Discussion
\(\{X_{n}, n \ge 1\}\), \(\{Y_{n}, n \ge 1\}\) and Y are random variables defined on the same probability space. Suppose as \(n \to \infty\), \(X_{n}\) converges in probability to 3 and \(Y_{n}\) converges in distribution to Y. Then which of the following is true as \(n \to \infty\) ?
Answer: D
\(X_{n}Y_{n}^{2}\) need not converge in probability but converges in distribution to \(3Y^{2}\).
Since \(X_{n} \to 3\), a constant, in probability, and \(Y_{n} \to Y\) in distribution, Slutsky's theorem gives \(X_{n}Y_{n}^{2} \xrightarrow{d} 3Y^{2}\) (using the continuous mapping theorem for \(y \mapsto y^{2}\)).
Convergence in probability to \(3Y^{2}\) would need \(Y_{n}\) to be close to Y on the probability space, not just in distribution. Take \(X_{n} = 3\), Y standard normal and \(Y_{n} = Z\) for every \(n\), where Z is an independent standard normal. Then \(Y_{n} \to Y\) in distribution, since the laws are identical, but \(X_{n}Y_{n}^{2} - 3Y^{2} = 3(Z^{2} - Y^{2})\) is a fixed non-degenerate random variable, so it does not tend to 0 in probability.
So option A is too strong, option C contradicts Slutsky, and option B denies the convergence in distribution that Slutsky guarantees.
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