Sampling Distributions and Limit Theorems Questions and Answers

Q: Suppose \(X_{n} \to X\) almost surely. Suppose \(Y_{n}\) and Y are defined as \(Y_{n} = X_{n}^{3} + 2X_{n}^{2} + 3X_{n} + 2\) and \(Y = X^{3} + 2X^{2} + 3X + 2\). The following are three statements :(I) \(Y_{n} \to Y\) almost surely(II) \(Y_{n} \to Y\) in probability(III) \(Y_{n} \to Y\) in law.Which of the following is true ?

All three are true.\(g(x) = x^{3} + 2x^{2} + 3x + 2\) is continuous, and \(Y_{n} = g(X_{n})\), \(Y = g(X)\). By the continuous mapping theorem, almost sure convergence of \(X_{n}\) gives almost sure convergence of \(g(X_{n})\): wherever \(X_{n}(\omega) \to X(\omega)\), \(g(X_{n}(\omega)) \to g(X(\omega))\). So (I) holds.Almost sure convergence implies convergence in probability, which in turn implies convergence in law, so (II) and (III) follow from (I).

Q: Suppose X is a real random variable on a probability space \((\Omega, A, P)\) and \(\{A_{n}, n \ge 1\}\) is a sequence of sets in A such that :\[ A_{n} \to \phi \text{ as } n \to \infty \]

The conclusion is \(XI_{A_{n}} \to 0\) in probability.\(A_{n} \to \varnothing\) means each \(\omega\) lies in only finitely many \(A_{n}\), so \(I_{A_{n}}(\omega) \to 0\) for every \(\omega\) and \(X I_{A_{n}} \to 0\) pointwise, hence almost surely. Almost sure convergence implies convergence in probability. Directly: for \(\varepsilon \gt 0\), \(P(|XI_{A_{n}}| \gt \varepsilon) \le P(A_{n}) \to 0\) by continuity of probability.The limit is 0, not X or 1, so options B and C fail unless X itself is 0. Convergence in law to X (option D) would need X to have the same law as the constant 0.

Q: Suppose as \(n \to \infty\), \(X_{n} \to 3\) in probability and \(Y_{n} \to Y\) in distribution. Which of the following is true as \(n \to \infty\) ?

The true statement is \(\frac{n(X_{n} - 4) + Y_{n}}{n} \to -1\) in probability.Split the expression:\[ \begin{aligned} \frac{n(X_{n} - 4) + Y_{n}}{n} &= (X_{n} - 4) \\ &\quad + \frac{Y_{n}}{n} \end{aligned} \]\(X_{n} - 4 \to -1\) in probability. \(Y_{n}\) converges in distribution, so it is bounded in probability (tight), and dividing by n sends \(\frac{Y_{n}}{n} \to 0\) in probability. The sum tends to \(-1\) in probability.Options A and C use Slutsky's theorem correctly for the limit but claim convergence in probability, when \(Y_{n}\) converges only in distribution, so only convergence in distribution is guaranteed. Option D gives \((Y_{n} - 4) + \frac{X_{n}}{n} \to Y - 4\) in distribution, not the constant \(-4\).

Q: Let \(\{X_{n}, n \ge 1\}\) be a sequence of random variables such that as \(n \to \infty\), \(X_{n} \to X\) in distribution. Which of the following is not always true as \(n \to \infty\) ?

The statement not always true is \(E[X_{n}^{2} - X^{2}] \to 0\).Convergence in distribution guarantees \(E\,h(X_{n}) \to E\,h(X)\) for every bounded continuous h. \(\cos x\), \(\frac{x^{2}}{1 + x^{2}}\) and \(e^{i(x + 2)}\) are all bounded and continuous, so options A, C and D always hold.\(x^{2}\) is unbounded, and the second moments need not converge. For example, let \(X_{n} = n\) with probability \(\frac{1}{n}\) and 0 otherwise. Then \(X_{n} \to 0\) in distribution, but \(E(X_{n}^{2}) = n \to \infty\).

31.

Suppose \(X_{n} \to X\) almost surely. Suppose \(Y_{n}\) and Y are defined as \(Y_{n} = X_{n}^{3} + 2X_{n}^{2} + 3X_{n} + 2\) and \(Y = X^{3} + 2X^{2} + 3X + 2\). The following are three statements :

(I) \(Y_{n} \to Y\) almost surely
(II) \(Y_{n} \to Y\) in probability
(III) \(Y_{n} \to Y\) in law.

Which of the following is true ?

A.Only (I) is true
B.Only (I) and (II) are true
C.Only (II) and (III) are true
D.All three are true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

All three are true.

\(g(x) = x^{3} + 2x^{2} + 3x + 2\) is continuous, and \(Y_{n} = g(X_{n})\), \(Y = g(X)\). By the continuous mapping theorem, almost sure convergence of \(X_{n}\) gives almost sure convergence of \(g(X_{n})\): wherever \(X_{n}(\omega) \to X(\omega)\), \(g(X_{n}(\omega)) \to g(X(\omega))\). So (I) holds.

Almost sure convergence implies convergence in probability, which in turn implies convergence in law, so (II) and (III) follow from (I).

Discuss in Forum

32.

Suppose X is a real random variable on a probability space \((\Omega, A, P)\) and \(\{A_{n}, n \ge 1\}\) is a sequence of sets in A such that :

\[ A_{n} \to \phi \text{ as } n \to \infty \]

A.\(XI_{A_{n}} \to 0\) in probability
B.\(XI_{A_{n}} \to X\) in probability
C.\(XI_{A_{n}} \to 1\) in probability
D.\(XI_{A_{n}} \to X\) in law

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The conclusion is \(XI_{A_{n}} \to 0\) in probability.

\(A_{n} \to \varnothing\) means each \(\omega\) lies in only finitely many \(A_{n}\), so \(I_{A_{n}}(\omega) \to 0\) for every \(\omega\) and \(X I_{A_{n}} \to 0\) pointwise, hence almost surely. Almost sure convergence implies convergence in probability. Directly: for \(\varepsilon \gt 0\), \(P(|XI_{A_{n}}| \gt \varepsilon) \le P(A_{n}) \to 0\) by continuity of probability.

The limit is 0, not X or 1, so options B and C fail unless X itself is 0. Convergence in law to X (option D) would need X to have the same law as the constant 0.

Discuss in Forum

33.

Suppose as \(n \to \infty\), \(X_{n} \to 3\) in probability and \(Y_{n} \to Y\) in distribution. Which of the following is true as \(n \to \infty\) ?

A.\((X_{n} + Y_{n}) \to 3 + Y\) in probability
B.\(\frac{n(X_{n} - 4) + Y_{n}}{n} \to -1\) in probability
C.\(X_{n}Y_{n} + Y_{n} \to 4Y\) in probability
D.\(\frac{n(Y_{n} - 4) + X_{n}}{n} \to -4\) in probability

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The true statement is \(\frac{n(X_{n} - 4) + Y_{n}}{n} \to -1\) in probability.

Split the expression:

\[ \begin{aligned} \frac{n(X_{n} - 4) + Y_{n}}{n} &= (X_{n} - 4) \\ &\quad + \frac{Y_{n}}{n} \end{aligned} \]

\(X_{n} - 4 \to -1\) in probability. \(Y_{n}\) converges in distribution, so it is bounded in probability (tight), and dividing by n sends \(\frac{Y_{n}}{n} \to 0\) in probability. The sum tends to \(-1\) in probability.

Options A and C use Slutsky's theorem correctly for the limit but claim convergence in probability, when \(Y_{n}\) converges only in distribution, so only convergence in distribution is guaranteed. Option D gives \((Y_{n} - 4) + \frac{X_{n}}{n} \to Y - 4\) in distribution, not the constant \(-4\).

Discuss in Forum

34.

Let \(\{X_{n}, n \ge 1\}\) be a sequence of random variables such that as \(n \to \infty\), \(X_{n} \to X\) in distribution. Which of the following is not always true as \(n \to \infty\) ?

A.\(E[\cos(X_{n}) - \cos(X)] \to 0\)
B.\(E[X_{n}^{2} - X^{2}] \to 0\)
C.\(E\left[\frac{X_{n}^{2}}{1 + X_{n}^{2}} - \frac{X^{2}}{1 + X^{2}}\right] \to 0\)
D.\(E\left[e^{i(X_{n}+2)} - e^{i(X+2)}\right] \to 0\), where \(i = \sqrt{-1}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The statement not always true is \(E[X_{n}^{2} - X^{2}] \to 0\).

Convergence in distribution guarantees \(E\,h(X_{n}) \to E\,h(X)\) for every bounded continuous h. \(\cos x\), \(\frac{x^{2}}{1 + x^{2}}\) and \(e^{i(x + 2)}\) are all bounded and continuous, so options A, C and D always hold.

\(x^{2}\) is unbounded, and the second moments need not converge. For example, let \(X_{n} = n\) with probability \(\frac{1}{n}\) and 0 otherwise. Then \(X_{n} \to 0\) in distribution, but \(E(X_{n}^{2}) = n \to \infty\).

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