Finding optimal decision threshold for binary comm system

Question

I'm trying to solve the following exercise: Given a system that transmits bit $b$ with probability $p_b$ and $-b$ with $p_{-b}$ and the transmission is degraded by AWGN $N(0,\sigma^2)$. What is the optimal decision threshold $\mu$ to minimize the probability of error? Use Q functions to express the probability of error.

My approach:

$P_e = P_b P(-b|b) + P_{-b} P(b|-b)$, where $P(-b|b)$ means probability of detecting $-b$ given that $b$ was transmitted.

$P(-b|b)$ is the left tail of the distribution ($\Phi$) since I need to receive a value smaller than $\mu$ after sending $b$, but since the exercise asks in $Q$ -> $Q(-(\frac{\mu-b}{\sigma}))$

$P(b|-b)$ is already the right tail as I need to get something greater than $\mu$ after sending $-b$, so $Q(\frac{\mu+b}{\sigma})$.

To find the optimal $\mu$, I took the derivate of $P_e$ with respect to $\mu$ and set it equal to 0. To facilitate notation, I will use $X = \frac{-(\mu-b)}{\sigma}$ and $Y = \frac{\mu+b}{\sigma}$.

Now I have $P_1 e^{-\frac{X^2}{2}} = -P_{-1} e^{-\frac{Y^2}{2}}$.

After some manipulation and taking $ln$ on both sides: $ln(\frac{p_1}{p_{-1}}) = \frac{\frac{Y^2}{2}}{\frac{-X^2}{2}} = \frac{2\mu b}{\sigma^2}$.

So, $\mu = ln(\frac{p_1}{p_{-1}}) \frac{\sigma^2}{2b}$ but the solutions I have don't have the $\sigma^2$ term, so I cannot find where I made a mistake.

As for the derivative of the $Q$ function, it should be $-\Phi'$, right? Which ends up being the pdf, so $e^{-\frac{z^2}{2}}$ (omitting the normalization factor as they will cancel out on this particular question)

Answer 1

Let P(0) ( P(1) ) is a probability of transmitting bit zero (one); P(e|0) ( P(e|1) ) is a probability of error when detecting bit zero (one).

The probability of erroneous detection is $$ P(e) = P(e|0)P(0) + P(e|1)P(1) \tag {1} $$ Let $V_0$ ( $V_1$ ) be a nominal signal voltage of bit zero ( one ) signal at the transmitter. $$ P(e|0) = \int_T^{\infty}{{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(\nu-V_0)^2/2 {\sigma}^2\right) d\nu } \\ P(e|1) = \int_{-\infty}^T{{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(\nu-V_1)^2/2 {\sigma}^2\right) d\nu } $$ where $T$ is a detection threshold (OP's $\mu$).

Differentiating $P(e)$ of eq.1 w.r.t. $T$, we arrive at $$ -P(0){{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(T-V_0)^2/2 {\sigma}^2\right)} + P(1){{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(T-V_1)^2/2 {\sigma}^2\right)} \tag {2} $$

To find an optimal threshold, we equate the expression in eq.2 to zero: $$ P(0)\exp\left(-{\frac {(T-V_0)^2} {2 {\sigma}^2}}\right) = P(1)\exp\left(-{\frac {(T-V_1)^2} {2 {\sigma}^2}}\right) \\ T = {\frac {V_0+V_1} {2}} + {\sigma}^2{\ln{(\frac {P(1)} {P(0)})}} $$ that is, exactly the OP's result.

The OP's reference source may compute the optimal threshold for the system in which the probabilities of zero/one bit transmissions are equal. The variance ${\sigma} ^2$ is multipled by $\ln{(\frac {P(1)} {P(0)})}$, and it disappears from the expression for optimal threshold, when the zero/one probabilities in transmission are equal ($P(0)=P(1)=1/2$).