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Suppose the random variable Y has probability density $p_{0}(y)=\frac{1}{2}\exp(-|y|))$,$y\in{R}$.$$$$ How can I evaluate $$P_{0}(sgn(Y)>\gamma)$$ where $\gamma$ is a threshold value. Also, can any one help me derive this result: $$P_{0}(sgn(Y)>\gamma)= \begin{cases} 0 \text{ if $\gamma\geq{1};$ }\\ \frac{1}{2}\text{ if $-1\leq{\gamma}<1$ }\\1 \text{ if $\gamma<-1$ }\end{cases}$$

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Since sgn(Y) = -1,0,or 1, it is impossible for sgn(Y)>1 hence P(sgn(Y)>1) = 0. And clearly then P(sgn(Y) > g) = 1 if g<-1. If you look at the distribution p(y) you can see it is symmetric around 0. Then P(Y>=0) = P(Y<=0) = 1/2. So for 1>g>=-1, P(sgn(Y) > g) = P(Y>=0) = 1/2. – Apprentice Queue Nov 10 '12 at 18:08

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