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I have a decision (detection) problem trying to decide between symbols ${0,2}$. I have the two probability density functions: $ f(z|s=0) = \begin{cases} 0.25z + 0.5, & -2\le\ z <0 \\ -0.25z + 0.5, & 0\le\ z \le\ 2 \end{cases} $

and $ f(z|s=2) = \begin{cases} 0.25z, & 0\le\ z <2 \\ -0.25z + 1, & 2\le\ z \le\ 4 \end{cases} $

How can i mathematically prove that the optimal threshold value $T$ for that decision problem is equal to $1$?

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    @DilipSarwate: thank you, that is enough for me!2011-12-13

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One tries to maximize the probability $\frac12R(T)$ to guess right, where $ R(T)=\int_{-\infty}^Tf(z\mid s=0)\,\mathrm dz+\int_T^{+\infty}f(z\mid s=2)\,\mathrm dz. $ Thus, R'(T)=f(T\mid s=0)-f(T\mid s=2). The function R' is piecewise affine, R'(T)=0 if $T\lt-2$ or $T\gt4$, R'(T)=\frac14T+\frac12 if $-2\lt T\lt0$, R'(T)=-\frac12T+\frac12 if $0\lt T\lt2$, R'(T)=\frac14T-1 if $2\lt T\lt4$.

In particular, R'(T)\gt0 if $-2\lt T\lt1$ and R'(T)\lt0 if $1\lt T\lt4$. One sees that $R(T)$ is maximum at $T=1$.