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.

Solution. A sequence of random variables \(\{X_n\}_{n\geq 1}\) converges to \(X\) in probability, written \(X_n\underset {P}{\longrightarrow }X\), if for every \(\varepsilon >0\) \[\lim _{n\rightarrow \infty }P\left (|X_n-X|\geq \varepsilon \right ) = 0 ,\] or equivalently \[\lim _{n\rightarrow \infty }P\left (|X_n-X|<\varepsilon \right ) = 1 .\]

Solution. Every \(X_i\geq a\), so \(Y_n\geq a\): the density puts no mass below \(a\), and \(|Y_n-a| = Y_n-a\). For \(\varepsilon >0\), the minimum is at least \(a+\varepsilon \) exactly when every \(X_i\) is, so by independence \begin {align*} P\left (|Y_n-a|\geq \varepsilon \right ) &= P\left (Y_n\geq a+\varepsilon \right )\\ &= P\left (X_1\geq a+\varepsilon ,\ X_2\geq a+\varepsilon ,\ \dots ,\ X_n\geq a+\varepsilon \right )\\ &= \left [P\left (X_1\geq a+\varepsilon \right )\right ]^{n}. \end {align*}

Now \[P\left (X_1\geq a+\varepsilon \right ) = \int _{a+\varepsilon }^{\infty }e^{-(x-a)}\,dx = e^{a}\left [-e^{-x}\right ]_{a+\varepsilon }^{\infty } = e^{a}e^{-(a+\varepsilon )} = e^{-\varepsilon },\] so \[P\left (|Y_n-a|\geq \varepsilon \right ) = e^{-n\varepsilon }\longrightarrow 0 \qquad (n\rightarrow \infty ).\] Therefore \(Y_n\underset {P}{\longrightarrow }a\).

Solution. Let \(p_n = \frac {\lambda }{n+\lambda }\), so that \[1-p_n = \frac {n}{n+\lambda } = \frac {1}{1+\lambda /n}.\] \(X_n\) counts the trials up to the first success, so \(X_n>k\) exactly when the first \(k\) trials all fail: \[P(X_n>k) = (1-p_n)^{k},\qquad k=0,1,2,\dots \] Let \(Y_n = X_n/n\) and \(y>0\). Then \[P(Y_n>y) = P(X_n>ny) = (1-p_n)^{\lfloor ny\rfloor } = \left (1+\frac {\lambda }{n}\right )^{-\lfloor ny\rfloor }.\] Since \(ny-1<\lfloor ny\rfloor \leq ny\) and \(1+\lambda /n\rightarrow 1\), the floor makes no difference in the limit, and \[\lim _{n\rightarrow \infty }P(Y_n>y) = \lim _{n\rightarrow \infty }\left [\left (1+\frac {\lambda }{n}\right )^{n}\right ]^{-y} = \left (e^{\lambda }\right )^{-y} = e^{-\lambda y}.\] Hence \[F_{Y_n}(y) = 1-P(Y_n>y)\longrightarrow 1-e^{-\lambda y},\qquad y>0 ,\] which is the distribution function of the density \(\lambda e^{-\lambda y}\). So \(X_n/n\underset {D}{\longrightarrow }Y\) where \(Y\) is exponential with parameter \(\lambda \), that is, with mean \(1/\lambda \).