Expectation Properties - Statistics Graduate Level

1. Which statement about $E[Y\{E(X\mid G)-E(X\mid H)\}]$ is correct?
   Answer: It equals $0$ under the stated assumptions. Whenever you encounter expressions involving conditional expectations across nested sigma-fields, your instinct should be to apply the tower property and the pulling-out property of conditional expectation. Here, since $Y$ is $H$-measurable and $H \subseteq G$, $Y$ is also $G$-measurable. Consider expanding the expression: $E[Y\{E(X\mid G) - E(X\mid H)\}] = E[Y \cdot E(X\mid G)] - E[Y \cdot E(X\mid H)]$ For the first term, use the pulling-out property: since $Y$ is $G$-measurable, $E[Y \cdot E(X\mid G)] = E[E(YX\mid G)] = E[YX]$. For the second term, since $Y$ is $H$-measurable, $E[Y \cdot E(X\mid H)] = E[E(YX\mid H)] = E[YX]$. Both terms are equal, so their difference is exactly $0$. This confirms A is correct — no additional assumptions beyond those stated are needed. B is wrong because $E(YX) - E(Y)E(X)$ is the covariance $\text{Cov}(Y,X)$, which has no reason to equal zero in general, and the expression in question is always zero regardless of dependence structure. C is wrong because it imposes an unnecessary independence condition. The result holds purely from measurability and the tower property — independence of $X$ and $G$ is irrelevant. D is wrong for the same structural reason: $Y$ being constant is not required. Even a non-trivial bounded $H$-measurable $Y$ produces zero. As a study tip: whenever you see a bounded measurable $Y$ multiplied by a conditional expectation difference, immediately reach for the pulling-out property — it almost always telescopes to zero.
2. What is $E[(A-2)(X-9)]$?
   Answer: $12$ When you see an expression like $E[(A-2)(X-9)]$, your instinct should be to use the law of total expectation combined with iterated conditioning — specifically, conditioning on $A$ first to exploit what you know about $E(X \mid A)$. Start by writing $E[(A-2)(X-9)] = E\{E[(A-2)(X-9) \mid A]\}$. Since $(A-2)$ is a function of $A$, it factors out of the inner expectation: $E[(A-2)(X-9) \mid A] = (A-2)\cdot E[(X-9) \mid A] = (A-2)\cdot[E(X \mid A) - 9]$. Substituting $E(X \mid A) = 1 + 4A$ gives $(A-2)(1+4A-9) = (A-2)(4A-8) = 4(A-2)^2$. Taking the outer expectation: $E[4(A-2)^2] = 4\cdot E[(A-2)^2] = 4\cdot \text{Var}(A) = 4\cdot 3 = \mathbf{12}$, confirming C. For the wrong answers: A (0) tempts students who assume $A$ and $X$ are uncorrelated or independent — but the conditional mean structure directly creates dependence. B (9) likely comes from confusing $\text{Var}(A) = 3$ with $\text{Var}(A) = 9$, or misapplying the formula by forgetting the factor of 4. D (36) results from computing $4\cdot E(A-2)^2$ but using $E(A)^2 = 4$ instead of $\text{Var}(A) = 3$, conflating variance with squared mean. Your study tip: whenever a covariance-style expression involves a conditional mean, use iterated expectations and pull out functions of the conditioning variable — it almost always simplifies the problem cleanly.
3. Which conclusion is guaranteed?
   Answer: The series $Σ_{n=1}^{∞}X_n$ converges absolutely almost surely and in $L^1$, with expectation $Σ_{n=1}^{∞}E(X_n)$. When you see a condition like $\sum_{n=1}^{\infty} E(|X_n|) < \infty$, your instinct should immediately go to the dominated convergence theorem and its cousins in measure/integration theory. This condition is precisely what enables you to exchange limits and expectations — the heart of $L^1$ convergence theory. Here's the core reasoning for why C is correct. Define the partial sums $S_N = \sum_{n=1}^N X_n$ and consider $\sum_{n=1}^\infty E(|X_n|) < \infty$. By the monotone convergence theorem, $E\!\left(\sum_{n=1}^\infty |X_n|\right) = \sum_{n=1}^\infty E(|X_n|) < \infty$, which forces $\sum |X_n| < \infty$ almost surely — giving absolute a.s. convergence. Since the partial sums are dominated in $L^1$ by this integrable majorant, the dominated convergence theorem also yields $L^1$ convergence and justifies swapping the sum and expectation: $E\!\left(\sum X_n\right) = \sum E(X_n)$. Crucially, no independence assumption is needed — this holds for any integrable sequence. A is wrong because the convergence here is far stronger than mere convergence in probability; it's absolute a.s. and in $L^1$. B is wrong on two counts: the hypothesis says nothing about second moments, so $L^2$ convergence isn't guaranteed, and variance additivity requires uncorrelated (not just integrable) variables. D is wrong because the expectation identity follows from dominated convergence alone — independence is completely irrelevant here. Study tip: Whenever you see $\sum E(|X_n|) < \infty$, think "MCT gives an integrable majorant → DCT applies → swap sum and expectation freely, no independence needed."
4. What is $E(Z^TAZ)$?
   Answer: $17/2$ When you see a quadratic form involving a random vector, the key identity to recall is: for any random vector $Z$ and constant matrix $A$, $E(Z^T A Z) = \text{tr}(A\Sigma) + \mu^T A \mu$ This decomposes the expectation into a variance component and a mean component — a result derived from the identity $Z^T A Z = \text{tr}(AZZ^T)$ and linearity of expectation. Computing each piece: First, $\mu^T A \mu$ where $\mu = (1,-1)^T$: $A\mu = \begin{pmatrix}1(1)+2(-1)\\-1(1)+3(-1)\end{pmatrix} = \begin{pmatrix}-1\\-4\end{pmatrix}, \quad \mu^T A\mu = (1)(-1)+(-1)(-4) = 3$ Next, $\text{tr}(A\Sigma)$: $A\Sigma = \begin{pmatrix}1 & 2\\-1 & 3\end{pmatrix}\begin{pmatrix}2 & 1/2\\1/2 & 1\end{pmatrix} = \begin{pmatrix}3 & 5/2\\1/2 & 5/2\end{pmatrix}$ $\text{tr}(A\Sigma) = 3 + 5/2 = 11/2$ So $E(Z^T A Z) = 11/2 + 3 = 17/2$, confirming D. Choice A (3) gives only $\mu^T A \mu$, forgetting the trace term entirely. Choice B (11/2) gives only $\text{tr}(A\Sigma)$, ignoring the mean contribution. Choice C (9) likely results from computing $\text{tr}(A\Sigma) + \mu^T A \mu$ with an arithmetic error, possibly doubling $\mu^T A\mu$ incorrectly. A reliable study habit: memorize the quadratic form identity $E(Z^T A Z) = \text{tr}(A\Sigma) + \mu^T A\mu$ and practice splitting problems into these two components — this formula appears frequently in multivariate statistics and linear model theory.
5. What is $E[(2X-Y)(X+3Y)]$?
   Answer: $3/2$ When you see expressions involving products of random variables with zero means, your first move should be to expand everything algebraically and then apply linearity of expectation — reducing the problem to known quantities like $E(X^2)$, $E(Y^2)$, and $E(XY)$. Start by extracting $E(XY)$ from the given information. Expanding $E[(X+Y)^2] = E(X^2) + 2E(XY) + E(Y^2) = 1 + 2E(XY) + 1 = 3$, so $2E(XY) = 1$, giving $E(XY) = \frac{1}{2}$. Now expand the target expression: $(2X - Y)(X + 3Y) = 2X^2 + 6XY - XY - 3Y^2 = 2X^2 + 5XY - 3Y^2$. Applying linearity of expectation: $E[(2X-Y)(X+3Y)] = 2E(X^2) + 5E(XY) - 3E(Y^2) = 2(1) + 5\left(\frac{1}{2}\right) - 3(1) = 2 + \frac{5}{2} - 3 = \frac{3}{2}$. This confirms the answer is A. As for the distractors: B (1/2) likely comes from stopping after computing $E(XY) = \frac{1}{2}$ and forgetting to complete the full expansion. C (-1) may result from a sign error during expansion — for example, treating the middle terms as $-5XY$ instead of $+5XY$. D (7/2) could arise from adding rather than subtracting the $3E(Y^2)$ term, i.e., computing $2 + \frac{5}{2} + 3$ instead. The key study tip: always fully expand bilinear products before taking expectations, and remember that $E[(X+Y)^2]$ is the standard tool for uncovering $E(XY)$ when means are zero.
6. What is $E[Σ_{i=1}^{4}(X_i-\bar X)^2]$?
   Answer: $9$ When you see a question asking for the expected value of a sum of squared deviations from the sample mean, think immediately about the bias-variance decomposition and how individual variances contribute to that sum. The key identity is: $E\left[\sum_{i=1}^{4}(X_i - \bar{X})^2\right] = \sum_{i=1}^{4} \text{Var}(X_i) - \frac{1}{n}\left(\sum_{i=1}^{4} \text{Var}(X_i)\right)$ This comes from writing $(X_i - \bar{X}) = X_i - \frac{1}{n}\sum_j X_j$, computing the variance of each deviation, and summing. More precisely, $E\left[\sum_{i=1}^{4}(X_i - \bar{X})^2\right] = \left(1 - \frac{1}{n}\right)\sum_{i=1}^{4}\sigma_i^2$ Here $\sum \sigma_i^2 = 1 + 2 + 3 + 6 = 12$, and with $n = 4$: $\left(1 - \frac{1}{4}\right)(12) = \frac{3}{4}(12) = 9$ So the correct answer is C. A (3) likely comes from computing $\frac{1}{n}\sum \sigma_i^2 = 3$, which is just the variance of $\bar{X}$ scaled up — a piece of the calculation, not the whole thing. B (8) has no clean derivation and likely reflects an arithmetic error. D (12) is the raw sum of variances $\sum \sigma_i^2$, ignoring the correction factor that accounts for the mean being estimated from the same data — this is the most tempting trap. Your study tip: remember that $\sum(X_i - \bar{X})^2$ is a biased estimator of $\sum \sigma_i^2$ precisely because $\bar{X}$ "absorbs" one degree of freedom. The factor $(1 - 1/n)$ always appears, even when variances are unequal.
7. Random variables $X$ and $Y$ satisfy $E(X)=1$, $E(Y)=-1$, $Var(X)=4$, $Var(Y)=9$, and $Cov(X,Y)=-2$. What is $E[(X-2Y+1)^2]$?
   Answer: $64$ When you see a question asking for $E[g(X,Y)^2]$, your instinct should be to use the identity $E[Z^2] = Var(Z) + [E(Z)]^2$. This cleanly separates the problem into computing a mean and a variance. Let $Z = X - 2Y + 1$. First, find $E(Z)$: $E(Z) = E(X) - 2E(Y) + 1 = 1 - 2(-1) + 1 = 4$ Next, find $Var(Z)$. Since the constant 1 doesn't affect variance: $Var(X - 2Y) = Var(X) + 4\,Var(Y) - 2(2)\,Cov(X,Y)$ $= 4 + 4(9) - 4(-2) = 4 + 36 + 8 = 48$ Now apply the identity: $E(Z^2) = Var(Z) + [E(Z)]^2 = 48 + 16 = 64$ The correct answer is D. Tracing the wrong answers reveals exactly where students stumble. Choice A (16) is simply $[E(Z)]^2 = 4^2$, forgetting to add the variance — a classic error of treating expectation as linear through squaring. Choice B (48) is just $Var(Z)$, ignoring the mean term entirely. Choice C (56) likely comes from a sign error on the covariance term: if you treat $Cov(X,Y) = -2$ as $+2$, you get $Var(Z) = 40$, and $40 + 16 = 56$. Study tip: Always decompose $E[Z^2] = Var(Z) + [E(Z)]^2$ before computing. Also, memorize the bilinear variance formula carefully — the covariance term picks up a factor of $2 \cdot a \cdot b$ for $aX + bY$, and sign errors there are the most common trap on these problems.
8. Without assuming independent inclusion indicators, what is the expected number of sampled units belonging to the designated subgroup?
   Answer: $24/5$ When a sample size is itself random, you need a tool that handles randomness at two levels — and that tool is the law of total expectation. Whenever you see "random sample size" paired with "expected count in a subgroup," think: condition on the sample size first, then average over it. Here's the core reasoning. Let $M$ be the number of sampled units from the subgroup of 20. For any fixed sample size $K = k$, exchangeability guarantees every unit has equal inclusion probability. This means the expected number from the subgroup, given $K = k$, is simply proportional: $E(M \mid K = k) = k \cdot \frac{20}{50} = \frac{2k}{5}$. Critically, this proportionality holds regardless of whether inclusion indicators are independent — exchangeability alone is sufficient to establish the marginal inclusion probability as $k/50$. Now apply the law of total expectation: $E(M) = E\!\left[E(M \mid K)\right] = E\!\left[\frac{2K}{5}\right] = \frac{2}{5} E(K) = \frac{2}{5} \cdot 12 = \frac{24}{5}$. That confirms B. As for the distractors: A ($18/5$) would correspond to a subgroup fraction of $30/50$ — a wrong proportion, perhaps from miscounting. C ($6$) comes from applying the subgroup fraction to only half the expected sample size, or confusing $E(K) = 12$ with some intermediate step. D ($12$) is the trap of simply reporting $E(K)$ itself, ignoring that only 20 of 50 units belong to the subgroup. Your strategy: on questions with random sample sizes, always decompose using $E(M) = \frac{\text{subgroup size}}{\text{population size}} \cdot E(K)$. Independence of indicators is irrelevant — exchangeability is all you need.
9. What is $E[Σ_{i=1}^{N}X_i]$?
   Answer: $1/√(2π)$ This question tests your ability to work with random sums (compound random variables), where both the number of terms and the terms themselves involve randomness. The key tool is the law of total expectation: condition on the value of $N$. Since $N$ is determined by $X_1$, you need to split the expectation into two cases. When $N = 0$ (i.e., $X_1 \leq 0$), the sum is empty and equals $0$ by convention, contributing nothing. When $N = 1$ (i.e., $X_1 > 0$), the sum equals $X_1$ itself. So: $E\!\left[\sum_{i=1}^{N} X_i\right] = E[X_1 \cdot \mathbf{1}_{X_1 > 0}]$ This is the expectation of a truncated standard normal. Computing directly: $E[X_1 \cdot \mathbf{1}_{X_1 > 0}] = \int_0^{\infty} x \cdot \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\, dx = \frac{1}{\sqrt{2\pi}}\left[-e^{-x^2/2}\right]_0^{\infty} = \frac{1}{\sqrt{2\pi}}$ This confirms C is correct. Choice B ($0$) is the most seductive trap — you might think symmetry of the standard normal implies the answer is zero, but symmetry only applies to $E[X_1]$ over the full real line. Here you're restricting to the positive half. Choice A ($-1/\sqrt{2\pi}$) would correspond to $E[X_1 \cdot \mathbf{1}_{X_1 < 0}]$, the negative-half analog — a sign error. Choice D ($1/2$) confuses this with $P(X_1 > 0)$, a probability, not an expectation. When you see random sums, always condition on the index first — symmetry arguments are only valid when the domain of integration is symmetric.
10. Which value of $c$ makes $T_c$ unbiased for $θ$?
    Answer: $-1/4$ When a problem mixes a random "switching" variable like $M$ with a biased estimator, your first move should be computing the unconditional bias using the law of total expectation: $E(T_c) = E(T_c \mid M=1)P(M=1) + E(T_c \mid M=0)P(M=0)$. Notice that $2M - 1$ equals $+1$ when $M=1$ and $-1$ when $M=0$, so $T_c = T + c(2M-1)$ shifts $T$ by $+c$ or $-c$ depending on the group. The conditional expectations become $E(T_c \mid M=1) = (\theta + 2) + c$ and $E(T_c \mid M=0) = (\theta - 1) - c$. Plugging into total expectation with $P(M=1)=0.3$ and $P(M=0)=0.7$: $E(T_c) = 0.3[(\theta+2)+c] + 0.7[(\theta-1)-c] = \theta + 0.3(2) + 0.3c + 0.7(-1) - 0.7c$ $= \theta + 0.6 - 0.7 + c(0.3 - 0.7) = \theta - 0.1 - 0.4c$ For unbiasedness, set this equal to $\theta$: $-0.1 - 0.4c = 0 \Rightarrow c = -0.1/0.4 = -1/4$. That confirms B. Choice A ($c = -1/2$) overcorrects — it likely comes from ignoring the probability weights and solving $2c + (-1)(-c) = 0$ incorrectly. Choice C ($c = 1/4$) gets the magnitude right but the sign wrong, a common sign error when misreading $2M-1$. Choice D ($c = 1/2$) compounds both errors. Study tip: Whenever you see a correction term involving an indicator variable, always write out $2M - 1$ explicitly in each case before computing total expectation — sign errors and missed probability weights are the two classic traps here.

Start this quiz