Example 1

A simple, but interesting, example that results in a computable answer occurs when the mean of Gaussian random variables is either zero (model 0) or is ±m ± m with equal probability (hypothesis 1). The second hypothesis means that a non-zero mean is present, but its sign is not known. 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 L statistically independent observations are made. ℳ 0 : r∼0σ2I ℳ 0 : r 0 σ 2 I ℳ 1 : r∼mσ2I ℳ 1 : r m σ 2 I m=m…mProb1/2-m…-mProb1/2 m m … m 12 m … m 12 The numerator of the likelihood ratio is the sum of two Gaussian densities weighted by 1/2 12 (the a priori probability values), one having a positive mean, the other negative. The likelihood ratio, after simple cancellation of common terms, becomes Λr=12ⅇ2m∑l=0L−1 r l −Lm22σ2+12ⅇ-2m∑l=0L−1 r l −Lm22σ2 Λ r 1 2 2 m l 0 L 1 r l L m 2 2 σ 2 1 2 -2 m l 0 L 1 r l L m 2 2 σ 2 and the decision rule takes the form coshmσ2∑l=0L−1 r l ≷ ℳ 0 ℳ 1 ηⅇLm22σ2 m σ 2 l 0 L 1 r l ≷ ℳ 0 ℳ 1 η L m 2 2 σ 2 where coshx x is the hyperbolic cosine given simply as ⅇx+ⅇ-x2 x x 2 . As this quantity is an even function, the sign of its argument has no effect on the result. The decision rule can be written more simply as |∑l=0L−1 r l | ≷ ℳ 0 ℳ 1 σ2|m|arccoshηⅇLm22σ2 l 0 L 1 r l ≷ ℳ 0 ℳ 1 σ 2 m η L m 2 2 σ 2 The sufficient statistic equals the magnitude of the sum of the observations in this case. While the right side of this expression, which equals γ γ, is complicated, it need only be computed once. Calculation of the performance probabilities can be complicated; in this case, the false-alarm probability is easy to find and the others more difficult.