Solution.

In probability

For \(0<\varepsilon <1\), \(P\left (\left |X_n\right |\geq \varepsilon \right ) = \tfrac 1n\rightarrow 0\), so \(X_n\underset {P}{\longrightarrow }0\).

In quadratic mean

\(E\left (X_n^{2}\right ) = 1^{2}\cdot \tfrac 1n = \tfrac 1n\rightarrow 0\), so \(X_n\underset {QM}{\longrightarrow }0\) as well. Note the contrast with the example in the text, where the value was \(n\) rather than \(1\): there the second moment diverged. Boundedness is what makes the difference.

In distribution

It follows from convergence in probability.

Almost surely

This depends on information not given. If the \(X_n\) are independent then \(\sum _n P(X_n=1) = \sum _n \tfrac 1n\) diverges, so by the second Borel–Cantelli lemma \(X_n=1\) infinitely often with probability one and there is no almost sure convergence. If instead \(X_n = I\left (U<\tfrac 1n\right )\) for a single uniform \(U\), the sequence is eventually zero for every \(U>0\) and converges almost surely.

The moral is that the first three modes are determined by the marginal distributions alone, while almost sure convergence is not — it is a property of the joint behaviour, which is why it sits apart in the hierarchy.

Solution. Both are instances of the continuous mapping theorem. The function \(g(x)=\sqrt {x}\) is continuous at every \(x>0\), and \(\mu >0\), so \(g\left (\overline {X}_n\right )\underset {P}{\longrightarrow }g(\mu )\). The same argument with \(h(x)=1/x\), continuous at every \(x\neq 0\), gives the second.

The hypothesis \(\mu >0\) is doing real work: at \(\mu =0\) the reciprocal is not continuous and the conclusion fails. This is the same proviso that appears in part (c) of Slutsky’s theorem, and for the same reason.