Loss Functions & Decision Theory - Statistics Graduate Level

1. Which action is Bayes, and what are the posterior expected losses of the two actions?
   Answer: Choose $a_0$; the posterior losses are $3.80$ for $a_0$ and $5.25$ for $a_1$. When you encounter a Bayesian decision problem, your goal is to compute the posterior expected loss for each action and choose the action that minimizes it. The posterior expected loss for action $a$ is $\rho(a \mid x) = \sum_j L(a, H_j) \cdot P(H_j \mid x)$. With $P(H_0 \mid x) = 0.65$ and $P(H_1 \mid x) = 0.35$, apply this formula to each action: $\rho(a_0 \mid x) = L(a_0, H_0) \cdot P(H_0 \mid x) + L(a_0, H_1) \cdot P(H_1 \mid x) = 1(0.65) + 9(0.35) = 0.65 + 3.15 = 3.80$ $\rho(a_1 \mid x) = L(a_1, H_0) \cdot P(H_0 \mid x) + L(a_1, H_1) \cdot P(H_1 \mid x) = 7(0.65) + 2(0.35) = 4.55 + 0.70 = 5.25$ Since $3.80 < 5.25$, the Bayes action is $a_0$ — retain $H_0$. This confirms answer A. Answer B incorrectly selects $a_1$ despite its higher posterior expected loss, and swaps the two computed values between the actions. Answer C uses entirely incorrect calculations — perhaps from misassigning which loss corresponds to which hypothesis-action pair. Answer D produces values close enough to look plausible but reflects arithmetic errors likely from mixing up posterior probabilities or loss entries. A useful strategy: always build a small table pairing each action with each hypothesis state, fill in the losses, weight by posterior probabilities, and sum. The action with the smaller weighted sum wins. On graduate exams, distractors often swap actions or swap probabilities — double-check your assignments before computing.
2. Which rejection rule minimizes Bayes risk?
   Answer: Reject $H_0$ when $x>0.431$, approximately. When you see a Bayesian hypothesis testing problem with asymmetric losses and prior probabilities, your goal is to find the threshold where the posterior expected loss of rejecting equals the posterior expected loss of retaining — in other words, minimize Bayes risk by comparing weighted loss terms. The Bayes-optimal rule rejects $H_0$ when the likelihood ratio exceeds a threshold determined by the priors and losses. Specifically, reject $H_0$ when: $\frac{f(x|H_1)}{f(x|H_0)} > \frac{L_{01} \cdot P(H_0)}{L_{10} \cdot P(H_1)}$ where $L_{01} = 2$ (loss for false rejection) and $L_{10} = 5$ (loss for false retention). The threshold is: $\frac{0.7 \times 2}{0.3 \times 5} = \frac{1.4}{1.5} = \frac{14}{15} \approx 0.9333$ With $N(0,1)$ vs $N(1,1)$, the log-likelihood ratio simplifies to $x - \frac{1}{2}$, so you reject when $e^{x - 1/2} > 0.9333$, giving $x - 0.5 > \ln(0.9333) \approx -0.0690$, so $x > 0.431$. This confirms answer A. Answer B ($x > 0.500$) ignores the asymmetric losses and priors entirely — it's the naive midpoint between the two means. Answer C ($x > 1.347$) inverts the threshold ratio, over-weighting the cost of false rejection and making the test far too conservative. Answer D ($x > -0.347$) likely results from a sign error in computing the log-threshold, shifting the boundary in the wrong direction. Study tip: Always set up the likelihood ratio inequality first, then plug in priors and losses — don't guess the threshold from the means alone. The asymmetry of losses and priors almost always shifts the boundary away from the midpoint.
3. Which transformation is guaranteed to preserve the set of Bayes actions for every prior and every observed dataset?
   Answer: $L^*(a,\theta)=3L(a,\theta)+b(\theta)$ for an arbitrary finite function $b$. When a transformation is applied to a loss function, the Bayes action under prior $\pi$ is the action minimizing the posterior expected loss $\mathbb{E}_{\theta|x}[L^*(a,\theta)]$. A transformation preserves Bayes actions if and only if it preserves the ordering of posterior expected losses across all actions — meaning the minimizer cannot change. The key insight is that adding a function $b(\theta)$ to the loss is always safe: since $b(\theta)$ doesn't depend on $a$, the term $\mathbb{E}_{\theta|x}[b(\theta)]$ is a constant with respect to the action and shifts all posterior expected losses equally, leaving the minimizer unchanged. Multiplying by a positive constant also preserves ordering. Answer D, $L^*(a,\theta) = 3L(a,\theta) + b(\theta)$, applies both: multiplication by $3 > 0$ and addition of $b(\theta)$. This guarantees the Bayes action is preserved for every prior and every dataset. Answer A fails because squaring is a nonlinear transformation. Even with nonneg­ative losses, squaring can reorder posterior expected losses — the action minimizing $\mathbb{E}[L]$ need not minimize $\mathbb{E}[L^2]$, since the latter penalizes variance as well as mean. Answer B adds $c(a)$, a function of the action alone. Taking expectations, $c(a)$ remains as-is and shifts different actions by different constants, directly distorting the ranking of posterior expected losses. Answer C multiplies by $-2$, a negative constant, which reverses all inequalities and turns the minimizer into a maximizer, completely destroying the Bayes action. Study tip: When evaluating loss transformations, ask two questions: does it preserve action-ordering, and does the additive term depend on $a$? Any additive $b(\theta)$ is harmless; any additive $c(a)$ is dangerous; any negative scalar flips everything.
4. Which statement correctly relates this Bayes test to the Neyman–Pearson framework?
   Answer: It is a most powerful level-$\alpha_B$ test, up to null-probability differences at the boundary. When you see a question connecting Bayesian decision theory to Neyman–Pearson hypothesis testing, the key insight to anchor on is the Neyman–Pearson Lemma: any test that rejects based on a likelihood ratio threshold is automatically most powerful at whatever size that threshold induces. A Bayes test with asymmetric losses and a prior rejects when the likelihood ratio $\Lambda$ exceeds some critical value $c_B$. Because this is a likelihood-ratio test, the Neyman–Pearson Lemma guarantees it is the most powerful test among all level-$\alpha_B$ tests, where $\alpha_B = P_{H_0}(\Lambda > c_B)$. The phrase "up to null-probability differences at the boundary" acknowledges that with a continuous distribution, $P_{H_0}(\Lambda = c_B) = 0$, so the boundary causes no ambiguity — making D correct. A is wrong because calling the test "prior-free" misrepresents it. The threshold $c_B$ itself is determined by the prior and loss function; only after fixing $c_B$ does $\alpha_B$ depend solely on the null distribution. The test is decidedly prior-dependent. B is wrong because most powerful level-$\alpha$ tests correspond to thresholds $c_\alpha \leq c_B$, meaning larger rejection regions. The Bayes test is most powerful at level $\alpha_B$, not at every $\alpha > \alpha_B$, which would require different (lower) thresholds. C is wrong because minimax optimality is a separate property requiring a specific prior (the least favorable prior). An arbitrary prior and asymmetric losses do not generally yield a minimax rule. Study tip: Remember that any likelihood-ratio test is a Neyman–Pearson test — Bayesian or not. The Lemma cares about the form of the test, not its motivation.
5. Which value of $q$ gives the minimax randomized test, and what is its maximum risk?
   Answer: Use $q=0.25$, giving maximum risk $0.25$. When you encounter a minimax problem with randomized tests, your goal is to find the mixture that minimizes the worst-case risk across all states of nature. A randomized test using weight $q$ produces risks: $R(\theta_0, \delta_q) = 0.10q + 0.30(1-q) = 0.30 - 0.20q$ $R(\theta_1, \delta_q) = 0.40q + 0.20(1-q) = 0.20 + 0.20q$ Notice $R(\theta_0)$ decreases in $q$ while $R(\theta_1)$ increases in $q$. The minimax solution sets these equal — the point where neither state is "worse" than the other: $0.30 - 0.20q = 0.20 + 0.20q \implies 0.10 = 0.40q \implies q = 0.25$ Substituting back: $R(\theta_0) = R(\theta_1) = 0.30 - 0.20(0.25) = 0.25$. The maximum risk is therefore 0.25, confirming answer A. Choice B uses $q = 0.50$, giving risks $0.20$ and $0.30$ — the maximum is $0.30$, which is worse than $0.25$. Choice C uses $q = 0.75$, giving risks $0.15$ and $0.35$ — maximum risk $0.35$, even worse. Choice D takes $q = 1.00$ (pure $\delta_1$), yielding a maximum risk of $0.40$ — the worst possible option here. Each of these fails because they don't balance the two risk functions, leaving one state of nature with unnecessarily high risk. A useful study tip: in minimax randomization problems, always write both risk functions as linear functions of $q$, then solve for the intersection. That crossing point is always your minimax solution when the risks move in opposite directions.
6. Ignoring ties at the boundary points, which Bayes decision regions are correct?
   Answer: Retain for $p<0.15$, defer for $0.15<p<0.70$, and reject for $p>0.70$. When facing a three-action Bayes decision problem, your goal is to find the posterior probability thresholds where switching actions minimizes expected loss. Compute the expected loss for each action as a function of $p = P(H_1 \mid x)$: - **Retain $H_0$:** Loss = $0 \cdot (1-p) + 4 \cdot p = 4p$ - **Reject $H_0$:** Loss = $2 \cdot (1-p) + 0 \cdot p = 2(1-p)$ - Defer: Loss = $0.6(1-p) + 0.6p = 0.6$ Now find the crossover points. Set Retain = Defer: $4p = 0.6 \Rightarrow p = 0.15$. Set Defer = Reject: $0.6 = 2(1-p) \Rightarrow p = 0.70$. This gives: retain when $p < 0.15$, defer when $0.15 < p < 0.70$, and reject when $p > 0.70$, which is exactly answer C. Choice A uses $p = 1/3$ as the retain/defer threshold, which would come from incorrectly equating retain and reject losses directly ($4p = 2(1-p) \Rightarrow p = 1/3$) while ignoring the deferral option entirely — a classic mistake of forgetting there are three competing actions. Choice B uses symmetric thresholds of 0.20 and 0.80, which don't correspond to any correct loss-equating calculation; they may reflect a misapplication of equal-weight intuition. Choice D correctly finds 0.15 as one threshold and 1/3 as the other, but assigns the wrong actions to those regions — reject and defer are swapped, which would happen if you confused which loss function crosses which. Your strategy: always write out all three expected loss functions explicitly, then solve pairwise crossover equations. Forgetting one action or mislabeling regions after finding thresholds are the two most common errors on Bayes decision problems.
7. Which conclusion follows from these risks?
   Answer: $\delta_A$ has smaller Bayes risk for this prior, but neither procedure dominates the other. When comparing decision procedures in Bayesian decision theory, you need to evaluate two distinct concepts separately: Bayes risk (a weighted average of risks under a prior) and admissibility/dominance (a pointwise comparison across all parameter values). Start by computing the Bayes risk for each procedure. Given $\pi(\theta_0) = 0.80$ and $\pi(\theta_1) = 0.20$: $r(\pi, \delta_A) = 0.80(0.20) + 0.20(0.60) = 0.16 + 0.12 = 0.28$ $r(\pi, \delta_B) = 0.80(0.35) + 0.20(0.40) = 0.28 + 0.08 = 0.36$ So $\delta_A$ has strictly smaller Bayes risk under this prior. Now check dominance: $\delta_A$ dominates $\delta_B$ only if $R(\theta, \delta_A) \leq R(\theta, \delta_B)$ for all $\theta$, with strict inequality somewhere. At $\theta_0$: $0.20 < 0.35$ ✓, but at $\theta_1$: $0.60 > 0.40$ ✗. Since $\delta_A$ is worse at $\theta_1$, neither procedure dominates the other. This confirms answer A. Answer B is wrong because smaller Bayes risk under one prior does not imply dominance — dominance is a frequentist, prior-free concept requiring pointwise superiority. Answer C incorrectly claims $\delta_B$ has smaller Bayes risk, which the calculation above refutes. Answer D is wrong on both counts: the Bayes risks are 0.28 vs. 0.36 (not equal), and the maximum risks are $\max(0.60) = 0.60$ for $\delta_A$ vs. $\max(0.40) = 0.40$ for $\delta_B$, so while $\delta_B$ does have smaller minimax risk, the premise about equal Bayes risk is false. Remember: Bayes risk and dominance measure different things. A procedure can win on one criterion and lose on the other — always evaluate them independently.
8. What is the Bayes decision based on the observed data?
   Answer: Reject $H_0$ because the posterior odds are about $0.895$, exceeding the loss threshold $0.75$. When you encounter a Bayes decision problem under composite hypotheses, your goal is to compute posterior odds and compare them against a loss-ratio threshold that determines when switching decisions becomes costly enough to matter. Start by computing each hypothesis's marginal likelihood — a weighted average of component likelihoods using the conditional prior weights. For $H_0$: $0.75(0.12) + 0.25(0.02) = 0.09 + 0.005 = 0.095$. For $H_1$: $0.25(0.08) + 0.75(0.20) = 0.02 + 0.15 = 0.17$. Next, apply Bayes' theorem using the prior odds $P(H_1)/P(H_0) = 1/2$: $\text{Posterior odds} = \frac{P(H_1)}{P(H_0)} \cdot \frac{m_1}{m_0} = \frac{1}{2} \cdot \frac{0.17}{0.095} \approx \frac{1}{2}(1.789) \approx 0.895$ Now find the loss-ratio threshold. Under the Bayes decision rule, reject $H_0$ when posterior odds exceed $L_{H_0}/L_{H_1} = 3/4 = 0.75$, where $L_{H_0} = 3$ (loss for false rejection) and $L_{H_1} = 4$ (loss for false retention). Since $0.895 > 0.75$, you **reject $H_0$ — confirming answer C** is correct. Choice A is wrong because it ignores the prior odds and uses only the largest component likelihoods — there's no principled rule based on a single component ratio. Choice B uses the correct framework but calculates posterior odds as $0.447$, likely forgetting to include the marginal likelihood ratio correctly. Choice D computes a marginal likelihood ratio of $0.559$ but misidentifies the decision threshold as $4/3$ rather than $3/4$. Strategy tip: Always build the full marginal likelihoods first, then multiply by prior odds for posterior odds — and remember the threshold is loss of false-rejection over loss of false-retention, not the reverse.
9. Which comparison of the Bayes and minimax criteria is correct?
   Answer: The Bayes criterion selects Test B, whereas the minimax criterion selects Test A. When comparing decision rules under uncertainty, two fundamental criteria often diverge: the Bayes criterion minimizes expected loss using prior probabilities, while the minimax criterion minimizes the maximum possible loss across states of nature (ignoring priors). To apply the Bayes criterion, compute each test's expected loss: $E[L] = \pi_0 \cdot \alpha \cdot L_I + \pi_1 \cdot \beta \cdot L_{II}$, where $\pi_0 = 0.20$, $\pi_1 = 0.80$, $L_I = 4$, $L_{II} = 1$, and $\beta = 1 - \text{power}$. - Test A: $0.20(0.05)(4) + 0.80(0.30)(1) = 0.04 + 0.24 = 0.28$ - Test B: $0.20(0.10)(4) + 0.80(0.15)(1) = 0.08 + 0.12 = 0.20$ Test B has lower Bayes risk, so the Bayes criterion selects Test B. For the minimax criterion, find each test's worst-case (maximum) loss across the two states: - Test A: $\max(0.05 \times 4,\ 0.30 \times 1) = \max(0.20, 0.30) = 0.30$ - Test B: $\max(0.10 \times 4,\ 0.15 \times 1) = \max(0.40, 0.15) = 0.40$ Test A's worst-case loss (0.30) is smaller, so the minimax criterion selects Test A. This confirms answer B. Answer A reverses the assignments. Answers C and D both claim the criteria agree — they don't here, which is precisely what makes this problem instructive. Answer C incorrectly assigns both to Test A; D incorrectly assigns both to Test B. A useful pattern: Bayes and minimax often disagree when the prior heavily weights one hypothesis, because the minimax criterion ignores that weighting entirely. Always compute both separately and resist assuming they'll align.
10. What randomization probability attains size $0.10$, and what is the resulting power?
    Answer: Use $\gamma=0.1875$; the resulting power is $0.70$. When you encounter a Neyman-Pearson hypothesis testing problem with a discrete distribution, your first instinct should be to set up the size constraint and solve for the randomization probability that makes the test exactly attain the nominal level. Here, $X \sim \text{Binomial}(2, p)$. Under $H_0: p = 0.2$, the probabilities are $P(X=0) = 0.64$, $P(X=1) = 0.32$, and $P(X=2) = 0.04$. The likelihood ratio favors larger $X$, so the test always rejects when $X=2$ and randomizes when $X=1$. The type I error (size) is: $\alpha = P(X=2 \mid p=0.2) + \gamma \cdot P(X=1 \mid p=0.2) = 0.04 + \gamma(0.32)$ Setting $\alpha = 0.10$: $0.04 + 0.32\gamma = 0.10 \Rightarrow \gamma = \frac{0.06}{0.32} = 0.1875$ Now compute power under $H_1: p = 0.8$, where $P(X=1) = 0.32$ and $P(X=2) = 0.64$: $\text{Power} = 0.64 + 0.1875(0.32) = 0.64 + 0.06 = 0.70$ This confirms B is correct. Choice A uses $\gamma = 0.3125$, which would set $\alpha = 0.04 + 0.32(0.3125) = 0.14$, exceeding the 0.10 constraint. Choice C appears to miscalculate $\gamma$ by misidentifying $P(X=1)$ under $H_0$. Choice D uses no randomization ($\gamma = 0$), which gives size $0.04 < 0.10$—a valid but suboptimal test that wastes power by not fully exploiting the size budget. The key study tip: always check that you're using the correct null distribution to set size, and remember that the most powerful test exhausts the size allowance—leaving $\alpha < 0.10$ is wasteful.

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