Statistics Graduate Level Quiz: Bayesian Inference Basics

10 questions · exam conditions

0:00

Bayesian Inference BasicsQuestion 1 of 10

Suppose $Y_1,Y_2,Y_3$ are conditionally independent with $Y_i\mid\theta\sim N(\theta,1)$ . An analyst uses the improper prior $\pi(\theta)\propto1$ on the real line.

Which statement correctly describes the resulting Bayesian analysis?

The posterior is

N(\bar Y,1/3)

, but an ordinary marginal likelihood retains an arbitrary prior scaling constant.

The posterior remains improper because multiplying a proper likelihood by an improper prior cannot produce a density.

The posterior is

N(\bar Y,1)

, and its normalization also makes the original prior a proper distribution.

The posterior is proper, and the marginal likelihood is uniquely defined because the likelihood determines the missing prior constant.

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Bayesian Inference Basics

Practice Bayesian Inference Basics 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 Bayesian Inference Basics, 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

Suppose $Y_1,Y_2,Y_3$ are conditionally independent with $Y_i\mid\theta\sim N(\theta,1)$ . An analyst uses the improper prior $\pi(\theta)\propto1$ on the real line.

Which statement correctly describes the resulting Bayesian analysis?

The posterior is $N(\bar Y,1/3)$ , but an ordinary marginal likelihood retains an arbitrary prior scaling constant. (correct answer)
The posterior remains improper because multiplying a proper likelihood by an improper prior cannot produce a density.
The posterior is $N(\bar Y,1)$ , and its normalization also makes the original prior a proper distribution.
The posterior is proper, and the marginal likelihood is uniquely defined because the likelihood determines the missing prior constant.

Explanation: When working with improper priors in Bayesian analysis, you need to track two separate questions: (1) does the posterior normalize to a proper density, and (2) does the marginal likelihood have a well-defined value? With the flat prior

\pi(\theta) \propto 1

and likelihood

L(\theta) \propto \exp\!\left(-\frac{3}{2}(\theta - \bar{Y})^2\right)

, the posterior is proportional to a Gaussian in

\theta

. Completing the square gives a proper

N(\bar{Y}, 1/3)

posterior — the integral over

\theta

converges even though the prior was improper. So far so good. However, the marginal likelihood requires integrating the joint

\pi(\theta)L(\theta)

over

\theta

. Since

\pi(\theta) \propto c

for an unspecified constant

c

, the marginal likelihood inherits that same arbitrary scaling factor. The posterior is fine because

c

cancels in the ratio

\pi(\theta \mid y) \propto \pi(\theta)L(\theta)

, but the marginal likelihood itself does not cancel it. This makes answer A correct. Answer B is wrong because a likelihood that is integrable in

\theta

can absolutely normalize with an improper prior — the posterior here is a textbook example of this. Answer C incorrectly claims the posterior's normalizability "fixes" the prior into a proper distribution. Normalizing the posterior doesn't retroactively assign a finite total mass to

\pi(\theta) \propto 1

\mathbb{R}

. Answer D is the subtlest trap: the likelihood does not determine the missing constant in

\pi(\theta)

, because the prior constant never appears in the likelihood function. Remember: improper priors can yield proper posteriors, but the marginal likelihood remains scale-ambiguous — a critical distinction for Bayes factor comparisons.

Question 2

Independent counts satisfy $Y_i\mid\lambda\sim\operatorname{Poisson}(t_i\lambda)$ , where the exposures $t_i$ are known. The prior for $\lambda$ is gamma with shape $1/2$ and rate $1$ . Across all observations, the total count is $2$ and the total exposure is $4$ .

What is the posterior mode of $\lambda$ ?

$\frac{1}{2}$ , using the mean of the posterior gamma distribution
$\frac{3}{10}$ , using the shape-minus-one formula with the posterior rate (correct answer)
$\frac{3}{8}$ , using the posterior shape but omitting the prior rate
$\frac{5}{8}$ , using the posterior shape divided only by total exposure

Explanation: When you see a Bayesian inference problem with a Poisson likelihood and Gamma prior, your first move should be identifying the conjugate posterior and then carefully applying the mode formula. Here's the framework: if

Y_i \mid \lambda \sim \text{Poisson}(t_i \lambda)

with a

\text{Gamma}(\alpha_0, \beta_0)

prior (shape, rate), the posterior is

\text{Gamma}(\alpha_0 + \sum y_i,\ \beta_0 + \sum t_i)

. With prior shape

\alpha_0 = 1/2

, prior rate

\beta_0 = 1

, total count

\sum y_i = 2

, and total exposure

\sum t_i = 4

, the posterior is

\text{Gamma}(5/2,\ 5)

. The mode of a

\text{Gamma}(\alpha, \beta)

distribution is

\frac{\alpha - 1}{\beta}

, valid when

\alpha > 1

. This gives

\frac{5/2 - 1}{5} = \frac{3/2}{5} = \frac{3}{10}

, confirming B is correct. Choice A is wrong because it uses the posterior mean (

\alpha/\beta = 1/2

) rather than the mode, conflating two distinct summary statistics. Choice C divides

\alpha - 1 = 3/2

only by the total exposure

4

, forgetting to include the prior rate in the denominator — a classic bookkeeping error when accumulating the rate parameter. Choice D divides the full posterior shape

5/2

by the exposure alone, making two errors simultaneously: ignoring the prior rate and forgetting the "minus one" in the mode formula. Your study tip: memorize the Gamma mode as

(\alpha - 1)/\beta

and always verify both updated parameters — shape and rate — before plugging in. Missing either one is the most common trap in conjugate Bayesian problems.

Question 3

A coin is modeled by either $H_0:\theta=1/2$ or $H_1:\theta\sim\operatorname{Uniform}(0,1)$ , with equal prior probabilities on the two hypotheses. In $10$ tosses, the observed result is $9$ heads and $1$ tail.

What is the posterior probability of $H_1$ ?

$\frac{55}{567}$ , obtained by reversing the Bayes factor so that $H_0$ rather than $H_1$ appears in the numerator
$\frac{5}{6}$ , obtained by using the posterior mean of $\theta$ under $H_1$ as a model probability
$\frac{0.9^9(0.1)}{0.9^9(0.1)+0.5^{10}}$ , obtained by substituting the MLE of $\theta$ under $H_1$ instead of integrating
$\frac{512}{567}$ , obtained by integrating the likelihood over the uniform prior under $H_1$ (correct answer)

Explanation: When a question mixes a point hypothesis with a composite (prior-weighted) hypothesis, your core tool is Bayes' theorem on the hypothesis level. You need to compute the marginal likelihood of the data under each hypothesis, weight by the prior on each hypothesis, and normalize. Under

H_0

, the likelihood is simply

\theta = 1/2

plugged in:

P(X \mid H_0) = \binom{10}{9}(1/2)^{10} = 10/1024

. Under

H_1

\theta

is unknown with a Uniform(0,1) prior, so you must integrate out

\theta

P(X \mid H_1) = \int_0^1 \binom{10}{9}\theta^9(1-\theta)\,d\theta = 10\cdot B(10,2) = 10\cdot\frac{9!\,1!}{11!} = \frac{10}{110} = \frac{1}{11}.

With equal priors on

H_0

and

H_1

, the posterior probability of

H_1

is:

P(H_1 \mid X) = \frac{1/11}{1/11 + 10/1024} = \frac{1/11}{1/11 + 5/512} = \frac{512}{512 + 55} = \frac{512}{567}.

This confirms D is correct. A is wrong because it inverts the Bayes factor, putting

H_0

in the numerator and yielding

55/567

, which is actually

P(H_0 \mid X)

, not

P(H_1 \mid X)

. B is wrong because it substitutes the posterior mean of

\theta

under

H_1

(which is

10/12 = 5/6

) as if it were a probability of

H_1

itself — a category error confusing a parameter estimate with a hypothesis probability. C is wrong because it plugs in the MLE (

\hat\theta = 0.9

) rather than integrating over the prior. Skipping the integral is the classic Bayesian composite-hypothesis mistake. Study tip: Whenever one hypothesis has a free parameter with a prior, always marginalize — integrate the likelihood against the prior. Never substitute a point estimate for the integral.

Question 4

A quality-control analyst assigns a $\operatorname{Beta}(2,3)$ prior to the unknown probability $\theta$ that an item is defective. In a conditionally independent sample of $10$ items, $7$ are defective.

What is the posterior predictive probability that both of the next two items will be defective?

$\frac{9}{25}$ , obtained by treating the two outcomes as independent at the posterior mean
$\frac{3}{8}$ , obtained by integrating the two-outcome probability over the posterior (correct answer)
$\frac{7}{15}$ , obtained by updating the prior predictive probability using the observed proportion
$\frac{9}{16}$ , obtained by using the posterior success count after one additional trial

Explanation: When you see a question about posterior predictive probabilities, your instinct should be to integrate over uncertainty in

\theta

rather than plug in a point estimate. The posterior predictive distribution accounts for the fact that

\theta

is unknown even after observing data. With a

\text{Beta}(2,3)

prior and 7 defectives in 10 trials, the posterior is

\text{Beta}(2+7, 3+3) = \text{Beta}(9, 6)

. To find the probability that the next two items are both defective, you compute

P(X_1=1, X_2=1 \mid \text{data}) = E[\theta^2 \mid \text{data}]

, since the two future draws are conditionally independent given

\theta

. For a

\text{Beta}(\alpha, \beta)

distribution,

E[\theta^2] = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)} = \frac{9 \cdot 10}{15 \cdot 16} = \frac{90}{240} = \frac{3}{8}

. That confirms B. A is wrong because it squares the posterior mean:

\left(\frac{9}{15}\right)^2 = \frac{9}{25}

. This ignores posterior variance — squaring the mean underestimates

E[\theta^2]

by Jensen's inequality (since

\theta^2

is convex). C is a fabricated procedure with no Bayesian foundation; "updating the prior predictive with the observed proportion" is not a standard operation. D imagines conditioning on one more hypothetical success to get a

\text{Beta}(10,6)

posterior mean of

\frac{10}{16} = \frac{5}{8}

, then squaring — an ad hoc approach with no justification. The key study tip: whenever computing the posterior predictive for multiple future observations, always compute

E[\theta^k \mid \text{data}]

using the Beta moment formula — never just raise the posterior mean to the

k

-th power.

Question 5

The same Bernoulli outcome sequence contains $r$ successes and $f$ failures, with the final observation being a success. Analyst I uses a design with fixed sample size $r+f$ . Analyst II uses a design that stops upon observing the $r$ th success. Both analysts use the same proper prior for the success probability $\theta$ .

Which statement about their posterior distributions for $\theta$ is correct?

They differ because the stopping rule changes the powers of $\theta$ and $1-\theta$ in the likelihood.
They agree only when the common prior is uniform, because otherwise the stopping rule interacts with the prior.
They differ because the negative-binomial design contributes one fewer failure to the likelihood than the fixed-size design.
They agree because the two likelihoods differ only by a factor that does not depend on $\theta$ . (correct answer)

Explanation: Whenever you see a question involving different sampling designs applied to the same data, your first instinct should be to write out both likelihoods and compare them term by term — specifically, check whether they differ only by a multiplicative constant that is free of

\theta

. For the fixed-sample design (Analyst I), the likelihood is binomial:

L_I(\theta) \propto \binom{r+f}{r} \theta^r (1-\theta)^f

. For the negative-binomial design (Analyst II), which stops at the

r

th success, the likelihood is:

L_{II}(\theta) \propto \binom{r+f-1}{r-1} \theta^r (1-\theta)^f

. Notice that both likelihoods share the same kernel

\theta^r(1-\theta)^f

. The only difference is the combinatorial prefactor —

\binom{r+f}{r}

versus

\binom{r+f-1}{r-1}

— and neither depends on

\theta

. Because the posterior is proportional to likelihood times prior, and these constants cancel in normalization, both analysts arrive at identical posteriors for any proper prior. This is the essence of the Likelihood Principle: inference about

\theta

depends only on the

\theta

-dependent part of the likelihood. Choice A is wrong because the powers of

\theta

and

1-\theta

are identical in both likelihoods — the stopping rule only affects the combinatorial coefficient. Choice B is wrong because posterior agreement holds for any proper prior, not just the uniform; the prior is irrelevant to this cancellation. Choice C is wrong because both likelihoods involve exactly

f

failures — there is no "one fewer failure" discrepancy between the two designs. As a study tip: when comparing likelihoods across designs, factor out everything that doesn't involve the parameter. If what remains is identical, the posteriors must agree regardless of the prior chosen.

Question 6

Let $\theta$ have a uniform prior on $(0,1)$ . Given $\theta$ , one observation satisfies $X\sim\operatorname{Bernoulli}(\theta)$ , and the observed value is $X=1$ . Define the transformed parameter $\phi=\theta^2$ .

Which is the posterior density of $\phi$ on $(0,1)$ ?

$f(\phi\mid X=1)=2\phi$ , obtained by applying the Bernoulli likelihood directly to $\phi$
$f(\phi\mid X=1)=1$ , obtained after including the transformation Jacobian (correct answer)
$f(\phi\mid X=1)=\frac{1}{2\sqrt{\phi}}$ , obtained by transforming only the original prior
$f(\phi\mid X=1)=2\sqrt{\phi}$ , obtained by substituting $\theta=\sqrt{\phi}$ without a Jacobian

Explanation: When transforming parameters in a Bayesian setting, you must apply the full change-of-variables procedure to the posterior density, not just the prior or just the likelihood. The posterior of

\phi = \theta^2

requires substituting

\theta = \sqrt{\phi}

and multiplying by the Jacobian

\left|\frac{d\theta}{d\phi}\right| = \frac{1}{2\sqrt{\phi}}

. Start with Bayes' theorem. The prior is

\pi(\theta) = 1

(0,1)

, and the likelihood given

X=1

L(\theta) = \theta

. So the posterior is

\pi(\theta \mid X=1) \propto \theta \cdot 1 = \theta

, which normalizes to

f(\theta \mid X=1) = 2\theta

(0,1)

. Now transform:

f(\phi \mid X=1) = f(\theta \mid X=1)\big|_{\theta=\sqrt{\phi}} \cdot \frac{1}{2\sqrt{\phi}} = 2\sqrt{\phi} \cdot \frac{1}{2\sqrt{\phi}} = 1

. This confirms B — the posterior of

\phi

is Uniform

(0,1)

. Choice A mistakenly plugs

\phi

directly into the likelihood as if

\phi

were itself a Bernoulli parameter, skipping both the prior transformation and the Jacobian. Choice C applies the Jacobian only to the prior, ignoring the posterior's likelihood component — it transforms the wrong density. Choice D substitutes

\theta = \sqrt{\phi}

into the posterior correctly but forgets to include the Jacobian factor, leaving the density unnormalized and incorrect. The key strategy: whenever you change variables in a posterior, always transform the complete posterior density and never forget the Jacobian. A quick self-check is verifying your result integrates to 1.

Question 7

Suppose $X_1,\ldots,X_4$ are conditionally independent with $X_i\mid\mu\sim N(\mu,9)$ . The prior is $\mu\sim N(1,4)$ , and the observed sample mean is $\bar X=4$ .

Which posterior distribution for $\mu$ is correct?

$N\left(\frac{25}{13},\frac{36}{13}\right)$ , treating the sample mean as having variance $9$
$N\left(\frac{52}{25},\frac{36}{25}\right)$ , attaching each precision to the other source's mean
$N\left(\frac{73}{25},\frac{36}{25}\right)$ , combining prior and sample-mean precisions (correct answer)
$N\left(\frac{73}{25},\frac{25}{36}\right)$ , reporting the posterior precision as its variance

Explanation: Bayesian updating with a normal-normal conjugate model is a classic graduate topic. Whenever you see a normal likelihood and normal prior, remember the precision-weighted posterior mean formula: the posterior mean is a weighted average of the prior mean and the data mean, where weights are precisions (reciprocals of variances), and the posterior variance is the reciprocal of the total precision. Here's the key mechanics. With

n=4

observations and individual variance

\sigma^2=9

, the sample mean

\bar{X}

has variance

\sigma^2/n = 9/4

, giving it precision

4/9

. The prior

\mu \sim N(1,4)

has precision

1/4

. The posterior precision is the sum:

1/4 + 4/9 = 9/36 + 16/36 = 25/36

, so the posterior variance is

36/25

. The posterior mean is

\frac{(1/4)(1)+(4/9)(4)}{25/36} = \frac{9/36+64/36}{25/36} = \frac{73/36}{25/36} = 73/25

. This confirms C is correct:

N\!\left(\tfrac{73}{25},\tfrac{36}{25}\right)

. Choice A makes the error of using variance

9

for

\bar{X}

rather than

9/4

, ignoring that averaging four observations reduces variance by

n

. Choice B swaps the means: it attaches the prior precision to the data mean and vice versa, a simple but fatal bookkeeping error. Choice D computes the right posterior mean but reports the precision

25/36

as if it were the variance — a classic sign/inverse confusion. Your study tip: always track whether you're working with variance or precision. Label each quantity explicitly, because the most common errors on conjugate-normal problems come from accidentally inverting or misassigning these reciprocal quantities.

Question 8

A normal-means model has prior probability $0.4$ that $\theta=0$ and prior probability $0.6$ that $\theta\sim N(0,1)$ . Conditional on $\theta$ , an observation satisfies $Y\mid\theta\sim N(\theta,1)$ . The observed value is $Y=0$ .

What is the posterior probability that $\theta=0$ ?

$\frac{2}{5}$ , because the observation is equally centered under both prior components
$\frac{2}{2+3\sqrt{2}}$ , obtained by reversing the two components' predictive density ratio
$\frac{\sqrt{2}}{1+\sqrt{2}}$ , obtained by comparing predictive densities without prior component weights
$\frac{4}{4+3\sqrt{2}}$ , obtained by weighting the two component-specific predictive densities (correct answer)

Explanation: When a prior mixes a point mass with a continuous distribution, posterior inference requires Bayes' theorem applied carefully to each component's predictive density — the marginal likelihood of the data under that component. For the point-mass component (

\theta = 0

), the predictive density of

Y = 0

is simply

f(Y=0 \mid \theta=0) = \phi(0) = \frac{1}{\sqrt{2\pi}}

, where

\phi

is the standard normal density. For the continuous component (

\theta \sim N(0,1)

), you must marginalize:

Y \sim N(0, 1+1) = N(0,2)

, giving predictive density

\frac{1}{\sqrt{4\pi}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2\pi}}

. Applying Bayes' theorem with prior weights 0.4 and 0.6:

P(\theta=0 \mid Y=0) = \frac{0.4 \cdot \frac{1}{\sqrt{2\pi}}}{0.4 \cdot \frac{1}{\sqrt{2\pi}} + 0.6 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2\pi}}} = \frac{0.4}{0.4 + \frac{0.6}{\sqrt{2}}} = \frac{4}{4 + 3\sqrt{2}}

This confirms D is correct. Choice A ignores that the continuous component's predictive variance is 2, not 1 — the observation

Y=0

is actually more concentrated under

\theta=0

than under the diffuse component, so the point mass gains posterior weight. Choice B inverts the likelihood ratio, swapping numerator and denominator. Choice C applies equal prior weights (0.5/0.5) instead of the given 0.4/0.6 split. Your study tip: in spike-and-slab or mixed priors, always compute the marginal likelihood for each component separately before combining with prior weights — conflating the prior and likelihood is the most common error here.

Question 9

Before observing data, a researcher assigns probability $0.20$ to hypothesis $H_1$ and probability $0.80$ to hypothesis $H_0$ . After observing the data, the posterior probability of each hypothesis is $0.50$ .

What is the Bayes factor in favor of $H_1$ over $H_0$ ?

$4$ , the posterior odds divided by the prior odds (correct answer)
$5$ , the posterior odds divided by the prior probability of $H_1$
$\frac{5}{2}$ , the posterior probability divided by the prior probability
$\frac{3}{4}$ , the increase in probability expressed relative to the prior odds

Explanation: Whenever you encounter a Bayes factor question, anchor yourself to one core relationship: the Bayes factor measures how much the data shift your beliefs, defined precisely as the ratio of posterior odds to prior odds. The prior odds in favor of

H_1

are

\frac{0.20}{0.80} = \frac{1}{4}

. After seeing the data, the posterior odds are

\frac{0.50}{0.50} = 1

. The Bayes factor is therefore:

BF_{10} = \frac{\text{posterior odds}}{\text{prior odds}} = \frac{1}{1/4} = 4

This confirms A is correct. The Bayes factor of 4 tells you the data made

H_1

four times more credible relative to

H_0

than it was before. B is wrong because dividing the posterior odds (1) by the prior probability of

H_1

(0.20) gives

5

, which mixes odds with a raw probability — these are not interchangeable. The formula requires odds in both numerator and denominator. C divides the posterior probability of

H_1

(0.50) by its prior probability (0.20), yielding

2.5

. This ratio describes how much the probability of

H_1

alone changed, but ignores the corresponding shift in

H_0

's probability — the Bayes factor must account for both hypotheses simultaneously through odds. D constructs an arbitrary quantity unrelated to any standard definition of the Bayes factor. A reliable study tip: always convert probabilities to odds before computing the Bayes factor. Writing out prior odds and posterior odds explicitly before dividing will prevent the unit-mixing errors that traps B and C represent.

Question 10

A disease has prevalence $0.10$ . Two diagnostic tests each have sensitivity $0.80$ and specificity $0.90$ . Conditional on disease status, the two test results are independent. A patient receives a positive result on the first test and a negative result on the second.

What is the posterior probability that the patient has the disease?

$\frac{8}{17}$ , based only on the likelihood ratio from the positive result
$\frac{2}{11}$ , based only on the likelihood ratio from the negative result
$\frac{16}{97}$ , based on multiplying both conditional likelihood ratios (correct answer)
$\frac{8}{89}$ , based on treating the discordant results as canceling one another

Explanation: Bayesian updating with multiple independent tests is the core concept here. When conditional independence holds, you can multiply likelihood ratios sequentially — each test result updates the odds independently. Start with prior odds:

\frac{P(D)}{P(\bar{D})} = \frac{0.10}{0.90} = \frac{1}{9}

. Now compute each test's likelihood ratio. For a positive result:

LR_+ = \frac{\text{sensitivity}}{1 - \text{specificity}} = \frac{0.80}{0.10} = 8

. For a negative result:

LR_- = \frac{1 - \text{sensitivity}}{\text{specificity}} = \frac{0.20}{0.90} = \frac{2}{9}

. Because the tests are conditionally independent, multiply both ratios against the prior odds:

\text{Posterior odds} = \frac{1}{9} \times 8 \times \frac{2}{9} = \frac{16}{81}

. Converting to probability:

\frac{16}{81 + 16} = \frac{16}{97}

. That confirms C. Choice A applies only the positive test's likelihood ratio, giving posterior odds

\frac{8}{9}

, hence

\frac{8}{17}

— it ignores the negative result entirely. Choice B applies only the negative test's likelihood ratio, ignoring the positive result, yielding

\frac{2}{11}

. Both A and B commit the error of discarding information from one of the two tests. Choice D assumes the two discordant results "cancel out" and reduces back toward the prior — this is intuitive but mathematically wrong; the results don't cancel, they both independently update the odds in their respective directions. Your takeaway: whenever tests are conditionally independent, posterior odds = prior odds × product of all individual likelihood ratios. Never discard a test result, even when results conflict — discordant results still each carry real evidential weight.