Independence & Conditioning - Statistics Graduate Level

1. Which statement about the transformed variables and the conditional mean is correct?
   Answer: $R$ and $S$ are independent, and $E(X\mid S=s)=2s/5$. Whenever you see a sum and ratio constructed from independent Gamma random variables, your first instinct should be to recall the Gamma-Beta decomposition theorem: if $X \sim \text{Gamma}(\alpha_1, \theta)$ and $Y \sim \text{Gamma}(\alpha_2, \theta)$ are independent, then $S = X+Y \sim \text{Gamma}(\alpha_1+\alpha_2, \theta)$ and $R = X/S \sim \text{Beta}(\alpha_1, \alpha_2)$, and crucially, **$R$ and $S$ are independent**. This independence is a classical result that follows from the Jacobian transformation of the joint density. Here, $\alpha_1 = 2$ and $\alpha_2 = 3$, so $S \sim \text{Gamma}(5, \theta)$ and $R \sim \text{Beta}(2,3)$, and they are independent. To find $E(X \mid S = s)$, write $X = R \cdot S$. Since $R$ and $S$ are independent, conditioning on $S = s$ treats $s$ as a constant: $E(X \mid S=s) = E(R \cdot s) = s \cdot E(R)$. The mean of a $\text{Beta}(\alpha_1, \alpha_2)$ is $\alpha_1/(\alpha_1+\alpha_2) = 2/5$, giving $E(X \mid S=s) = 2s/5$. This confirms answer A is correct. Answer B incorrectly uses $3/5$ for the Beta mean — that would be $E(Y/S)$, not $E(X/S)$, confusing which shape parameter belongs to $X$ versus $Y$. Answers C and D both claim $R$ and $S$ are dependent, which contradicts the fundamental Gamma-Beta independence theorem — a common misconception among students who assume ratios and sums must share information. Study tip: Memorize the Gamma-Beta decomposition as a single package — sum is Gamma, ratio is Beta, and they are always independent when the scale parameters match. This trio of facts solves an entire class of exam problems.
2. For distinct indices $i$ and $j$, what is $\operatorname{Cov}(X_i,X_j\mid K=k)$?
   Answer: $-\dfrac{k(n-k)}{n^2(n-1)}$ When you condition on the sum $K = k$ of $n$ independent Bernoulli trials, the joint conditional distribution of $(X_1, \ldots, X_n)$ becomes exchangeable and uniform over all binary vectors with exactly $k$ ones. This is precisely the hypergeometric structure: each $X_i \mid K=k$ is marginally Bernoulli with success probability $k/n$, and the pair $(X_i, X_j)$ follows a hypergeometric-type joint distribution. To find the covariance, use $\operatorname{Cov}(X_i, X_j \mid K=k) = E[X_i X_j \mid K=k] - E[X_i \mid K=k]\,E[X_j \mid K=k]$. Since $X_i X_j = 1$ only when both are 1, and by symmetry $P(X_i=1, X_j=1 \mid K=k) = \frac{k(k-1)}{n(n-1)}$, we get: $\operatorname{Cov} = \frac{k(k-1)}{n(n-1)} - \left(\frac{k}{n}\right)^2 = \frac{k(k-1)}{n(n-1)} - \frac{k^2}{n^2} = -\frac{k(n-k)}{n^2(n-1)}$ This confirms D. The covariance is negative because fixing the sum creates a competition: if $X_i = 1$, there are fewer "successes" available for $X_j$. A is wrong because conditioning on $K=k$ destroys the independence of the $X_i$'s — they are no longer uncorrelated. B is wrong because it omits one factor of $n$ in the denominator, missing the $n^2$ term. C is wrong in sign — the covariance must be negative under this fixed-sum constraint, not positive. As a strategy, whenever you see conditioning on a sufficient statistic like a sum, remember it typically induces negative pairwise correlations through an exchangeable structure — this is the hallmark of sampling without replacement and hypergeometric distributions.
3. Which statement describes the relationship between $A$ and $B$ after conditioning on $C$?
   Answer: They become positively associated because $P(A\cap B\mid C)=0.4$, exceeding $1/4$. This question tests a subtle but important phenomenon called Berkson's paradox (or collider bias): conditioning on a common effect of two independent variables induces a dependence between them. Start by computing $P(C)$. Since $A$ and $B$ are independent with equal probabilities, each of the four joint events has probability $1/4$. So: $P(C) = \frac{1}{4}(0.8) + \frac{1}{4}(0.2) + \frac{1}{4}(0.2) + \frac{1}{4}(0.8) = 0.5$ Now find $P(A \cap B \mid C)$ using Bayes' theorem: $P(A \cap B \mid C) = \frac{P(C \mid A \cap B)\,P(A \cap B)}{P(C)} = \frac{0.8 \times 0.25}{0.5} = 0.4$ By symmetry, $P(A \mid C) = P(B \mid C) = 1/2$, so if $A$ and $B$ were conditionally independent, we'd need $P(A \cap B \mid C) = 1/4$. Instead, we get $0.4 > 1/4$, confirming positive association — answer B is correct. Answer A is tempting because the marginals $P(A \mid C) = P(B \mid C) = 1/2$ look like independence, but equal marginals alone never guarantee independence; you must check the joint. Answer C gets the direction backwards — the joint probability exceeds $1/4$, making the association positive, not negative. Answer D confuses the conditional selection probability with an independence criterion, which has no statistical basis. The key study tip: marginal conditionals can look symmetric even when variables are strongly dependent. Whenever a "collider" variable $C$ is influenced by two variables, always check the joint conditional probability, not just the marginals.
4. Which statement correctly describes the independence structure and the indicated conditional probability?
   Answer: The variables are pairwise independent but not mutually independent, and $P(X=1\mid Y=1,Z=1)=0$. Questions like this test your ability to distinguish pairwise independence from mutual independence — a subtle but critical distinction in probability theory. Start by building the joint distribution. Since $X, Y \in \{0,1\}$ each with probability $1/2$, there are four equally likely outcomes: $(0,0), (0,1), (1,0), (1,1)$. By definition, $Z=1$ when $X \neq Y$, so $Z=1$ for outcomes $(0,1)$ and $(1,0)$, and $Z=0$ for $(0,0)$ and $(1,1)$. Pairwise independence holds for all three pairs. For example, $P(Z=1 \mid X=1) = P(Y=0 \mid X=1) = 1/2 = P(Z=1)$, and similarly for the other pairs. However, mutual independence requires $P(X=1, Y=1, Z=1) = P(X=1)P(Y=1)P(Z=1)$. But $P(X=1, Y=1, Z=1) = 0$ since $Z=1$ is impossible when $X=Y$. Since $1/8 \neq 0$, mutual independence fails. For the conditional probability: $P(X=1 \mid Y=1, Z=1)$ asks for the probability that $X=1$ given $Y=1$ and $Z=1$. The only outcome with $Y=1$ and $Z=1$ is $(0,1)$, where $X=0$. Therefore $P(X=1 \mid Y=1, Z=1) = 0$. This confirms answer A. Answer B is wrong because mutual independence fails as shown. Answer C incorrectly claims only $X$ and $Y$ are independent, ignoring the other pairwise relationships. Answer D correctly identifies pairwise-but-not-mutual independence but gets the conditional probability wrong — $X$ must equal 0 when $Y=1$ and $Z=1$. Study tip: When you see a derived variable like $Z$, always construct the full joint table explicitly — it makes both independence checks and conditional probabilities mechanical rather than guesswork.
5. Random elements $X$, $Y$, and $Z$ satisfy $X\perp Y\mid Z$ and $Y\perp Z$. Which independence conclusion necessarily follows?
   Answer: $Y\perp (X,Z)$ Conditional independence and marginal independence interact in subtle ways, and this question tests whether you can carefully apply their definitions rather than assume they "combine" intuitively. Start with what's given: $X \perp Y \mid Z$ means $X$ and $Y$ are independent once you know $Z$, and $Y \perp Z$ means $Y$ and $Z$ are marginally independent. Now ask what these two conditions together imply about $Y$'s joint relationship with $(X, Z)$. To show $Y \perp (X, Z)$, you need $f(y, x, z) = f(y) \cdot f(x, z)$. Using the chain rule: $f(x, y, z) = f(y \mid x, z) \cdot f(x, z)$. From $X \perp Y \mid Z$, we have $f(y \mid x, z) = f(y \mid z)$. From $Y \perp Z$, we have $f(y \mid z) = f(y)$. Therefore $f(x, y, z) = f(y) \cdot f(x, z)$, which is exactly $Y \perp (X, Z)$. So C follows necessarily. Choice A, $X \perp Z \mid Y$, does not follow — the given conditions say nothing about the conditional relationship between $X$ and $Z$ given $Y$, and counterexamples exist. Choice B, $X \perp (Y, Z)$, is too strong; you'd need $X \perp Z$ marginally, which isn't guaranteed. Choice D claims full mutual independence, which is even stronger than B and similarly unsupported — knowing $Y \perp Z$ and $X \perp Y \mid Z$ says nothing definitive about $X$'s marginal relationship with $Z$. When you see a mix of conditional and marginal independence conditions, chain-rule the joint density step by step — don't guess that independence "transfers" without verification.
6. For any real $s$, which statement about the conditional relationship between $X$ and $Y$ given $S=s$ is correct?
   Answer: $\operatorname{Cov}(X,Y\mid S=s)=-1/2$, and the conditional correlation is $-1$. When you condition on a linear combination of jointly normal variables, you're working in the realm of conditional distributions for multivariate normals — a favorite testing ground for graduate statistics exams. The key insight is that fixing $S = s$ creates a deterministic constraint $Y = s - X$, which induces a strong negative dependence between $X$ and $Y$. To find the conditional distribution of $(X, Y)$ given $S = s$, note that the joint vector $(X, S)$ is bivariate normal with $\operatorname{Var}(X) = 1$, $\operatorname{Var}(S) = 2$, and $\operatorname{Cov}(X, S) = 1$. By standard multivariate normal conditioning, $X \mid S = s \sim N(s/2,\ 1/2)$, and by symmetry $Y \mid S = s \sim N(s/2,\ 1/2)$. Since $Y = s - X$ deterministically given $S = s$, we have $\operatorname{Cov}(X, Y \mid S = s) = \operatorname{Cov}(X, s - X) = -\operatorname{Var}(X \mid S = s) = -1/2$. The conditional correlation is then $\rho = \frac{-1/2}{\sqrt{1/2}\cdot\sqrt{1/2}} = \frac{-1/2}{1/2} = -1$, confirming answer D. Answer A gets the correlation right direction but miscalculates the covariance as $-1$ (forgetting to condition properly) and misreports the correlation as $-1/2$. Answer B reflects the unconditional independence of $X$ and $Y$ — a classic trap of ignoring that conditioning on their sum destroys independence. Answer C inverts the sign entirely, perhaps from confusing $\operatorname{Cov}(X, S)$ with $\operatorname{Cov}(X, Y \mid S)$. A powerful study tip: conditioning on a sum always creates perfect negative correlation between the summands when they're exchangeable — the constraint $X + Y = s$ forces them to move in opposite directions, yielding $\rho = -1$ regardless of the marginal variances.
7. For a randomly selected person, which statement about a positive result on Test 2 conditional on a positive result on Test 1 is correct?
   Answer: The probability is $9/26$, and the two test results are not marginally independent. When two tests are conditionally independent given disease status, they are generally not marginally independent — and this question tests exactly that distinction. Start by finding the marginal probability of a positive on Test 1. Using the law of total probability: $P(T_1^+) = (0.80)(0.10) + (0.20)(0.90) = 0.08 + 0.18 = 0.26$. Similarly, $P(T_2^+) = (0.90)(0.10) + (0.10)(0.90) = 0.09 + 0.09 = 0.18$. Now compute the joint probability $P(T_1^+ \cap T_2^+)$. Since the tests are conditionally independent given disease status, condition on disease state: $P(T_1^+ \cap T_2^+) = (0.80)(0.90)(0.10) + (0.20)(0.10)(0.90) = 0.072 + 0.018 = 0.090$. The desired conditional probability is: $P(T_2^+ \mid T_1^+) = \frac{0.090}{0.26} = \frac{9}{26}$. To check marginal independence, ask whether $P(T_1^+ \cap T_2^+) = P(T_1^+) \cdot P(T_2^+)$: $0.26 \times 0.18 = 0.0468 \neq 0.090$. The tests are not marginally independent — disease status is a common cause that induces correlation between the two test results in the population. This confirms A is correct. Choice B gets the probability right but incorrectly claims marginal independence. Choice C arrives at $9/50 = 0.18$, which simply equals $P(T_2^+)$ — the error of mistakenly treating the tests as marginally independent when computing the conditional. Choice D's value $4/13$ doesn't follow from any standard calculation here. Key takeaway: Conditional independence does not imply marginal independence when conditioning variables (like disease status) are shared. Always verify independence claims by checking whether the joint equals the product of marginals.
8. For $a>0$ and $b>0$, what is $P(X>t+a,\ Y>t+b\mid \min(X,Y)>t)$?
   Answer: $\exp\{-\lambda(a+b)\}$ When you see conditioning on the minimum of independent exponentials, your first instinct should be to apply the memoryless property. For an exponential random variable, $P(X > t + a \mid X > t) = P(X > a) = e^{-\lambda a}$. This holds because the exponential "forgets" how long it has already waited. The key insight here is that conditioning on $\min(X,Y) > t$ is equivalent to conditioning on both $X > t$ and $Y > t$ simultaneously. Since $X$ and $Y$ are independent, this joint conditioning separates cleanly. By the memoryless property applied to each variable independently: $P(X > t+a,\ Y > t+b \mid \min(X,Y)>t) = P(X > t+a \mid X > t)\cdot P(Y > t+b \mid Y > t)$ $= e^{-\lambda a} \cdot e^{-\lambda b} = e^{-\lambda(a+b)}$ This confirms C is correct. Choice A, $e^{-\lambda \min(a,b)}$, would be the answer to a different question — specifically $P(\min(X,Y) > t + \min(a,b) \mid \min(X,Y) > t)$, which involves the minimum of the shifted variables, not the joint exceedance. Choice B, $e^{-\lambda \max(a,b)}$, has no natural probabilistic interpretation in this context and likely tempts students who confuse min/max operations. Choice D, $e^{-\lambda(t+a+b)}$, is the classic memoryless property trap — it ignores that conditioning on $\min(X,Y) > t$ already "cancels" the $t$, leaving only the incremental distances $a$ and $b$. As a study tip: whenever you condition an exponential on survival past time $t$, immediately replace it with a fresh exponential — the $t$ disappears entirely.
9. Let $A$, $B$, and $C$ be events with $P(C)>0$. Which equality is exactly equivalent to conditional independence of $A$ and $B$ given $C$?
   Answer: $P(A\cap B\cap C)P(C)=P(A\cap C)P(B\cap C)$ Conditional independence is one of those concepts where precision in notation makes all the difference. When we say $A$ and $B$ are conditionally independent given $C$, the formal definition is $P(A \cap B \mid C) = P(A \mid C)P(B \mid C)$. Your job on a question like this is to algebraically manipulate that definition into an equivalent form using only intersection and unconditional probabilities. Starting from $P(A \cap B \mid C) = P(A \mid C)P(B \mid C)$, expand each term using the definition of conditional probability: $\frac{P(A \cap B \cap C)}{P(C)} = \frac{P(A \cap C)}{P(C)} \cdot \frac{P(B \cap C)}{P(C)}$. Multiply both sides by $P(C)^2$ to clear denominators, and you get $P(A \cap B \cap C) \cdot P(C) = P(A \cap C) \cdot P(B \cap C)$, which is exactly answer A. The distractors each represent a distinct misconception. B describes mutual independence of three events — the condition $P(A \cap B \cap C) = P(A)P(B)P(C)$ — which is an entirely different concept from conditional independence. C looks structurally similar to A but replaces $P(B \cap C)$ with $P(A \cap B)$, which has no grounding in the definition and mixes events incorrectly. D is dimensionally wrong: the left side is a probability of an intersection (a number), while the right side omits $P(C)$ from the denominator, making it not a valid probability expression for the joint event. The key study tip: always return to the formal definition and manipulate algebraically. Questions like this test whether you can translate conditional probability statements into intersection form — so practice that substitution until it's automatic.
10. What is $P(A=1\mid C=1)$?
    Answer: $111/202$ When you see a Markov chain problem asking for a "reversed" conditional probability like $P(A=1\mid C=1)$, your instinct should be Bayes' theorem combined with the law of total probability. The Markov property $A\perp C\mid B$ means you must route all influence through $B$. Start by computing $P(C=1)$ using total probability over both values of $B$, and $B$ over both values of $A$: $P(B=1) = P(B=1\mid A=1)P(A=1) + P(B=1\mid A=0)P(A=0) = 0.8(0.3)+0.2(0.7) = 0.24+0.14 = 0.38$ $P(C=1) = P(C=1\mid B=1)P(B=1)+P(C=1\mid B=0)P(B=0) = 0.9(0.38)+0.1(0.62) = 0.342+0.062 = 0.404$ Now apply Bayes' theorem. Because $A\perp C\mid B$, you compute $P(C=1\mid A=1)$ by conditioning through $B$: $P(C=1\mid A=1) = 0.9(0.8)+0.1(0.2) = 0.72+0.02 = 0.74$ $P(A=1\mid C=1) = \frac{P(C=1\mid A=1)P(A=1)}{P(C=1)} = \frac{0.74 \times 0.3}{0.404} = \frac{0.222}{0.404} = \frac{111}{202}$ That confirms B is correct. Choice A ($3/10$) is simply $P(A=1)$, ignoring the evidence from $C$ entirely — a classic trap of forgetting to update. Choice C ($37/50 = 0.74$) is $P(C=1\mid A=1)$, confusing the direction of conditioning. Choice D ($9/20$) likely arises from using $P(C=1\mid B=1)$ alone without properly propagating through the chain. Your study tip: whenever a chain reversal is required, always build the full joint machinery first — compute marginals bottom-up, then apply Bayes top-down. Don't shortcut the direction of conditioning.

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