The input X to the Binary Symmetric Channel (BSC) shown in the figure is ‘1’ with probability 0.8. The cross-over probability is 1/7. If the received bit Y = 0, the conditional probability that ‘1’ was transmitted is _______.

A source emits bit 0 with probability $${1 \over 3}$$ and bit 1 with probability $${2 \over 3}$$. The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R are as
$${f_{\left. R \right|o}}\,(r) = \left\{ {\matrix{ {{1 \over 4},} & { - \,3\,\, \le \,\,x\,\, \le \,\,1,\,} \cr 0 & {otherwise,} \cr } } \right.and$$
$${f_{R/o}}\,(r) = \left\{ {\matrix{ {{1 \over 6},} & { - \,1\,\, \le \,\,x\,\, \le \,\,5\,,} \cr 0 & {otherwise.} \cr } } \right.$$

The minimum decision error orobability is

Consider a discrete-time channel Y = X + Z, where the additive noise Z is signal- dependent. In particular, given the trasmitted symbol $$X\, \in \,\{ \, - \,a,\,\, + \,a\} $$ at any instant, the noise sample Z is chosen independently from a Gaussian distribution with mean $$\beta X$$ and unit variance. Assume a threshold detector with zero threshold at the receiver. When $$\beta $$ = 0 the BER was found to be $$Q\,(a) = 1\, \times \,{10^{ - 8}}$$. $$\left( {Q\,\,(v)\, = {1 \over {\sqrt {2\,\pi } }}\,\int\limits_v^\infty {{e^{ - {u^2}/2}}} } \right.$$ du, and for v > 1,
use $$Q\,(v) \approx \,{e^{ - {v^2}/2}}$$
When $$\beta = - \,0.3,\,$$ the BER is closed to

Consider a communication scheme where the binary valued signal X satisfies P{X = + 1} = 0.75 and P {X = - 1} = 0.25. The received signal Y = X + Z, where Z is a Gaussian random variable with zero mean and variance $${\sigma ^2}$$. The received signal Y is fed to the threshold detector. The output of the threshold detector $${\hat X}$$ is: $$$\hat X:\left\{ {\matrix{ { + \,1,} & {Y\, > \tau } \cr { - \,1,} & {Y\, \le \,\,\tau .} \cr } } \right.$$$ To achieve a minimum probability of error $$P\{ \hat X\, \ne \,X\} $$, the threshold $$\tau $$ should be