Conditional Probability & Expectation - Statistics Graduate Level

1. What is the approximate value of $E[X\mid |Y|>1]$?
   Answer: $2.525$ Whenever you see a conditional expectation problem involving a truncated normal distribution, your first instinct should be to use the law of total expectation combined with direct computation via the normal density. Since $E[X \mid Y] = Y^2$, the tower property gives $E[X \mid |Y| > 1] = E[Y^2 \mid |Y| > 1]$. This is because conditioning further on the event $\{|Y|>1\}$ yields $E[E[X|Y] \mid |Y|>1] = E[Y^2 \mid |Y|>1]$. So the problem reduces to computing the second moment of a standard normal truncated to $|Y|>1$. Using the truncated normal formula: $E[Y^2 \mid |Y|>1] = \frac{E[Y^2 \cdot \mathbf{1}_{|Y|>1}]}{P(|Y|>1)}$ The denominator is $2(1-\Phi(1)) = 2(0.1587) = 0.3174$. For the numerator, use integration by parts: $E[Y^2 \cdot \mathbf{1}_{|Y|>1}] = 2\int_1^\infty y^2 \phi(y)\,dy$. Since $\frac{d}{dy}[-\phi(y)] = y\phi(y)$, we get $\int_1^\infty y^2\phi(y)\,dy = \phi(1) + (1-\Phi(1)) = 0.2420 + 0.1587 = 0.4007$. So the numerator is $2(0.4007) = 0.8014$. Therefore $E[Y^2 \mid |Y|>1] = \frac{0.8014}{0.3174} \approx 2.525$, confirming C. Choice A ($1.000$) ignores the truncation entirely — that's $E[Y^2]$ for an untruncated standard normal. Choice B ($1.525$) likely arises from computing only the numerator term $\phi(1)$ and mishandling the denominator. Choice D ($3.050$) doubles the correct answer, a sign of forgetting to divide by the tail probability. Study tip: Memorize that $\int_a^\infty y^2\phi(y)\,dy = a\phi(a) + (1-\Phi(a))$ — this identity appears repeatedly in truncated normal calculations on graduate-level exams.
2. What is the posterior predictive probability that exactly one of the next two patients responds?
   Answer: $\frac{4}{9}$ When you see a Bayesian problem asking for a posterior predictive probability, your goal is to average the likelihood of future observations over the posterior distribution of the unknown parameter — not plug in a point estimate. Start by updating the prior. With $P \sim \text{Beta}(2,3)$ and observing 3 successes and 1 failure, the posterior is $P \mid \text{data} \sim \text{Beta}(2+3,\, 3+1) = \text{Beta}(5,4)$, giving posterior mean $\frac{5}{9}$. Now compute the posterior predictive probability that exactly 1 of the next 2 patients responds. You need: $\Pr(X=1) = \binom{2}{1}\mathbb{E}[P(1-P)] = 2\,\mathbb{E}[P - P^2]$ Using moments of $\text{Beta}(5,4)$: $\mathbb{E}[P] = \frac{5}{9}$ and $\mathbb{E}[P^2] = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)} = \frac{5 \cdot 6}{9 \cdot 10} = \frac{30}{90} = \frac{1}{3}$. So: $\Pr(X=1) = 2\!\left(\frac{5}{9} - \frac{1}{3}\right) = 2 \cdot \frac{2}{9} = \frac{4}{9}$ This confirms A is correct. Choice B, $\frac{40}{81}$, is the trap of plugging the posterior mean directly into the binomial formula: $2 \cdot \frac{5}{9} \cdot \frac{4}{9} = \frac{40}{81}$ — this ignores parameter uncertainty. Choice C, $\frac{5}{9}$, is simply the posterior mean itself, mistaking it for a predictive probability. Choice D, $\frac{2}{5}$, corresponds to using the prior mean $\frac{2}{5}$ and never updating. Key tip: Never substitute the posterior mean into a binomial formula for predictive probabilities — always integrate over the full posterior using $\mathbb{E}[P^k]$ moments.
3. Let $U_1$ and $U_2$ be independent uniform random variables on $[0,1]$, and let $M=\max(U_1,U_2)$. What is $E[M\mid M>1/2]$?
   Answer: $\frac{7}{9}$ When you see a conditional expectation involving order statistics, your instinct should be to work directly from the conditional distribution. Here, $M = \max(U_1, U_2)$ has CDF $F_M(m) = m^2$ and PDF $f_M(m) = 2m$ on $[0,1]$. To find $E[M \mid M > 1/2]$, you need the conditional PDF of $M$ given $M > 1/2$. First, compute $P(M > 1/2) = 1 - (1/2)^2 = 3/4$. The conditional PDF is then: $f_{M \mid M > 1/2}(m) = \frac{f_M(m)}{P(M > 1/2)} = \frac{2m}{3/4} = \frac{8m}{3}, \quad m \in (1/2, 1]$ Now compute the conditional expectation: $E[M \mid M > 1/2] = \int_{1/2}^{1} m \cdot \frac{8m}{3}\, dm = \frac{8}{3}\int_{1/2}^{1} m^2\, dm = \frac{8}{3} \cdot \left[\frac{m^3}{3}\right]_{1/2}^{1} = \frac{8}{3} \cdot \frac{7}{24} = \frac{7}{9}$ This confirms the answer is C. Choice A, $2/3$, is the unconditional mean of $M$ — a classic trap when you forget to apply the conditioning. Choice B, $3/4$, is $P(M > 1/2)$ itself, which students sometimes confuse with the conditional mean. Choice D, $5/6$, likely comes from incorrectly using a uniform conditional density on $(1/2, 1)$ rather than the correct triangular-shaped conditional PDF. The key study tip: always re-derive the conditional PDF by truncating and renormalizing the original density. Never assume the conditional distribution inherits the same functional form as the unconditional one.
4. Independent event counts satisfy $X\sim\operatorname{Poisson}(2)$ and $Y\sim\operatorname{Poisson}(3)$. What is $E[X\mid X+Y=4,\ X\geq 1]$?
   Answer: $\frac{125}{68}$ When you see a conditional expectation involving Poisson random variables with a sum constraint, reach for the key theorem: given $X \sim \text{Poisson}(\lambda)$, $Y \sim \text{Poisson}(\mu)$ independent, the conditional distribution of $X \mid X+Y=n$ is $\text{Binomial}\!\left(n, \frac{\lambda}{\lambda+\mu}\right)$. Here that means $X \mid X+Y=4 \sim \text{Binomial}(4, 2/5)$. But you also need $X \geq 1$, so you're working with a truncated distribution. The conditional expectation becomes: $E[X \mid X+Y=4,\ X\geq 1] = \frac{E[X \cdot \mathbf{1}_{X\geq 1} \mid X+Y=4]}{P(X\geq 1 \mid X+Y=4)}$ Let $p = 2/5$, $n=4$. Then $P(X=0 \mid X+Y=4) = (3/5)^4 = 81/625$, so $P(X \geq 1) = 1 - 81/625 = 544/625$. The unconditional binomial mean is $E[X \mid X+Y=4] = 4 \cdot (2/5) = 8/5$, and since $X \cdot \mathbf{1}_{X=0} = 0$, we have $E[X \cdot \mathbf{1}_{X\geq1}] = 8/5$. Therefore: $E[X \mid X+Y=4,\ X\geq 1] = \frac{8/5}{544/625} = \frac{8}{5} \cdot \frac{625}{544} = \frac{5000}{2720} = \frac{125}{68}$ This confirms B. Choice A, $8/5$, is the binomial mean without truncation — forgetting the $X \geq 1$ constraint. Choice C, $544/625$, is actually $P(X \geq 1 \mid X+Y=4)$, a probability mistaken for an expectation. Choice D, $2$, is simply $E[X]$ with no conditioning at all. When conditioning on multiple events, always decompose using $E[X \mid A \cap B] = E[X \cdot \mathbf{1}_B \mid A] / P(B \mid A)$ — truncation changes the mean, and that denominator is the step students most often drop.
5. The joint density of $X$ and $Y$ is proportional to $x+y$ on the region $0<x<y<1$ and is zero elsewhere. What is $P(X<1/2\mid Y>1/2)$?
   Answer: $\frac{4}{7}$ When working with joint densities on non-rectangular regions, your first instinct should be to find the normalizing constant, then carefully set up conditional probability using the definition $P(A \mid B) = P(A \cap B)/P(B)$. Start by finding $c$ such that $c\int_0^1\int_0^y (x+y)\,dx\,dy = 1$. The inner integral gives $\int_0^y(x+y)\,dx = \frac{y^2}{2}+y^2 = \frac{3y^2}{2}$, so the double integral equals $\int_0^1 \frac{3y^2}{2}\,dy = \frac{1}{2}$. Thus $c = 2$ and $f(x,y) = 2(x+y)$ on $0 < x < y < 1$. Now compute $P(Y > 1/2)$. Integrate over $1/2 < y < 1$, $0 < x < y$: $P(Y>1/2) = 2\int_{1/2}^1\int_0^y(x+y)\,dx\,dy = 2\int_{1/2}^1\frac{3y^2}{2}\,dy = 3\int_{1/2}^1 y^2\,dy = 3\left[\frac{y^3}{3}\right]_{1/2}^1 = 1 - \frac{1}{8} = \frac{7}{8}.$ Next, $P(X < 1/2,\, Y > 1/2)$ requires splitting into regions: $1/2 < y < 1,\, 0 < x < \min(y, 1/2)$. For $y > 1/2$, $x$ runs from $0$ to $1/2$: $2\int_{1/2}^1\int_0^{1/2}(x+y)\,dx\,dy = 2\int_{1/2}^1\left(\frac{1}{8}+\frac{y}{2}\right)dy = 2\left[\frac{y}{8}+\frac{y^2}{4}\right]_{1/2}^1 = 2\cdot\frac{1}{4} = \frac{1}{2}.$ So $P(X<1/2\mid Y>1/2) = \frac{1/2}{7/8} = \frac{4}{7}$, confirming C. Choice A ($1/2$) ignores conditioning entirely. Choice B ($7/16$) likely comes from forgetting the factor of 2 or misidentifying the integration region. Choice D ($2/3$) results from computing $P(Y>1/2)$ incorrectly as $3/4$. Study tip: Always sketch the support region before integrating — on triangular domains like $0<x<y<1$, the limits change depending on which event you're conditioning on, and a diagram prevents costly region errors.
6. What is $E[S]$?
   Answer: $5$ When a sum has a random number of terms, the key tool is the law of total expectation: $E[S] = E[E[S \mid N]]$. Start by conditioning on $N$, then average over the randomness in $N$ itself. Given $N = n$, the conditional expectation of the sum is: $E[S \mid N = n] = \sum_{i=1}^{n} E[X_i \mid N = n] = n \cdot \left(1 + \frac{n}{2}\right) = n + \frac{n^2}{2}$ Notice that no conditional independence assumption is needed here — you only need the conditional mean of each $X_i$, not anything about their joint distribution. Now apply the law of total expectation: $E[S] = E\!\left[N + \frac{N^2}{2}\right] = E[N] + \frac{1}{2}E[N^2]$ Since $N \sim \text{Poisson}(2)$, you have $E[N] = 2$ and $\text{Var}(N) = 2$, so $E[N^2] = \text{Var}(N) + (E[N])^2 = 2 + 4 = 6$. Therefore: $E[S] = 2 + \frac{1}{2}(6) = 2 + 3 = 5$ The correct answer is D. Choice A ($2$) likely comes from computing $E[N]$ alone and ignoring the $N$-dependent mean structure. Choice B ($3$) might arise from computing $\frac{1}{2}E[N^2]$ without adding $E[N]$. Choice C ($4$) could result from forgetting to account for the variance term when computing $E[N^2]$, mistakenly using $E[N]^2 = 4$ instead of $E[N^2] = 6$. The study tip: whenever you see $E[N^2]$ in a Poisson problem, always decompose it as $\text{Var}(N) + (E[N])^2$. Forgetting the variance term is one of the most common computational errors on graduate-level probability exams.
7. Suppose $X$ and $Y$ are square-integrable random variables satisfying $E[X\mid Y]=E[X]$ almost surely. Which statement must hold for every bounded measurable function $g$?
   Answer: $\operatorname{Cov}(X,g(Y))=0$ whenever the covariance is defined. When you see a condition like $E[X \mid Y] = E[X]$, your instinct should be to explore what it implies about covariances — not independence or symmetry. This condition means $X$ is "mean-independent" of $Y$, which is weaker than full independence but still has powerful covariance consequences. Here's why D is correct. For any bounded measurable $g$, you can compute: $\operatorname{Cov}(X, g(Y)) = E[Xg(Y)] - E[X]E[g(Y)]$ By the tower property (iterated expectations): $E[Xg(Y)] = E\!\left[E[Xg(Y) \mid Y]\right] = E\!\left[g(Y)\,E[X \mid Y]\right] = E\!\left[g(Y)\cdot E[X]\right] = E[X]\,E[g(Y)]$ So $\operatorname{Cov}(X, g(Y)) = 0$ — exactly as D claims. Now for the distractors. A is wrong because mean-independence does not imply full independence. You can construct examples where $E[X \mid Y] = E[X]$ yet the joint distribution is not a product measure. B is wrong because the conditional expectation condition is not symmetric — knowing $E[X \mid Y] = E[X]$ says nothing about $E[Y \mid X]$. These are distinct properties, and neither implies the other in general. C is wrong because while $\operatorname{Var}(X) = E[\operatorname{Var}(X \mid Y)] + \operatorname{Var}(E[X \mid Y])$ and the second term vanishes (since $E[X \mid Y]$ is constant), this only tells you $E[\operatorname{Var}(X \mid Y)] = \operatorname{Var}(X)$ — not that $\operatorname{Var}(X \mid Y)$ equals $\operatorname{Var}(X)$ almost surely. Study tip: The tower property is your best tool whenever conditional expectations appear — always ask whether you can condition inside an expectation to exploit a given constraint.
8. A random variable $X$ takes values $-2,-1,1,3$ with probabilities $0.1,0.2,0.3,0.4$, respectively. Let $\mathcal{G}$ be the sigma-field generated by the event $\{X>0\}$. What is $E[(X-E[X\mid\mathcal{G}])^2]$?
   Answer: $\frac{79}{105}$ When you see a conditional expectation with respect to a sigma-field generated by a single event, recognize that $\mathcal{G} = \sigma(\{X > 0\})$ partitions the sample space into exactly two atoms: $\{X > 0\} = \{X = 1 \text{ or } 3\}$ and $\{X \leq 0\} = \{X = -2 \text{ or } -1\}$. The conditional expectation $E[X \mid \mathcal{G}]$ is therefore a step function — constant on each atom. First, compute $E[X]$: $E[X] = (-2)(0.1) + (-1)(0.2) + (1)(0.3) + (3)(0.4) = -0.2 - 0.2 + 0.3 + 1.2 = 1.1$. On $\{X \leq 0\}$ (probability 0.3): $E[X \mid \mathcal{G}] = \frac{(-2)(0.1) + (-1)(0.2)}{0.3} = \frac{-0.4}{0.3} = -\frac{4}{3}$. On $\{X > 0\}$ (probability 0.7): $E[X \mid \mathcal{G}] = \frac{(1)(0.3) + (3)(0.4)}{0.7} = \frac{1.5}{0.7} = \frac{15}{7}$. Now compute $E[(X - E[X \mid \mathcal{G}])^2]$ by summing over all four values. For $X = -2$: $(-2 + \frac{4}{3})^2 = \frac{4}{9}$; for $X = -1$: $(-1 + \frac{4}{3})^2 = \frac{1}{9}$; for $X = 1$: $(1 - \frac{15}{7})^2 = \frac{64}{49}$; for $X = 3$: $(3 - \frac{15}{7})^2 = \frac{36}{49}$. Weighting: $(0.1)\frac{4}{9} + (0.2)\frac{1}{9} + (0.3)\frac{64}{49} + (0.4)\frac{36}{49} = \frac{6}{90} + \frac{33.6}{49 \cdot 10} = \frac{79}{105}$, confirming A. Choice B ($\frac{329}{100}$) is $E[X^2] - (E[X])^2 = \text{Var}(X)$, a common confusion — you're asked for variance around the conditional mean, not the unconditional variance. Choice C mixes intermediate calculations incorrectly. Choice D would only be correct if $\mathcal{G}$ were the full sigma-field generated by $X$ itself, making the residual identically zero. Remember: $E[(X - E[X\mid\mathcal{G}])^2]$ measures the unexplained variance after conditioning — it equals $\text{Var}(X) - \text{Var}(E[X\mid\mathcal{G}])$ by the law of total variance, which you can use as a check.
9. What is $E[T_1\mid T_1>1,\ T_2>1,\ T_1<T_2]$?
   Answer: $\frac{4}{3}$ When conditioning on multiple events simultaneously, your first instinct should be to identify the joint conditional distribution carefully — especially when competing exponentials are involved. Here, you need $E[T_1 \mid T_1 > 1,\ T_2 > 1,\ T_1 < T_2]$. By the memoryless property of exponentials, conditioning on $T_1 > 1$ and $T_2 > 1$ shifts both clocks: let $S_1 = T_1 - 1 \mid T_1 > 1 \sim \text{Exp}(2)$ and $S_2 = T_2 - 1 \mid T_2 > 1 \sim \text{Exp}(1)$, independently. The condition $T_1 < T_2$ becomes $S_1 < S_2$. So the problem reduces to finding $E[1 + S_1 \mid S_1 < S_2] = 1 + E[S_1 \mid S_1 < S_2]$. For independent exponentials with rates $\lambda_1 = 2$ and $\lambda_2 = 1$, the conditional expectation $E[S_1 \mid S_1 < S_2]$ equals $\frac{1}{\lambda_1 + \lambda_2} = \frac{1}{3}$. This is a standard result: given that $S_1$ "wins" the race, the winning time is $\text{Exp}(\lambda_1 + \lambda_2)$, giving mean $\frac{1}{3}$. Therefore $E[T_1 \mid \cdots] = 1 + \frac{1}{3} = \frac{4}{3}$, confirming B. Choice A ($\frac{1}{3}$) forgets to add back the shifted baseline of 1 after applying the memoryless property. Choice C ($\frac{3}{2}$) likely confuses the combined rate with $\lambda_1 + \lambda_2 = 3$ but adds $1$ incorrectly, or misapplies the unconditional mean of $T_1$. Choice D ($2$) is simply the unconditional mean $1/\lambda_1 = 1/2$... wait — it may reflect treating the problem as just $E[T_1] = 1/\lambda_1$ with the wrong rate, ignoring all conditioning. Study tip: Always apply the memoryless property first to simplify conditioning on survival past a threshold, then use the competing-exponentials "race" formula $E[\min] = \frac{1}{\sum \lambda_i}$ for the residual.
10. Given that the randomly selected participant tests positive, what is the probability that the participant came from Region A?
    Answer: $0.717$ Whenever you see a question involving conditional probability across multiple subgroups, reach for Bayes' theorem combined with the law of total probability. Here, you need to find the probability a positive test came from Region A — meaning you must work backward from an observed outcome. Start by computing the probability of testing positive in each region. For Region A: $P(+|A) = P(D)\cdot\text{Sens} + P(\bar{D})\cdot(1-\text{Spec}) = (0.10)(0.90) + (0.90)(0.05) = 0.09 + 0.045 = 0.135$. For Region B: $P(+|B) = (0.02)(0.80) + (0.98)(0.02) = 0.016 + 0.0196 = 0.0356$. Now apply total probability: $P(+) = (0.40)(0.135) + (0.60)(0.0356) = 0.054 + 0.02136 = 0.07536$. Finally, Bayes' theorem gives: $P(A|+) = \frac{(0.40)(0.135)}{0.07536} = \frac{0.054}{0.07536} \approx 0.717$, confirming A is correct. Choice B (0.400) is the trap of simply reporting the prior probability of Region A — ignoring the diagnostic information entirely. Choice C (0.790) likely results from omitting the false-positive term when computing $P(+|B)$, underestimating Region B's contribution to positive tests. Choice D (0.667) resembles an incorrect ratio using only the true-positive components without properly weighting by region priors. Strategy tip: In multi-group Bayes problems, always compute the full $P(+|\text{region})$ using both sensitivity and false-positive rate — forgetting false positives is the single most common error on these problems.

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