 '21 at 16:20
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Now I'm asked to explain why this result makes sense.

That's sort of a subjective question. One way to interpret this is to pick some cases where you already know the answer

  1. $\alpha =.5$ In this case everything is symmetric so your answer should be $\gamma = 0$. Check
  2. $\alpha = 1$. In this case your random variable is always 1 and you want $Y > \gamma$ for all values of $Y$. So the best choice for alpha is $\gamma = -\infty$ . Check.
  3. $\alpha = 0$. In this case your random variable is always 0 and you want $Y < \gamma$ for all values of $Y$. So the best choice for alpha is $\gamma = +\infty$ . Check.
  4. Low noise energy. If $\sigma$ is small, the noise has very little impact on your observation, so putting $\gamma = 0$ symmetrically between -1 and +1 makes sense. Check.
  5. It should be symmetric in $\alpha$ , i.e. $\gamma(\alpha) = -\gamma(1-\alpha)$. Check.

For anything else you can try to make up a plausibility argument or just run the numbers manually for one case.

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