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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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Didier Piau has used the result that the maximum-likelihood decision rule minimizes the sum of the false-alarm and missed-detection probabilities. If you were to sketch the two densities, and remember that the maximum-likelihood decision rule is that if the observation has value $\alpha$, then the decision is in favor of whichever hypothesis has larger likelihood $f(\alpha\mid H_i)$, you will see immediately that the threshold is $T = 1$. Decide $s = 0$ or $2$ according as the observed value $\alpha$ is smaller than or greater than $1$. But I suppose that is not a mathematical proof. – Dilip Sarwate Dec 13 '11 at 18:43
@DilipSarwate: thank you, that is enough for me! – nikos Dec 13 '11 at 19:42
up vote 1 down vote accepted

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 $T=1$.

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