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: Mean square check for $P$.
$X_n$ is $n$ with chance $1/n^2$, else $0$. So $E(X_n^2)=n^2\cdot\frac1{n^2}=1$ for every $n$.

Step 2: Read the verdict on $P$.
Since $E(X_n^2)$ stays at $1$ and does not go to $0$, there is no mean square convergence. $P$ fails.

Step 3: Probability check for $Q$.
For fixed $\epsilon>0$ and large $n$, $P(|X_n|>\epsilon)=P(X_n=n)=\frac1{n^2}$.

Step 4: Take the limit.
$\frac1{n^2}\to0$, so $X_n\to0$ in probability. $Q$ holds.

Step 5: Conclude.
$P$ wrong, $Q$ right, option (B).
\[ \boxed{(B)} \]