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)


1 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 $$ P(e) = P(e|0)P(0) + P(e|1)P(1) \tag {1} $$ P(e) 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 $\mu$). Threshold

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 ($P(0)=P(1)=1/2$).

  • $\begingroup$ Thank you! Yeah, on the next step of the problem it asks to calculate the threshold assuming the two priors are the same, so we end up with 0, so the $\sigma$ doesn't matter. But on the solution for this step, it is missing that term but maybe it is a mistake. $\endgroup$
    – tir_nor
    Mar 29, 2021 at 18:29
  • $\begingroup$ +1 Very nice! $\endgroup$ Mar 29, 2021 at 19:01

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