Exam MAS-II — Bayesian Analysis Practice Flashcards

**Answer: (C).** Beta is conjugate to the binomial: the posterior is $\text{Beta}(\alpha+x,\ \beta+n-x)=\text{Beta}(2+7,\ 3+10-7)=\text{Beta}(9,6)$. Under squared-error loss the Bayes estimate is the posterior **mean** $=\dfrac{\alpha^*}{\alpha^*+\beta^*}=\dfrac{9}{9+6}=\dfrac{9}{15}=0.60$. The prior mean was $\dfrac{2}{5}=0.40$ (distractor A) and the raw sample proportion is $\dfrac{7}{10}=0.70$ (distractor D); the posterior mean $0.60$ is a credibility-weighted blend of the two.

**Answer: (B).** Posterior $=\text{Beta}(\alpha+x,\ \beta+n-x)=\text{Beta}(1+8,\ 1+12-8)=\text{Beta}(9,5)$. Squared-error loss → posterior **mean** $=\dfrac{9}{9+5}=\dfrac{9}{14}\approx 0.643$. The MLE / raw proportion is $\dfrac{8}{12}=0.667$ (distractor C). With a uniform prior the posterior mean is $\dfrac{x+1}{n+2}$ (Laplace's rule of succession), which differs slightly from the MLE because the $\text{Beta}(1,1)$ prior contributes one pseudo-success and one pseudo-failure. The posterior **mode** equals $\dfrac{\alpha^*-1}{\alpha^*+\beta^*-2}=\dfrac{8}{12}=0.667$ (also option C) — the $0$-$1$ loss estimate — which here coincides with the MLE because the prior is flat; the round value $0.700$ (distractor D) and $\dfrac{9}{12}=0.75$ (distractor E, dividing $\alpha^*$ by $n$) are common slips.

**Answer: (D).** By Bayes' theorem, $P(\text{bad}\mid\text{claim})=\dfrac{P(\text{claim}\mid\text{bad})\,P(\text{bad})}{P(\text{claim}\mid\text{good})\,P(\text{good})+P(\text{claim}\mid\text{bad})\,P(\text{bad})}$. Numerator $=0.4(0.3)=0.12$. Denominator $=0.1(0.7)+0.4(0.3)=0.07+0.12=0.19$. $P(\text{bad}\mid\text{claim})=\dfrac{0.12}{0.19}\approx 0.632$. Observing a claim sharply raises the posterior weight on the high-frequency 'bad' class above its prior $0.3$ (distractor A). Forgetting to weight by the priors and using $\dfrac{0.4}{0.1+0.4}=0.8$ (distractor E) treats the two classes as equally likely a priori.

**Answer: (B).** Posterior $=\text{Beta}(2+3,\ 2+5-3)=\text{Beta}(5,4)$. Squared-error loss → posterior **mean** $=\dfrac{5}{5+4}=\dfrac{5}{9}\approx 0.556$. $0$-$1$ loss → posterior **mode** $=\dfrac{\alpha^*-1}{\alpha^*+\beta^*-2}=\dfrac{5-1}{5+4-2}=\dfrac{4}{7}\approx 0.571$. The mode exceeds the mean because $\text{Beta}(5,4)$ is slightly left-skewed. Confusing the two loss functions swaps the answers (distractor C); $0.600$ is the raw sample proportion $\tfrac{3}{5}$, not a posterior summary.

**Answer: (C).** Gamma is conjugate to the Poisson. With the rate parameterization the posterior is $\text{Gamma}\!\left(\alpha+\sum x_i,\ \beta+n\right)$. $\sum x_i = 1+3+0+2+4 = 10$, $n=5$, so the posterior is $\text{Gamma}(4+10,\ 2+5)=\text{Gamma}(14,7)$. Squared-error loss → posterior **mean** $=\dfrac{\alpha^*}{\beta^*}=\dfrac{14}{7}=2.00$. The sample mean is $\dfrac{10}{5}=2.0$ and the prior mean was $\dfrac{4}{2}=2.0$, so both agree. Dividing the total by $n+ \beta$ wrong (e.g. $\tfrac{14}{6}$) gives the $2.33$ distractor; the prior mean alone is $2.0$ here by coincidence — the credibility blend lands at $2.00$.

**Answer: (C).** The likelihood of $1$ success in $2$ trials is $\binom{2}{1}\theta(1-\theta)=2\theta(1-\theta)$. Form joint $=\pi(\theta)\times L$: $\theta=0.3:\ 0.6\cdot 2(0.3)(0.7)=0.6(0.42)=0.252$. $\theta=0.6:\ 0.4\cdot 2(0.6)(0.4)=0.4(0.48)=0.192$. Marginal $=0.252+0.192=0.444$. Posterior: $P(\theta=0.6\mid x)=\dfrac{0.192}{0.444}\approx 0.432$ (and $P(\theta=0.3\mid x)\approx 0.568$). The likelihood $2\theta(1-\theta)$ is maximized at $\theta=0.5$, so the symmetric outcome only modestly favors the smaller $\theta=0.3$; the posterior stays close to — but above — the prior weight $0.4$ on $\theta=0.6$ (distractor B is the unchanged prior).

**Answer: (D).** For the normal-normal model, precisions add: posterior precision $=\dfrac{1}{\tau^{2}}+\dfrac{n}{\sigma^{2}}=\dfrac{1}{20}+\dfrac{5}{80}=0.05+0.0625=0.1125$. Posterior variance $=\dfrac{1}{0.1125}\approx 8.889$. Posterior mean $=\sigma_n^{2}\!\left(\dfrac{\mu_0}{\tau^{2}}+\dfrac{n\bar{x}}{\sigma^{2}}\right)=8.889\!\left(\dfrac{60}{20}+\dfrac{5(72)}{80}\right)=8.889(3.0+4.5)=8.889(7.5)\approx 66.7$. The data precision ($0.0625$) slightly exceeds the prior precision ($0.05$), so the posterior mean is pulled past the midpoint $66.0$ (distractor C) toward $\bar{x}=72$.

**Answer: (C).** Posterior precision $=\dfrac{1}{\tau^{2}}+\dfrac{n}{\sigma^{2}}=\dfrac{1}{64}+\dfrac{16}{64}=\dfrac{17}{64}$, so posterior variance $=\dfrac{64}{17}\approx 3.765$. Posterior mean $=\dfrac{64}{17}\!\left(\dfrac{100}{64}+\dfrac{16(110)}{64}\right)=\dfrac{64}{17}\cdot\dfrac{100+1760}{64}=\dfrac{1860}{17}\approx 109.4$. With $n=16$ observations the data dominate the single prior 'observation,' so the posterior mean sits very close to $\bar{x}=110$, far from the prior mean $100$ and well past the simple midpoint $105$ (distractor B).