2.9 Random parameters

A simple, but interesting, example that results in a computable answer occurs when the mean of Gaussian randomvariables is either zero (model 0) or is $(m)$ with equal probability (hypothesis 1). The second hypothesis means that a non-zero mean is present, but its sign is notknown. We are therefore stating that if hypothesis one is in fact valid, the mean has fixed sign for each observation -what is random is its a priori value. As before, $L$ statistically independent observations are made. $${}_0:(r, (0, ^{2}I))$$ $${}_1:(r, (m, ^{2}I))$$ $$m=((\left(\begin{array}{c}m\\ \\ m\end{array}\right), 1/2), (\left(\begin{array}{c}-m\\ \\ -m\end{array}\right), 1/2))$$ The numerator of the likelihood ratio is the sum of two Gaussian densities weighted by $1/2$ (the a priori probability values), one having a positive mean, the other negative. The likelihood ratio,after simple cancellation of common terms, becomes $$(r)=\frac{1}{2}e^{\frac{2m\sum_{l=0}^{L-1} r_l()-Lm^2}{2^2}}+\frac{1}{2}e^{\frac{-2m\sum_{l=0}^{L-1} r_l()-Lm^2}{2^2}}$$ and the decision rule takes the form $$\cosh (\frac{m}{^2}\sum_{l=0}^{L-1} r_l)\underset{{}_0}{\overset{{}_1}{}}e^{\frac{Lm^2}{2^2}}$$ where $\cosh x$ is the hyperbolic cosine given simply as $\frac{e^x+e^{-x}}{2}$ . As this quantity is an even function, the sign of its argument has no effect on the result. The decision rulecan be written more simply as $$\left|\sum_{l=0}^{L-1} r_l\right|\underset{{}_0}{\overset{{}_1}{}}\frac{^2}{\left|m\right|}\mathrm{arccosh\,}(e^{\frac{Lm^2}{2^2}})$$ The sufficient statistic equals the magnitude of the sum of the observations in this case. While the right side of this expression, whichequals $$ , is complicated, it need only be computed once. Calculation of the performanceprobabilities can be complicated; in this case, the false-alarm probability is easy to find and the others moredifficult.