Posterior Summaries & Credible Intervals - Statistics Graduate Level

1. Which option correctly gives the Bayes estimate under squared-error loss, the Bayes estimate under absolute-error loss, and the central $50$ percent equal-tailed credible interval?
   Answer: Squared-error estimate $2.7$; absolute-error estimate $2$; interval $[0,2]$ When you encounter a Bayesian estimation question, think about three distinct tasks: the loss function determines the optimal point estimate, and the credible interval is built from the posterior CDF. Squared-error loss is minimized by the posterior mean. Computing it directly: $E[\theta] = 0(0.45) + 2(0.35) + 10(0.20) = 0 + 0.70 + 2.00 = 2.70$. So the Bayes estimate under squared-error loss is $2.7$. Absolute-error loss is minimized by the posterior median — the value where cumulative probability first reaches or exceeds $0.50$. The CDF gives $P(\theta \leq 0) = 0.45$ and $P(\theta \leq 2) = 0.80$. Since $0.45 < 0.50 \leq 0.80$, the median is $2$. For the central 50% equal-tailed credible interval, you need to cut off 25% from each tail. The lower bound satisfies $P(\theta < \text{lower}) \leq 0.25$: since $P(\theta < 0) = 0$, the lower bound is $0$. The upper bound satisfies $P(\theta > \text{upper}) \leq 0.25$: since $P(\theta > 2) = 0.20 \leq 0.25$, the upper bound is $2$. This gives the interval $[0, 2]$, confirming answer C. A is wrong because it correctly identifies the mean as $2.7$ but assigns the median incorrectly as $0$ (that's just $P(\theta=0)$, not the median). B swaps the mean and median entirely. D uses $2$ for both estimates and shifts the interval to $[2,10]$, which cuts the wrong tails. Remember: mean minimizes squared-error, median minimizes absolute-error — and always build credible intervals from cumulative probabilities, not raw probability masses.
2. Which of the following is the $90$ percent highest-posterior-density credible set?
   Answer: $[0,\ 1.151]$ When you see a question about highest-posterior-density (HPD) credible sets, remember the defining rule: an HPD region contains all parameter values with posterior density above some threshold, and it is the shortest interval achieving the target coverage. For a unimodal, monotonically decreasing posterior like $p(\theta\mid y) = 2e^{-2\theta}$, the density is maximized at $\theta = 0$ and falls off as $\theta$ increases. This means the HPD region always starts at $\theta = 0$ — you never exclude the highest-density region near zero. To find the 90% HPD interval $[0, c]$, integrate the posterior and set it equal to 0.90: $\int_0^c 2e^{-2\theta}\,d\theta = 1 - e^{-2c} = 0.90$ Solving: $e^{-2c} = 0.10$, so $c = \frac{\ln(10)}{2} \approx \frac{2.303}{2} \approx 1.151$. This confirms answer A, $[0,\ 1.151]$, is correct. Answer C, $[0,\ 1.498]$, would correspond to a 95% credible interval (since $1 - e^{-2(1.498)} \approx 0.95$), not 90% — a classic coverage-level mix-up. Answers B and D are equal-tailed or symmetric-style intervals that split probability in both tails. For a distribution supported on $[0, \infty)$ with a mode at zero, excluding any probability below $\theta = 0$ makes no sense, and cutting probability from the high-density region near zero violates the HPD principle entirely. Study tip: For any monotone posterior, the HPD interval always anchors at the mode. If the posterior is decreasing on $[0,\infty)$, the HPD set is always of the form $[0, c]$ — never a two-sided interval that excludes the boundary.
3. Which option gives, respectively, a central $95$ percent credible interval for $\mu$ and a central $95$ percent posterior predictive interval for $Y_{\mathrm{new}}$?
   Answer: For $\mu$: $[8.04,11.96]$; for $Y_{\mathrm{new}}$: $[5.62,14.38]$ Whenever you see a Bayesian prediction question, keep two distributions carefully separated: the posterior for the parameter and the posterior predictive for a future observation. These have different variances, and confusing them is exactly what the distractors exploit. Here, you're given $\mu \mid y \sim N(10, 1)$, where the second argument is the variance (so the posterior standard deviation is $\sqrt{1} = 1$). A central 95% credible interval for $\mu$ uses $10 \pm 1.96 \times 1$, giving $[8.04, 11.96]$. For $Y_{\text{new}}$, you must account for two sources of uncertainty: uncertainty about $\mu$ (variance = 1) and sampling variability around $\mu$ (variance = 4). The posterior predictive distribution is obtained by marginalizing out $\mu$: $Y_{\text{new}} \mid y \sim N(10,\ 1 + 4) = N(10, 5)$ The 95% predictive interval is then $10 \pm 1.96\sqrt{5} \approx 10 \pm 4.38$, giving $[5.62, 14.38]$. This confirms answer D is correct. A is wrong because it uses the wider predictive interval $[6.08, 13.92]$ (which corresponds to $\sqrt{4} = 2$, ignoring posterior uncertainty) for both quantities — a double error. B applies the narrow posterior interval to both, pretending future observations carry no sampling variability. C reverses the correct assignment — the wider interval belongs to $Y_{\text{new}}$, not to $\mu$. As a study tip: always remember that the predictive variance adds the parameter variance and the sampling variance. The predictive interval must be at least as wide as the credible interval for $\mu$.
4. Approximately what value of $z$ makes the posterior probability of this rectangular region equal to $0.95$?
   Answer: $z=2.24$, based on marginal probability about $0.9747$ When building a rectangular credible region for two independent parameters, the joint posterior probability is the product of the two marginal probabilities. If each marginal interval has probability $p$, then the rectangle has joint probability $p^2$. Your task is to find $z$ such that $p^2 = 0.95$, which means each marginal must satisfy $p = \sqrt{0.95} \approx 0.9747$. For a standard normal, you need $P(-z \leq \theta \leq z) = 0.9747$, meaning each tail has probability $\frac{1-0.9747}{2} \approx 0.0127$. Looking up the corresponding quantile gives $z \approx 2.24$. This confirms C is correct: you need marginal intervals at roughly the 97.47% level, not the familiar 95% level, to achieve 95% joint coverage. Choice A uses $z = 1.64$, which gives each marginal 90% coverage, so the joint probability is only $0.90^2 = 0.81$ — far below 0.95. Choice B is the most tempting trap: students often grab $z = 1.96$ because it's the standard 95% cutoff, but applying it to both dimensions gives $0.95^2 = 0.9025$, not 0.95. The rectangle is too small. Choice D uses $z = 2.58$ (99% marginal intervals), yielding $0.99^2 = 0.9801$ — a credible region that is actually too conservative, exceeding 95%. The key study tip: whenever a credible region is formed as the Cartesian product of marginal intervals, you must account for the multiplicative nature of joint probability. Don't default to the familiar univariate critical value — solve $p^d = \text{target}$ for the required marginal coverage $p$, where $d$ is the number of dimensions.
5. Which interval is the central $90$ percent posterior credible interval for $\mu$?
   Answer: $(-0.51,\ 2.43)$ When you see a Bayesian normal-normal conjugate model, your first task is computing the posterior mean and variance by combining the prior precision and the likelihood precision. Here, the prior is $\mu \sim N(0, \tau_0^2)$ with $\tau_0^2 = 4$, and the likelihood gives $\bar{y} = 1.2$ from $n = 9$ observations each with variance $\sigma^2 = 9$, so the sampling variance of $\bar{y}$ is $\sigma^2/n = 9/9 = 1$. The posterior precision is the sum of the prior and likelihood precisions: $\frac{1}{\tau_0^2} + \frac{n}{\sigma^2} = \frac{1}{4} + \frac{9}{9} = 0.25 + 1 = 1.25$ So the posterior variance is $\tau_n^2 = 1/1.25 = 0.8$, and the posterior mean is: $\mu_n = \tau_n^2\left(\frac{\mu_0}{\tau_0^2} + \frac{\bar{y}}{\sigma^2/n}\right) = 0.8\left(\frac{0}{4} + \frac{1.2}{1}\right) = 0.8 \times 1.2 = 0.96$ The central 90% credible interval uses $z_{0.95} = 1.645$, giving: $0.96 \pm 1.645\sqrt{0.8} = 0.96 \pm 1.645(0.894) \approx 0.96 \pm 1.47$ This yields approximately $(-0.51,\ 2.43)$, confirming A. Choice B uses the wrong posterior variance, likely forgetting to invert the total precision correctly. Choice C may use an incorrect $z$-value such as 1.96 (a 95% interval mistake) with the right posterior. Choice D resembles the prior's 90% interval ($0 \pm 1.645 \times 2$), ignoring the data entirely. Remember: always invert the sum of precisions to get the posterior variance — it's a common slip to add variances instead of precisions.
6. Which coverage statement follows without requiring any additional assumptions?
   Answer: Under the joint prior-predictive distribution of $\theta$ and $Y$, $P\{\theta\in C(Y)\}=0.95$. When you see a question about Bayesian credible sets, the key distinction to keep in mind is the difference between posterior coverage (conditional on observed data) and frequentist coverage (conditional on a fixed parameter). The passage guarantees one specific probability statement — and your job is to trace exactly what that implies without adding anything extra. The passage states that for every dataset $y$, $P(\theta \in C(y) \mid y) = 0.95$. This is a statement about the posterior. Now consider the joint distribution of $(\theta, Y)$ induced by the prior and sampling model — the prior-predictive distribution. By the law of total probability: $P(\theta \in C(Y)) = \int P(\theta \in C(y) \mid y)\, p(y)\, dy = \int 0.95 \cdot p(y)\, dy = 0.95.$ This integral collapses cleanly because the integrand is identically 0.95 for every $y$. No additional assumptions are needed — it follows immediately. That confirms B is correct. A is wrong because posterior coverage does not imply frequentist coverage. Frequentist coverage, $P(\theta \in C(Y) \mid \theta)$, depends on integrating over $Y$ at a fixed $\theta$, not over $\theta$. Without a matching prior structure or a conjugate argument, this need not equal 0.95. C is wrong because the result in B holds for any valid 0.95 posterior credible set — equal-tailed intervals are one choice, but HPD regions or any other construction work equally well. D is wrong for the same structural reason as C: symmetry of the posterior is irrelevant. The marginal coverage result only requires that the posterior probability equals 0.95 for each $y$, regardless of shape. The key study tip: whenever a Bayesian procedure guarantees a pointwise posterior probability, averaging over the marginal of $Y$ always preserves that probability — no extra conditions required.
7. What is the approximate central $95$ percent posterior credible interval for the contrast $\delta=\beta_1-\beta_2$?
   Answer: $[-2.81,\ 0.81]$ When working with linear combinations of jointly normal random variables in a Bayesian context, the key is recognizing that any linear combination of jointly normal posteriors is itself normal — and you need to correctly propagate the variance using the covariance structure. The contrast $\delta = \beta_1 - \beta_2$ has posterior mean $E(\delta \mid y) = 1 - 2 = -1$. For the posterior variance, recall that $\text{Var}(aX + bY) = a^2\text{Var}(X) + b^2\text{Var}(Y) + 2ab\,\text{Cov}(X,Y)$. Here $a=1, b=-1$, so: $\text{Var}(\delta \mid y) = 0.25 + 0.36 - 2(-0.12) = 0.25 + 0.36 + 0.24 = 0.85$ The posterior standard deviation is $\sqrt{0.85} \approx 0.922$. A central 95% credible interval is then $-1 \pm 1.96(0.922) \approx -1 \pm 1.807$, giving approximately $[-2.81,\ 0.81]$, which is D. Choice A $[-0.81, 2.81]$ uses the correct width but centers at $+1$ instead of $-1$ — a sign error in computing the posterior mean of $\delta$. Choice B $[-2.53, 0.53]$ likely results from ignoring the covariance term entirely, giving $\text{Var}(\delta) = 0.61$ and $\text{SD} \approx 0.781$, which underestimates the interval width. Choice C $[-2.19, 0.19]$ comes from incorrectly subtracting the covariance correction (treating $-2ab\,\text{Cov}$ as $-0.24$), which gets the sign of the covariance contribution wrong. The critical study tip: when computing variance of a difference, a negative covariance actually increases the variance — counterintuitive but essential. Always write out the full formula and plug in signs carefully before computing the interval.
8. Using the large-sample approximation $\operatorname{SE}(\hat q_p)\approx\sqrt{p(1-p)/n_{\mathrm{eff}}}/f(q_p)$, what is the Monte Carlo standard error of the lower endpoint?
   Answer: Approximately $0.130$, using the effective sample size and local density When estimating quantiles from MCMC output, you must account for two sources of imprecision: autocorrelation (which inflates effective variance) and the local posterior density (which controls how "steep" the distribution is at the quantile). The key formula here is $\operatorname{SE}(\hat{q}_p) \approx \frac{\sqrt{p(1-p)/n_{\text{eff}}}}{f(q_p)}$, where $p$ is the tail probability, $n_{\text{eff}}$ is the effective sample size, and $f(q_p)$ is the posterior density at the quantile. For the lower endpoint of a 95% central interval, $p = 0.025$. Plugging in: the numerator is $\sqrt{0.025 \times 0.975 / 400} = \sqrt{0.024375/400} = \sqrt{0.000060938} \approx 0.00781$. Dividing by the local density $f(q_p) = 0.06$ gives $0.00781 / 0.06 \approx 0.130$. This confirms B as correct. Choice A is wrong because it stops at the numerator (roughly $0.008$) and never divides by the density — it ignores how the local shape of the posterior amplifies quantile uncertainty when the density is small. Choice C uses the full $n = 4000$ raw draws instead of the effective sample size $n_{\text{eff}} = 400$, underestimating the standard error by a factor of $\sqrt{10}$; ignoring autocorrelation is a classic MCMC mistake. Choice D substitutes $p = 0.5$ (the median probability) rather than the correct tail probability $p = 0.025$, producing an inflated and meaningless result. A reliable study strategy: always identify which $p$ to use — it's the cumulative probability at that quantile, not 0.5. Then remember that small local density magnifies quantile Monte Carlo error significantly.
9. Which pair gives the posterior median of $\phi$ and its central $95$ percent credible interval, respectively?
   Answer: Median $1$; interval $[0.141,\ 7.10]$ When you apply a monotone transformation to a random variable, quantiles are preserved — meaning the median and credible interval boundaries transform the same way the variable does. This is the central insight for any question involving a change of parameterization in Bayesian inference. Here, $\eta = \log\phi$, so $\phi = e^\eta$. Since $\eta \mid y \sim N(0,1)$, the posterior median of $\eta$ is $0$ (the normal distribution is symmetric about its mean). Applying the exponential transformation, the posterior median of $\phi$ is $e^0 = 1$. For the 95% central credible interval, the endpoints of $\eta$ are the familiar $\pm 1.96$, giving $\phi$ bounds of $e^{-1.96} \approx 0.141$ and $e^{1.96} \approx 7.10$. This confirms answer A. Answer B is wrong because $1.649$ would be the posterior mean of $\phi$, not the median. The mean of a log-normal requires the formula $e^{\mu + \sigma^2/2} = e^{0 + 0.5} \approx 1.649$. Confusing mean with median after a nonlinear transformation is a classic trap — they are no longer equal once you leave the symmetric normal scale. Answer C uses incorrect interval endpoints. The values $0.383$ and $2.61$ correspond to $e^{\pm 1}$, as if the interval on the $\eta$ scale were $[-1, 1]$ rather than $[-1.96, 1.96]$. Answer D simply reports the posterior of $\eta$ directly — forgetting to back-transform to the $\phi$ scale entirely. Your study tip: whenever a posterior is given for a transformed parameter, always back-transform quantiles (not moments like the mean) using the inverse transformation. Medians and interval endpoints transform cleanly; means do not.
10. Using the highest-posterior-mass construction, which is a minimum-cardinality credible set having posterior probability at least $0.70$?
    Answer: $\{-2,2\}$ When building a highest-posterior-mass (HPM) credible set, your goal is to include parameter values in decreasing order of posterior probability until the cumulative mass meets or exceeds your target threshold — while using as few points as possible (minimum cardinality). Start by ranking the values by posterior probability: $\theta = -2: 0.36, \quad \theta = 2: 0.34, \quad \theta = 1: 0.16, \quad \theta = -1: 0.14$ Now greedily add values from highest to lowest mass. Taking $\theta = -2$ gives cumulative mass $0.36$, which falls short of $0.70$. Adding the next highest, $\theta = 2$, gives $0.36 + 0.34 = 0.70$. That meets the threshold exactly, so you stop. The resulting set is $\{-2, 2\}$, which is answer A — the correct choice. Answer B, $\{-2,-1,1,2\}$, includes all four values (cumulative mass $1.00$), which certainly covers $0.70$ but violates the minimum-cardinality requirement — you're adding unnecessary points. Answer C, $\{-2,-1,2\}$, has cumulative mass $0.36 + 0.14 + 0.34 = 0.84 \geq 0.70$, but it skips $\theta = 1$ (probability $0.16$) while including $\theta = -1$ (probability $0.14$), which violates the HPM rule of selecting by highest mass first. Answer D, $\{-2,1,2\}$, has mass $0.36+0.16+0.34 = 0.86$, but again uses three points when two suffice. Study tip: Always sort by posterior probability descending and accumulate greedily — stop the moment you hit the threshold. Any set with more elements than necessary, or that skips a higher-probability value to include a lower one, fails the HPM construction.

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