Question

Let \(X_1,X_2,\ldots\) be a sequence of independent random variables with

• A. \(P\) is correct and \(Q\) is NOT correct
• B. \(P\) is NOT correct and \(Q\) is correct
• C. Both \(P\) and \(Q\) are correct
• D. Neither \(P\) nor \(Q\) is correct

Hint:

Mean-square convergence requires \(E[(X_n-X)^2]\to 0\), while convergence in probability only requires \(P(|X_n-X|>\epsilon)\to 0\) for every \(\epsilon>0\).

Answer 1

The Correct Option is B

Step 1: Check convergence in mean-square.
A sequence \(X_n\) converges in mean-square to \(0\) if
\[ E[(X_n-0)^2]\to 0 \] That is,
\[ E(X_n^2)\to 0 \]
Here, \(X_n\) takes the value \(n\) with probability \(\frac{1}{n^2}\), and \(0\) with probability \(1-\frac{1}{n^2}\).
Therefore,
\[ E(X_n^2)=n^2\cdot \frac{1}{n^2}+0^2\cdot \left(1-\frac{1}{n^2}\right) \] \[ E(X_n^2)=1 \] Since
\[ E(X_n^2)=1 \] for every \(n\), we get
\[ E(X_n^2)\not\to 0 \] Hence, \(\{X_n\}\) does not converge in mean-square to \(0\).
Therefore, statement \(P\) is NOT correct.

Step 2: Check convergence in probability.
A sequence \(X_n\) converges in probability to \(0\) if for every \(\epsilon>0\),
\[ P(|X_n-0|>\epsilon)\to 0 \] That is,
\[ P(|X_n|>\epsilon)\to 0 \]
For any fixed \(\epsilon>0\), when \(n>\epsilon\), the event \(|X_n|>\epsilon\) occurs exactly when \(X_n=n\).
Thus,
\[ P(|X_n|>\epsilon)=P(X_n=n) \] \[ P(|X_n|>\epsilon)=\frac{1}{n^2} \] Now,
\[ \frac{1}{n^2}\to 0 \] Therefore,
\[ P(|X_n|>\epsilon)\to 0 \] Hence, \(\{X_n\}\) converges in probability to \(0\).
Therefore, statement \(Q\) is correct.

Step 3: Final conclusion.
Statement \(P\) is NOT correct and statement \(Q\) is correct.
Hence,
\[ \boxed{(B)} \]