For a random sample \(X_1,\ldots,X_n\), \(n\geq2\), from a population with distribution function \(F_X\), let \(Z_n(x)\) be the proportion of sample values less than or equal to \(x\), \(x\in\mathbb{R}\). Which of the following statements is/are true?
Show Hint
Write \(Z_n(x)\) as an average of Bernoulli\((F_X(x))\) indicators; its mean is \(F_X(x)\) and its variance is \(F_X(x)(1-F_X(x))/n\), not \(/n^2\), and values at different points are correlated.
\(\text{Cov}\big(Z_n(x),Z_n(y)\big)=0\), for all \(x\neq y\)
\(Z_n(x)\) is a consistent estimator of \(F_X(x)\)
Show Solution
The Correct Option isA, D
Solution and Explanation
Step 1: Recognise $Z_n(x)$ as the empirical distribution function. $Z_n(x)$ counts what fraction of the sample falls at or below $x$, so it can be written as an average of indicator variables: \[ Z_n(x)=\frac1n\sum_{i=1}^n I(X_i\le x) \] Each $I(X_i\le x)$ takes the value $1$ with probability $F_X(x)$ and $0$ otherwise, so it is a Bernoulli variable with success probability $F_X(x)$, and the $n$ of them are independent because the sample is random.
Step 2: Get the mean, statement (A). Averaging $n$ Bernoulli$(F_X(x))$ variables gives back the same probability, so \[ E(Z_n(x))=F_X(x) \] This is a direct match with statement (A), so it holds.
Step 3: Get the variance, statement (B). A single Bernoulli$(p)$ variable has variance $p(1-p)$. Summing $n$ independent copies scales the variance by $n$, and averaging (dividing by $n$) scales the variance by $1/n^2$. Multiplying these together, \[ \text{Var}(Z_n(x))=\frac{n\,F_X(x)(1-F_X(x))}{n^2}=\frac{F_X(x)(1-F_X(x))}{n} \] The $n$'s do not fully cancel to leave $n^2$ in the denominator, they leave a single $n$. So statement (B), which claims a denominator of $n^2$, does not hold.
Step 4: Test the covariance claim, statement (C). Pick $x<y$. The same observation $X_i$ feeds both $I(X_i\le x)$ and $I(X_i\le y)$, and being below $x$ forces being below $y$ too, so these two indicators move together rather than being unrelated. Working out the covariance of one pair and scaling by $n$ for the sum gives \[ \text{Cov}(Z_n(x),Z_n(y))=\frac{F_X(x)\big(1-F_X(y)\big)}{n} \] which is not zero except in boundary cases. So the claim of zero covariance at every pair of points fails.
Step 5: Test consistency, statement (D). $Z_n(x)$ has no bias, since its mean is exactly $F_X(x)$, and its variance shrinks toward $0$ as $n$ grows, since it carries a factor $1/n$. An estimator with vanishing bias and vanishing variance converges in probability to the true value. So $Z_n(x)$ is a consistent estimator of $F_X(x)$, which is just the familiar fact that the empirical distribution function estimates the true distribution function well for large samples. Statement (D) holds.
Step 6: Collect the results. Only the mean statement and the consistency statement survive scrutiny. \[ \boxed{\text{(A) and (D)}} \]
