Exercises

$A = \log\frac{1-0.05}{0.01} = \log(95) \approx 4.554$.

$B = \log\frac{0.05}{1-0.01} = \log(0.05/0.99) \approx -2.983$.

$\ell(y) = \log \frac{p_1(y)}{p_0(y)} = \frac{(\mu_1-\mu_0)}{\sigma^2}\bigl(y - \tfrac{\mu_0+\mu_1}{2}\bigr)$.

It is affine in $y$: the SPRT accumulates a shifted, scaled version of the observations and compares the sum with the Wald thresholds.

$\mathbb{E}_1[S_N] = \mathbb{E}_1[N] \cdot \mathbb{E}_1[\ell]$.

With probability $1-\beta$ the test accepts $\mathcal{H}_1$ and $S_N \approx A$; with probability $\beta$ it rejects and $S_N \approx B$. So $\mathbb{E}_1[S_N] \approx (1-\beta)A + \beta B$.

$\mathbb{E}_1[N] \approx \frac{(1-\beta)A+\beta B}{D(p_1\|p_0)}$.

$A = \log(99) \approx 4.595$, $B \approx -4.595$.

$\mathbb{E}_1[N] \approx (0.99 A + 0.01 B)/0.5 \approx 9.10$. By symmetry $\mathbb{E}_0[N] \approx 9.10$.

$Q^{-1}(0.01) \approx 2.326$, so $n \approx (2 \cdot 2.326)^2 \approx 21.6$, round to $n = 22$. The SPRT needs less than half as many samples on average.

Under the swap, $\ell' = -\ell$ and the thresholds $A', B' = -B, -A$. The stopping region is identical.

Sample paths and decisions are unchanged — the SPRT is symmetric under the hypothesis relabeling with swapped error probabilities.

$S_n/n \to D(p_1\|p_0) > 0$ a.s., so $S_n \to +\infty$.

$S_n/n \to -D(p_0\|p_1) < 0$ a.s., so $S_n \to -\infty$.

In either case $S_n$ leaves the bounded strip $[B,A]$ in finite time with probability one, so $N < \infty$ a.s.

$W_0 = 0 = S_0 - \min_{0 \leq k \leq 0} S_k$.

Assume $W_{n-1} = S_{n-1} - m_{n-1}$ where $m_{n-1} = \min_{k \leq n-1} S_k$. Then $W_n = \max(0, S_{n-1} - m_{n-1} + \ell(y_n)) = \max(0, S_n - m_{n-1}) = S_n - \min(m_{n-1}, S_n) = S_n - m_n$.

Started at $k$, the SPRT statistic is $S_n - S_k$ and it crosses upward when $S_n - S_k \geq h$.

$\max_{0 \leq k \leq n}(S_n - S_k) = S_n - \min_k S_k = W_n$. So CUSUM declares a change at the first $n$ with $W_n \geq h$, which is the first $n$ where at least one parallel SPRT crosses upward.

$\mathrm{ARL}_0 \approx e^5 / 0.5 \approx 296.8$ samples.

$\mathrm{ARL}_1 \approx 5/0.5 = 10$ samples.

$P_f^{-1/16} = 10^{6/16} = 10^{0.375} \approx 2.371$. $\alpha_{CA} = 16(2.371 - 1) \approx 21.94$.

The threshold is about 13.4 dB above the estimated noise power.

$y \sim \mathrm{Exp}(\lambda)$ under $\mathcal{H}_0$ (square-law detector output). $\hat\sigma^2 = \tfrac{1}{N}\sum_i y_i \sim \Gamma(N, \lambda/N)$.

$P_f = \Pr(y > \alpha_{CA}\hat\sigma^2) = \mathbb{E}_{\hat\sigma^2}[e^{-\lambda \alpha_{CA}\hat\sigma^2}]$. The MGF of the $\Gamma(N,\lambda/N)$ gives $P_f = (1+\alpha_{CA}/N)^{-N}$.

One or two interfering targets inflate $\hat\sigma^2_{CA}$ but leave the $k$th-smallest reference unchanged if $k \leq N-2$. OS-CFAR is therefore immune to a small number of target-like spikes.

The $k$th-order statistic is less efficient than the sample mean: OS-CFAR has $\approx 0.5$-$1$ dB additional SNR loss in homogeneous noise, and needs a larger threshold for the same $P_f$.

For small $\alpha = \beta$: $\mathbb{E}_1[N] \approx A/D \approx \log(1/\alpha)/D$.

For the LRT with $\alpha = \beta$: $n \approx ((Q^{-1}(\alpha))^2 \cdot 4)/d^2$ where $d^2 = (\mu_1-\mu_0)^2/\sigma^2 = 2D$. Using $Q^{-1}(\alpha) \sim \sqrt{2\log(1/\alpha)}$ gives $n \approx 2\log(1/\alpha)/D$.

$n/\mathbb{E}_1[N] \to 2$.

$10^{4/24} \approx 2.154$. $\alpha_{CA} = 24 \cdot 1.154 \approx 27.7$.

Solve $0.9 = (1 + 27.7/(24(1+\text{SNR})))^{-24}$. Take the $-1/24$ power: $0.9^{-1/24} \approx 1.00440$. So $27.7/(24(1+\text{SNR})) = 0.00440$, giving $1+\text{SNR} \approx 262$, SNR $\approx 261$, i.e., about 24.2 dB. (The CFAR loss of $\approx 0.7$ dB has been absorbed in the Monte-Carlo-like multiplier $\alpha_{CA}$.)

$\ell(y) = \tfrac{1}{2}\log\frac{\sigma^2}{\sigma^2(1+\text{SNR})} + \tfrac{y^2}{2\sigma^2}\bigl(1 - \tfrac{1}{1+\text{SNR}}\bigr)$. Simplify: $\ell(y) = -\tfrac{1}{2}\log(1+\text{SNR}) + \tfrac{y^2}{2\sigma^2}\cdot\tfrac{\text{SNR}}{1+\text{SNR}}$.

Accumulate $S_n = \sum_i \ell(y_i)$ and stop at the first $n$ with $S_n \geq A$ (declare occupied) or $S_n \leq B$ (idle).

CA-CFAR averages over both sides and lifts the threshold modestly, causing false alarms at the high side. GO-CFAR picks the maximum of the two half-window means, so at the high-side edge its threshold tracks the high side — false alarms are suppressed.

When a target straddles one half of the reference window, GO-CFAR picks the elevated half as its threshold, masking the CUT target. CA-CFAR is more tolerant here since it averages.

Use GO-CFAR in nonhomogeneous clutter with sharp power transitions and few targets per beam. Prefer OS-CFAR when multiple targets share the window.