Statistics Graduate Level Quiz: Posterior Summaries And Credible Intervals

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Posterior Summaries And Credible IntervalsQuestion 1 of 10

For every possible dataset $y$ , a Bayesian procedure constructs a set $C(y)$ satisfying $P(\theta\in C(y)\mid y)=0.95$ under a proper prior and the stated sampling model.

Which coverage statement follows without requiring any additional assumptions?

For every fixed value of

\theta

, repeated-sampling coverage of

C(Y)

is exactly

0.95

Under the joint prior-predictive distribution of

\theta

and

Y

P\{\theta\in C(Y)\}=0.95

The joint coverage is

0.95

only when every

C(y)

is an equal-tailed posterior interval.

The joint coverage is

0.95

only when the posterior distribution is symmetric for every dataset.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Posterior Summaries And Credible Intervals

Practice Posterior Summaries And Credible Intervals in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Posterior Summaries And Credible Intervals, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For every possible dataset $y$ , a Bayesian procedure constructs a set $C(y)$ satisfying $P(\theta\in C(y)\mid y)=0.95$ under a proper prior and the stated sampling model.

Which coverage statement follows without requiring any additional assumptions?

For every fixed value of $\theta$ , repeated-sampling coverage of $C(Y)$ is exactly $0.95$ .
Under the joint prior-predictive distribution of $\theta$ and $Y$ , $P\{\theta\in C(Y)\}=0.95$ . (correct answer)
The joint coverage is $0.95$ only when every $C(y)$ is an equal-tailed posterior interval.
The joint coverage is $0.95$ only when the posterior distribution is symmetric for every dataset.

Explanation: 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.

Question 2

In a Bayesian regression, the joint posterior distribution of two coefficients is bivariate normal with posterior means $E(\beta_1\mid y)=1$ and $E(\beta_2\mid y)=2$ , posterior variances $0.25$ and $0.36$ , and posterior covariance $-0.12$ .

What is the approximate central $95$ percent posterior credible interval for the contrast $\delta=\beta_1-\beta_2$ ?

$[-0.81,\ 2.81]$
$[-2.53,\ 0.53]$
$[-2.19,\ 0.19]$
$[-2.81,\ 0.81]$ (correct answer)

Explanation: 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}

-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.

Question 3

An analyst estimates the lower endpoint of a central $95$ percent posterior credible interval using $4000$ autocorrelated MCMC draws. The effective sample size for quantile estimation is $400$ . At the estimated lower quantile, the posterior density is approximately $0.06$ .

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?

Approximately $0.008$ , using the binomial quantile variance directly
Approximately $0.130$ , using the effective sample size and local density (correct answer)
Approximately $0.041$ , treating all MCMC draws as independent
Approximately $0.417$ , substituting the median probability for the tail probability

Explanation: 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.

Question 4

The posterior distribution of a scalar parameter assigns probabilities $0.45$ , $0.35$ , and $0.20$ to the values $0$ , $2$ , and $10$ , respectively.

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?

Squared-error estimate $2.7$ ; absolute-error estimate $0$ ; interval $[0,10]$
Squared-error estimate $2$ ; absolute-error estimate $2.7$ ; interval $[0,2]$
Squared-error estimate $2.7$ ; absolute-error estimate $2$ ; interval $[0,2]$ (correct answer)
Squared-error estimate $2$ ; absolute-error estimate $2$ ; interval $[2,10]$

Explanation: 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.

Question 5

The posterior distributions of $\theta_1$ and $\theta_2$ are independent standard normal distributions. A rectangular credible region is formed as $[-z,z]\times[-z,z]$ .

Approximately what value of $z$ makes the posterior probability of this rectangular region equal to $0.95$ ?

$z=1.64$ , based on two marginal $90$ percent intervals
$z=1.96$ , based on two marginal $95$ percent intervals
$z=2.24$ , based on marginal probability about $0.9747$ (correct answer)
$z=2.58$ , based on two marginal $99$ percent intervals

Explanation: 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.

Question 6

A normal sampling model has known observation variance. The prior is $\mu \sim N(0,4)$ , and the sample mean from $9$ observations is $\bar y=1.2$ . Each observation has variance $9$ .

Which interval is the central $90$ percent posterior credible interval for $\mu$ ?

$(-0.51,\ 2.43)$ (correct answer)
$(-0.45,\ 2.85)$
$(-0.67,\ 2.59)$
$(-3.73,\ 3.73)$

Explanation: 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}

\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.

Question 7

A discrete parameter has posterior probabilities $P(\theta=-2\mid y)=0.36$ , $P(\theta=-1\mid y)=0.14$ , $P(\theta=1\mid y)=0.16$ , and $P(\theta=2\mid y)=0.34$ .

Using the highest-posterior-mass construction, which is a minimum-cardinality credible set having posterior probability at least $0.70$ ?

$\{-2,2\}$ (correct answer)
$\{-2,-1,1,2\}$
$\{-2,-1,2\}$
$\{-2,1,2\}$

Explanation: 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.

Question 8

For a positive parameter $\phi$ , define $\eta=\log\phi$ . After observing the data, the posterior distribution is $\eta\mid y\sim N(0,1)$ .

Which pair gives the posterior median of $\phi$ and its central $95$ percent credible interval, respectively?

Median $1$ ; interval $[0.141,\ 7.10]$ (correct answer)
Median $1.649$ ; interval $[0.141,\ 7.10]$
Median $1$ ; interval $[0.383,\ 2.61]$
Median $0$ ; interval $[-1.96,\ 1.96]$

Explanation: 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

0

(the normal distribution is symmetric about its mean). Applying the exponential transformation, the posterior median of

\phi

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.

Question 9

The posterior density of a nonnegative parameter is $p(\theta\mid y)=2e^{-2\theta}$ for $\theta\geq0$ .

Which of the following is the $90$ percent highest-posterior-density credible set?

$[0,\ 1.151]$ (correct answer)
$[0.0256,\ 1.498]$
$[0,\ 1.498]$
$[0.0527,\ 1.151]$

Explanation: 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.

Question 10

For a normal sampling model, the posterior distribution of the unknown mean is $\mu\mid y\sim N(10,1)$ . Conditional on $\mu$ , a future observation satisfies $Y_{\mathrm{new}}\mid\mu\sim N(\mu,4)$ .

Which option gives, respectively, a central $95$ percent credible interval for $\mu$ and a central $95$ percent posterior predictive interval for $Y_{\mathrm{new}}$ ?

For $\mu$ : $[6.08,13.92]$ ; for $Y_{\mathrm{new}}$ : $[6.08,13.92]$
For $\mu$ : $[8.04,11.96]$ ; for $Y_{\mathrm{new}}$ : $[8.04,11.96]$
For $\mu$ : $[5.62,14.38]$ ; for $Y_{\mathrm{new}}$ : $[8.04,11.96]$
For $\mu$ : $[8.04,11.96]$ ; for $Y_{\mathrm{new}}$ : $[5.62,14.38]$ (correct answer)

Explanation: 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