Question

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?

• A. \(E\big(Z_n(x)\big)=F_X(x)\)
• B. \(\text{Var}\big(Z_n(x)\big)=\dfrac{F_X(x)(1-F_X(x))}{n^2}\)
• C. \(\text{Cov}\big(Z_n(x),Z_n(y)\big)=0\), for all \(x\neq y\)
• D. \(Z_n(x)\) is a consistent estimator of \(F_X(x)\)

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.

Answer 1

The Correct Option is A, D

Step 1: Write \(Z_n(x)\) in terms of indicator variables.
By definition, \(Z_n(x)\) is the fraction of sample points at or below \(x\), so
\[ Z_n(x)=\frac{1}{n}\sum_{i=1}^{n} I(X_i\leq x), \]
where \(I(X_i\leq x)\) is \(1\) if \(X_i\leq x\) and \(0\) otherwise. Since \(X_1,\ldots,X_n\) are a random sample, the indicators \(I(X_1\leq x),\ldots,I(X_n\leq x)\) are independent and identically distributed Bernoulli random variables with success probability \(P(X_i\leq x)=F_X(x)\).

Step 2: Check statement (A).
Taking expectation,
\[ E(Z_n(x))=\frac{1}{n}\sum_{i=1}^n E\big(I(X_i\leq x)\big)=\frac{1}{n}\sum_{i=1}^n F_X(x)=\frac{1}{n}\cdot nF_X(x)=F_X(x). \]
This matches statement (A) exactly, so (A) is TRUE.

Step 3: Check statement (B).
Each indicator is Bernoulli with parameter \(F_X(x)\), so its variance is \(F_X(x)(1-F_X(x))\). Since the \(n\) indicators are independent,
\[ \text{Var}\left(\sum_{i=1}^n I(X_i\leq x)\right)=nF_X(x)(1-F_X(x)). \]
Dividing by \(n^2\) because of the \(\frac{1}{n}\) factor in \(Z_n(x)\),
\[ \text{Var}(Z_n(x))=\frac{1}{n^2}\times nF_X(x)(1-F_X(x))=\frac{F_X(x)(1-F_X(x))}{n}. \]
The statement in (B) writes the denominator as \(n^2\) instead of \(n\), which is wrong. So (B) is FALSE.

Step 4: Check statement (C).
Take \(x<y\) without loss of generality. The indicators \(I(X_i\leq x)\) and \(I(X_i\leq y)\) are built from the same observation \(X_i\), and whenever \(X_i\leq x\) it is automatically true that \(X_i\leq y\). This dependence gives \[ \text{Cov}\big(I(X_i\leq x),I(X_i\leq y)\big)=F_X(x)-F_X(x)F_X(y)=F_X(x)\big(1-F_X(y)\big)\neq0 \] in general. Summing over the \(n\) independent observations, \[ \text{Cov}(Z_n(x),Z_n(y))=\frac{F_X(x)(1-F_X(y))}{n}\neq0 \] whenever \(0<F_X(x)<1\) and \(0<F_X(y)<1\). So the claim that this covariance is always \(0\) is FALSE.

Step 5: Check statement (D).
From Step 2, \(Z_n(x)\) is unbiased for \(F_X(x)\), and from Step 3, its variance equals \(\dfrac{F_X(x)(1-F_X(x))}{n}\), which goes to \(0\) as \(n\to\infty\). An estimator whose bias is \(0\) and whose variance shrinks to \(0\) converges in probability, in fact in mean square, to the parameter it estimates. So \(Z_n(x)\) is a consistent estimator of \(F_X(x)\), matching the Glivenko-Cantelli type behaviour of the empirical distribution function. Statement (D) is TRUE.

Final Answer:
\(Z_n(x)\) is unbiased and consistent for \(F_X(x)\), but its variance has \(n\), not \(n^2\), in the denominator, and values of \(Z_n\) at different points are correlated, not uncorrelated.\[ \boxed{\text{(A) and (D)}} \]