Computing via Conditioning - Statistics Graduate Level

1. If the aggregate loss is $S=\sum_{i=1}^{N}X_i$, what is $Var(S)$?
   Answer: $35$ When you see aggregate loss variance with a random claim intensity, reach for the law of total variance combined with the compound Poisson variance formula. The key insight is that $\Lambda$ itself is random, so you need an extra layer of conditioning. Start by conditioning on $\Lambda$. Given $\Lambda$, $S$ is a compound Poisson sum, so: $Var(S|\Lambda) = \Lambda \cdot E[X_i^2] = \Lambda(Var(X_i) + E[X_i]^2) = \Lambda(5+4) = 9\Lambda$ $E[S|\Lambda] = \Lambda \cdot E[X_i] = 2\Lambda$ Now apply the law of total variance: $Var(S) = E[Var(S|\Lambda)] + Var(E[S|\Lambda])$ $= E[9\Lambda] + Var(2\Lambda)$ $= 9E[\Lambda] + 4\,Var(\Lambda)$ $= 9(3) + 4(2) = 27 + 8 = 35$ So C) 35 is correct. A) 23 ignores the extra variance from the random intensity entirely — it's roughly what you'd get treating $\Lambda$ as fixed at its mean, computing $9(3) = 27$ minus an arithmetic error. B) 27 is a common trap: it computes $9E[\Lambda] = 27$ but forgets the second term $Var(E[S|\Lambda]) = 4\,Var(\Lambda) = 8$. This is the mistake of omitting the between-group variance component. D) 45 likely confuses $Var(\Lambda)$ or $E[X_i^2]$ with incorrect values, overstating one of the components. Study tip: Memorize the two-step structure — condition on the mixing variable, then apply total variance — because this exact framework appears frequently in actuarial and graduate-level insurance problems whenever the Poisson mean is itself random.
2. Which pair gives $E[X]$ and $Var(X)$, respectively?
   Answer: $(3,3.36)$ When you see a random count $N$ feeding into a binomial-like selection process, you're in the territory of compound distributions — specifically, the law of total expectation and the law of total variance. For the mean, the law of total expectation gives $E[X] = E[E[X \mid N]]$. Since conditional on $N$, $X \sim \text{Binomial}(N, 0.3)$, we have $E[X \mid N] = 0.3N$. Therefore, $E[X] = 0.3 \cdot E[N] = 0.3 \times 10 = 3$. For the variance, the law of total variance states $\text{Var}(X) = E[\text{Var}(X \mid N)] + \text{Var}(E[X \mid N])$. The two components are: $E[\text{Var}(X \mid N)] = E[N \cdot 0.3 \cdot 0.7] = 0.21 \cdot E[N] = 0.21 \times 10 = 2.10$, and $\text{Var}(E[X \mid N]) = \text{Var}(0.3N) = 0.09 \cdot \text{Var}(N) = 0.09 \times 14 = 1.26$. Adding these: $\text{Var}(X) = 2.10 + 1.26 = 3.36$. This confirms answer A, $(3, 3.36)$. Answer B gives variance $3.00$, which likely comes from treating $N$ as fixed at its mean (ignoring the extra variability from $N$ being random). Answer C reports only the first component $E[\text{Var}(X \mid N)] = 2.10$, forgetting to add the $\text{Var}(E[X \mid N])$ term entirely. Answer D uses an incorrect mean of $4.2$, perhaps confusing the retention probability with something else. The key study tip: whenever $N$ is random, never plug $E[N]$ directly into the binomial variance formula — always apply the full law of total variance, which adds an extra term proportional to $\text{Var}(N)$.
3. Suppose $\Theta\sim N(0,4)$ and, conditional on $\Theta$, an observation satisfies $Y|\Theta\sim N(\Theta,5)$. What is $Var(E[\Theta|Y])$?
   Answer: $16/9$ When you see a question involving a normal prior and normal likelihood, your first instinct should be Bayesian updating and the Law of Total Variance: $Var(Y) = E[Var(Y|X)] + Var(E[Y|X])$. Here, the question asks for $Var(E[\Theta|Y])$, which is precisely one piece of that decomposition — but applied to $\Theta$ given $Y$. Start by finding the posterior $E[\Theta|Y]$. With $\Theta \sim N(0,4)$ (prior variance $\tau^2 = 4$) and $Y|\Theta \sim N(\Theta, 5)$ (likelihood variance $\sigma^2 = 5$), the posterior mean in a normal-normal model is: $E[\Theta|Y] = \frac{\tau^2}{\tau^2 + \sigma^2} Y = \frac{4}{4+5}Y = \frac{4}{9}Y$ Since $Y \sim N(0, \tau^2 + \sigma^2) = N(0, 9)$ marginally, we get: $Var(E[\Theta|Y]) = \left(\frac{4}{9}\right)^2 Var(Y) = \frac{16}{81} \cdot 9 = \frac{16}{9}$ That confirms B is correct. Now for the distractors: A ($4/9$) is the shrinkage weight itself, not squared and multiplied through — a partial-computation trap. C ($20/9$) likely comes from incorrectly using $\sigma^2 = 5$ in the numerator instead of $\tau^2 = 4$, confusing prior and likelihood variances. D ($4$) is simply the prior variance $Var(\Theta)$, which would equal $Var(E[\Theta|Y])$ only if $Y$ were a perfect signal — forgetting that shrinkage reduces variance. A reliable strategy: in normal-normal Bayesian problems, memorize that $E[\Theta|Y] = \frac{\tau^2}{\tau^2+\sigma^2}Y$ and always remember to compute the variance of this linear function of $Y$ using the marginal variance of $Y$.
4. What is $Var(T)$?
   Answer: $8$ When a parameter is itself random, you need the Law of Total Variance: $\text{Var}(T) = E[\text{Var}(T\mid P)] + \text{Var}(E[T\mid P])$. This decomposition separates within-group variance from between-group variance, and forgetting either piece is exactly the trap this question sets. For a Geometric$(p)$ distribution, recall $E[T\mid P=p] = 1/p$ and $\text{Var}(T\mid P=p) = (1-p)/p^2$. Since $P = 1/2$ or $P = 1/4$ each with probability $1/2$: $E[T\mid P] = \begin{cases} 2 & P=1/2 \\ 4 & P=1/4 \end{cases}, \quad \text{Var}(T\mid P) = \begin{cases} 2 & P=1/2 \\ 12 & P=1/4 \end{cases}$ First term: $E[\text{Var}(T\mid P)] = \frac{1}{2}(2) + \frac{1}{2}(12) = 7$ Second term: $E[T\mid P]$ takes values $2$ and $4$ with equal probability, so $E[E[T\mid P]] = 3$ and $\text{Var}(E[T\mid P]) = \frac{1}{2}(2-3)^2 + \frac{1}{2}(4-3)^2 = 1$ Therefore $\text{Var}(T) = 7 + 1 = \mathbf{8}$, confirming D. Choice C (7) is the most seductive trap — it equals only $E[\text{Var}(T\mid P)]$, forgetting the between-group variance term entirely. Choice B (6) likely arises from averaging the conditional variances incorrectly or using wrong geometric formulas. Choice A (40/9) may come from computing $\text{Var}(T)$ as if $P$ were its mean value $3/8$, plugging a single averaged $p$ into the formula instead of applying total variance. Your study tip: whenever a parameter is randomly selected, never plug in its average value directly — always apply the Law of Total Variance, since the $\text{Var}(E[\cdot\mid\theta])$ term accounts for uncertainty in the parameter itself.
5. What is the marginal variance $Var(Y)$?
   Answer: $22$ Whenever you see a question asking for a marginal variance involving a conditional mean and conditional variance, your instinct should be the Law of Total Variance: $Var(Y) = E[Var(Y|X)] + Var(E[Y|X])$. This decomposition is essential in random-design regression, where both the conditional mean and conditional variance depend on a random predictor. Start with the two pieces. For the first term, $E[Var(Y|X)] = E[1 + X^2] = 1 + E[X^2]$. Since $Var(X) = E[X^2] - (E[X])^2$, we get $E[X^2] = Var(X) + (E[X])^2 = 2 + 1 = 3$. So $E[Var(Y|X)] = 1 + 3 = 4$. For the second term, $Var(E[Y|X]) = Var(3X) = 9 \cdot Var(X) = 9 \cdot 2 = 18$. Adding them: $Var(Y) = 4 + 18 = 22$, confirming that A) 22 is correct. Choice B) 20 likely comes from forgetting to include the 1 in $E[1 + X^2]$, computing $E[Var(Y|X)] = E[X^2] = 3$ instead of 4, giving $3 + 18 = 21$, or from a similar arithmetic slip. Choice C) 18 reflects only computing $Var(3X)$ and ignoring $E[Var(Y|X)]$ entirely — a common mistake of confusing marginal variance with just the variance of the conditional mean. Choice D) 31 likely results from incorrectly using $E[X]$ instead of $E[X^2]$ somewhere, inflating one of the components. Your study tip: always write out the Law of Total Variance formula explicitly before computing — students most often lose points by dropping one of the two terms entirely.
6. What is $Var(Z_2)$?
   Answer: $7.5$ Galton–Watson branching processes require a specific variance formula that accounts for how randomness compounds across generations. When you see a question asking for $\text{Var}(Z_t)$, your first instinct should be to recall the recursive variance formula rather than trying to reason from scratch. For a supercritical process with $m > 1$, the variance of $Z_t$ is given by: $\text{Var}(Z_t) = v \cdot m^{t-1} \cdot \frac{m^t - 1}{m - 1}$ where $v$ is the offspring variance and $m$ is the offspring mean. For $t = 2$, $m = 1.5$, and $v = 2$: $\text{Var}(Z_2) = 2 \cdot m^{1} \cdot \frac{m^2 - 1}{m - 1} = 2 \cdot 1.5 \cdot \frac{2.25 - 1}{0.5} = 3 \cdot \frac{1.25}{0.5} = 3 \cdot 2.5 = 7.5$ This confirms D is correct. Choice A (3.0) likely comes from computing $v \cdot m = 2 \cdot 1.5$, which only captures one layer of variability and ignores the compounding from generation 1 to 2. Choice B (4.5) may reflect multiplying $v \cdot m^2 = 2 \cdot 2.25$, misapplying the exponent without the correct summation structure. Choice C (6.0) could arise from using $v \cdot (m + m^2) = 2(1.5 + 2.25) = 7.5$ but making an arithmetic error, or from an incorrect simplified formula altogether. A reliable study tip: memorize the Galton–Watson variance formula and practice applying it at $t = 1, 2, 3$ to build fluency — exam questions almost always test $t = 2$ because that's where compounding first becomes nontrivial.
7. Let $P$ have a beta distribution with parameters $\alpha=2$ and $\beta=3$. Conditional on $P$, the random variables $X$ and $Y$ are independent and each has a binomial distribution with parameters $6$ and $P$. What is $Var(X-Y)$?
   Answer: $2.40$ When you see a problem mixing a Beta-distributed random variable with conditional Binomials, your framework should be the Law of Total Variance: $Var(X-Y) = E[Var(X-Y|P)] + Var(E[X-Y|P])$. Since $X$ and $Y$ are conditionally independent given $P$, we have $Var(X-Y|P) = Var(X|P) + Var(Y|P) = 2 \cdot 6P(1-P) = 12P(1-P)$, and $E[X-Y|P] = 6P - 6P = 0$. For the Beta(2,3) distribution, the key moments are: $E[P] = \frac{\alpha}{\alpha+\beta} = \frac{2}{5} = 0.4$ $E[P^2] = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)} = \frac{6}{30} = 0.2$ $Var(P) = E[P^2] - (E[P])^2 = 0.2 - 0.16 = 0.04$ Now apply total variance: $E[Var(X-Y|P)] = 12E[P(1-P)] = 12(E[P] - E[P^2]) = 12(0.4 - 0.2) = 2.40$ $Var(E[X-Y|P]) = Var(0) = 0$ So $Var(X-Y) = 2.40 + 0 = \mathbf{2.40}$, confirming answer B. Choice A (1.20) likely comes from forgetting that $Var(X-Y|P) = Var(X|P) + Var(Y|P)$ — using only one binomial's variance instead of summing both. Choice C (2.88) results from incorrectly including a $Var(E[X-Y|P])$ term using $Var(6P)$ rather than recognizing the conditional mean is identically zero. Choice D (5.28) reflects adding $Var(6P)$ twice, double-counting the second term of total variance. Your study tip: always check whether the conditional mean of the quantity of interest is a constant — if so, the second term of total variance vanishes, simplifying your calculation dramatically.
8. What is the posterior predictive variance of $Y_1+Y_2$?
   Answer: $16$ When you need to find the variance of a posterior predictive quantity, reach for the law of total variance: $\text{Var}(Y) = E[\text{Var}(Y|\Lambda)] + \text{Var}(E[Y|\Lambda])$. This decomposes uncertainty into two sources — randomness in the outcome given the parameter, and uncertainty about the parameter itself. Here, let $S = Y_1 + Y_2$. Conditional on $\Lambda$, both $Y_1$ and $Y_2$ are independent Poisson$(\Lambda)$, so $S|\Lambda \sim \text{Poisson}(2\Lambda)$, meaning $E[S|\Lambda] = 2\Lambda$ and $\text{Var}(S|\Lambda) = 2\Lambda$. Applying the law of total variance: $\text{Var}(S) = E[\text{Var}(S|\Lambda)] + \text{Var}(E[S|\Lambda])$ $= E[2\Lambda] + \text{Var}(2\Lambda)$ $= 2E[\Lambda] + 4\text{Var}(\Lambda)$ $= 2(4) + 4(2) = 8 + 8 = 16$ So C) 16 is correct. Choice A) 8 captures only the first term — the expected within-$\Lambda$ variance — ignoring the additional uncertainty from not knowing $\Lambda$ exactly. Choice B) 12 likely comes from adding $E[\Lambda]$ terms incorrectly or misapplying the variance of a sum. Choice D) 20 may result from using $\text{Var}(2\Lambda) = 2\cdot\text{Var}(\Lambda) = 4$ instead of $4\cdot\text{Var}(\Lambda) = 8$, forgetting to square the constant when computing variance. Your key takeaway: whenever a predictive variance problem involves an unknown parameter, always apply the law of total variance — ignoring the $\text{Var}(E[Y|\Lambda])$ term is the single most common mistake on Bayesian prediction problems.
9. Let $X\sim N(0,1)$. Conditional on $X$, a random variable $Y$ has conditional mean $E[Y|X]=X^2$ and conditional variance $Var(Y|X)=1+X^2$. What is $Var(Y)$?
   Answer: $4$ Whenever you see a question involving a conditional mean and conditional variance, your instinct should be to reach for the Law of Total Variance: $\text{Var}(Y) = E[\text{Var}(Y|X)] + \text{Var}(E[Y|X])$. This decomposition splits total variance into two pieces — the average within-group variance and the variance of the group means. Here, you're given $E[Y|X] = X^2$ and $\text{Var}(Y|X) = 1 + X^2$, with $X \sim N(0,1)$. Compute each piece separately. For the first term: $E[\text{Var}(Y|X)] = E[1 + X^2] = 1 + E[X^2] = 1 + 1 = 2$, since $E[X^2] = \text{Var}(X) = 1$ for a standard normal. For the second term: $\text{Var}(E[Y|X]) = \text{Var}(X^2)$. Since $X \sim N(0,1)$, $X^2 \sim \chi^2(1)$, which has variance $2$. So $\text{Var}(X^2) = 2$. Adding both pieces: $\text{Var}(Y) = 2 + 2 = 4$, confirming the answer is C. Choice A ($2$) comes from computing only the first term $E[\text{Var}(Y|X)]$ and forgetting the second term entirely. Choice B ($3$) likely arises from correctly computing $E[\text{Var}(Y|X)] = 2$ but mistakenly using $\text{Var}(X^2) = 1$ instead of $2$. Choice D ($5$) may result from using $E[X^4] = 3$ directly as $\text{Var}(X^2)$ without subtracting $(E[X^2])^2 = 1$, giving $3$ for the second term instead of $2$. Study tip: Always memorize that for $X \sim N(0,1)$, $\text{Var}(X^2) = 2$ (not 1!), and never skip the second term in the Law of Total Variance — it's the most common trap on questions like this.
10. Using conditioning on $K$, what is $Var(X)$?
    Answer: $1.0500$ When a problem involves a random variable whose distribution depends on another random variable, the Law of Total Variance is your essential tool: $Var(X) = E[Var(X \mid K)] + Var(E[X \mid K])$. Given $K$ successes in the urn, $X$ follows a Hypergeometric distribution with parameters $N=20$, $K$, and $n=5$. Its conditional moments are: $E[X \mid K] = \frac{5K}{20} = \frac{K}{4}, \quad Var(X \mid K) = \frac{5 \cdot K \cdot (20-K)}{20^2} \cdot \frac{20-5}{19} = \frac{5K(20-K)}{400} \cdot \frac{15}{19}$ Since $K \sim Bin(20, 0.3)$, we have $E[K]=6$, $Var(K)=4.2$, and $E[K^2]=Var(K)+[E(K)]^2=4.2+36=40.2$. First term: $E[Var(X\mid K)] = \frac{75}{400 \cdot 19}E[K(20-K)] = \frac{75}{7600}(20E[K]-E[K^2]) = \frac{75}{7600}(120-40.2) = \frac{75 \times 79.8}{7600} \approx 0.7875$ Second term: $Var(E[X\mid K]) = Var\!\left(\frac{K}{4}\right) = \frac{Var(K)}{16} = \frac{4.2}{16} = 0.2625$ Adding both: $Var(X) = 0.7875 + 0.2625 = 1.0500$, confirming D. Choice A (0.2625) is only the second term — forgetting $E[Var(X\mid K)]$. Choice B (0.7875) is only the first term — forgetting $Var(E[X\mid K])$. Choice C (0.8290) likely arises from misapplying the hypergeometric variance formula, perhaps omitting the finite-population correction. Always decompose variance using both components of the Law of Total Variance — dropping either piece is the most common trap on conditioning problems.

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