Probability has several convergence notions because random variables can converge pointwise, in probability, in distribution, or in mean. Knowing the implications prevents many false arguments.
Four convergence modes
The common modes compare random variables at different levels of strength.
Definition. For random variables \(X_n,X\), write \(X_n\to X\) a.s. if \(P(\lim_nX_n=X)=1\), in probability if \(P(|X_n-X|\gt\varepsilon)\to0\) for every \(\varepsilon\gt0\), in \(L^p\) if \(\mathbb E|X_n-X|^p\to0\), and in distribution if \(\mathbb E f(X_n)\to\mathbb E f(X)\) for every bounded continuous \(f\).
Basic implications
Almost-sure and \(L^p\) convergence imply convergence in probability; convergence in probability implies convergence in distribution.
Theorem. If \(X_n\to X\) a.s. or in \(L^p\), then \(X_n\to X\) in probability. If \(X_n\to X\) in probability, then \(X_n\Rightarrow X\).
Proof. Almost-sure convergence gives \(1_{\{|X_n-X|\gt\varepsilon\}}\to0\) a.s.; dominated convergence gives the probabilities tend to zero. \(L^p\)-convergence gives \(P(|X_n-X|\gt\varepsilon)\le\varepsilon^{-p}\mathbb E|X_n-X|^p\). For convergence in distribution, take bounded continuous \(f\); convergence in probability implies \(f(X_n)\to f(X)\) in probability, and boundedness plus subsequence arguments or dominated convergence along almost-sure subsequences gives convergence of expectations.
Subsequence principle
Convergence in probability is exactly almost-sure convergence along subsequences after further extraction.
Proposition. \(X_n\to X\) in probability if and only if every subsequence has a further subsequence converging to \(X\) almost surely.
Proof. If \(X_n\to X\) in probability, choose a subsubsequence \(n_k\) with \(P(|X_{n_k}-X|\gt2^{-k})\le2^{-k}\). Borel-Cantelli gives almost-sure convergence along \(n_k\). The converse follows because if convergence in probability failed, some subsequence would keep \(P(|X_n-X|\gt\varepsilon)\) bounded away from zero, preventing any almost-sure convergent further subsequence.
Proposition. If \(X_n\Rightarrow c\) for a constant \(c\), then \(X_n\to c\) in probability.
Proof. For \(\varepsilon>0\), the closed set \[ F_\varepsilon=(-\infty,c-\varepsilon]\cup[c+\varepsilon,\infty) \] has limiting probability \(\delta_c(F_\varepsilon)=0\). Portmanteau gives \[ \limsup_nP(|X_n-c|\ge\varepsilon)\le0. \]
Theorem (Slutsky). If \(X_n\Rightarrow X\) and \(Y_n\to c\) in probability, then \[ X_n+Y_n\Rightarrow X+c,\qquad X_nY_n\Rightarrow cX. \] If \(c\ne0\), then \(X_n/Y_n\Rightarrow X/c\).
Proof. It suffices to prove \((X_n,Y_n)\Rightarrow(X,c)\). For bounded Lipschitz \(f\), \[ \begin{aligned} |\mathbb Ef(X_n,Y_n)-\mathbb Ef(X_n,c)| &\le \mathbb E[\min\{2\|f\|_\infty,L|Y_n-c|\}] \longrightarrow0 \end{aligned} \] by convergence in probability and boundedness. The second term converges by \(X_n\Rightarrow X\). Apply the continuous mapping theorem to addition, multiplication, and division away from \(c=0\).
Counterexamples to reverse implications. Let \(U\sim\operatorname{Unif}(0,1)\) and \(X_n=\mathbf1_{\{U\le1/n\}}\). Then \(X_n\to0\) almost surely but \(\|X_n\|_\infty=1\), so a.s. convergence need not be uniform. The spike \(Y_n=n\mathbf1_{\{U\le1/n\}}\) converges in probability to \(0\) but \(\mathbb E|Y_n|=1\), so convergence in probability need not imply \(L^1\). Independent identically distributed nondegenerate \(Z_n\) all have the same law and hence \(Z_n\Rightarrow Z_1\), but they do not converge in probability to \(Z_1\).