When Does Convergence Pass to Expectations?
Convergence in probability says that large deviations from a limiting random variable become unlikely. It does not, by itself, prevent those deviations from having very large magnitude. Since expectation accounts for both magnitude and probability, a small-probability event can still make a substantial contribution to an expectation.
Uniform integrability supplies the missing control: it rules out a significant amount of expected absolute value hiding in increasingly remote tails. This tutorial develops a practical test. First check the mode of convergence; then check uniform integrability; finally use Vitali’s theorem to conclude convergence in \(L^1\), and hence convergence of expectations.
Here \(\mathbf 1_A\) denotes the indicator of an event \(A\). The defining limit controls the expected absolute contribution beyond a large threshold, uniformly over the whole sequence. Uniform integrability is stronger than merely having finite expectations, and stronger than a uniform bound on the expectations alone.
Convergence in probability of \(X_n\) to \(X\) means that for every \(\varepsilon>0\), \(\mathbb P(|X_n-X|>\varepsilon)\to0\). Almost-sure convergence implies convergence in probability. In particular, if \(X_n\to X\) almost surely, then for each fixed \(\varepsilon>0\) the indicators \(\mathbf 1_{\{|X_n-X|>\varepsilon\}}\) tend to zero almost surely; bounded convergence gives convergence of their expectations, which are the probabilities in question.
Vitali’s Theorem and the Expectation Conclusion
The \(L^1\) conclusion is the key: it says the expected absolute error itself vanishes. It is stronger than convergence of signed expectations, because cancellation cannot make \(\mathbb E|X_n-X|\) small. Here is a proof that also explains the role of the tail condition.
From convergence in probability, choose a subsequence \((X_{n_j})\) such that $$ \mathbb P(|X_{n_j}-X|>2^{-j})<2^{-j}. $$ The first Borel–Cantelli lemma implies \(X_{n_j}\to X\) almost surely. Fatou’s lemma now gives $$ \mathbb E|X|\le \liminf_{j\to\infty}\mathbb E|X_{n_j}| \le K_0+1, $$ so \(X\) is integrable.
For every \(K>0\), on the event \(\{|X|>K\}\), almost-sure convergence along this subsequence implies that eventually \(|X_{n_j}|>K/2\), and \(|X_{n_j}|\to|X|\). Fatou’s lemma therefore gives $$ \mathbb E\!\left[|X|\mathbf 1_{\{|X|>K\}}\right] \le \liminf_{j\to\infty} \mathbb E\!\left[|X_{n_j}|\mathbf 1_{\{|X_{n_j}|>K/2\}}\right] \le \delta(K/2). $$
Define the clipping function \(T_K(x)=\max(-K,\min(x,K))\). It satisfies $$ |x-T_K(x)|\le |x|\mathbf 1_{\{|x|>K\}}. $$ The function \(T_K\) is continuous, so \(T_K(X_n)\to T_K(X)\) in probability. Also \(|T_K(X_n)-T_K(X)|\le2K\). For any \(\eta>0\), $$ \mathbb E|T_K(X_n)-T_K(X)| \le \eta+2K\,\mathbb P(|T_K(X_n)-T_K(X)|>\eta). $$ Taking the limit superior as \(n\to\infty\), then letting \(\eta\downarrow0\), shows that this expectation tends to zero for each fixed \(K\).
By the triangle inequality and the clipping error bounds, $$ \limsup_{n\to\infty}\mathbb E|X_n-X| \le \delta(K)+\delta(K/2). $$ Letting \(K\to\infty\) proves \(\mathbb E|X_n-X|\to0\). Finally, $$ |\mathbb E X_n-\mathbb E X|\le \mathbb E|X_n-X|\longrightarrow0, $$ which proves convergence of expectations. \(\square\)
The proof splits the error into two parts. Clipping makes the variables bounded, where convergence in probability suffices to control expected error. Uniform integrability makes the clipping errors small, uniformly in \(n\). This is why convergence in probability alone is not enough.
Worked Examples: Applying the Criterion
Worked Example: A Uniform Moment Bound Gives Convergence of Expectations
Let the probability space be \([0,1]\) with Lebesgue probability measure, and set $$ X_n(\omega)=2+\sqrt n\,\mathbf 1_{(0,1/n)}(\omega). $$ For each \(\omega>0\), the indicator is eventually zero, so \(X_n(\omega)\to2\); at \(\omega=0\) it is zero for every \(n\) as well. Thus \(X_n\to2\) almost surely.
For \(p=3/2\), the inequality \((a+b)^p\le 2^{p-1}(a^p+b^p)\) for nonnegative \(a,b\) gives $$ \mathbb E|X_n|^{3/2} \le 2^{1/2}\left(2^{3/2}+\frac{n^{3/4}}{n}\right) \le 2^{1/2}(2^{3/2}+1). $$ For any \(K>0\), \(|x|\mathbf 1_{\{|x|>K\}}\le |x|^{3/2}/K^{1/2}\). The displayed moment bound therefore implies that the tail expectations tend uniformly to zero as \(K\to\infty\). The sequence is uniformly integrable. Vitali’s theorem applies, and direct calculation confirms its conclusion: $$ \mathbb E X_n=2+\sqrt n\,\mathbb P(0<\omega<1/n) =2+\frac{1}{\sqrt n}\longrightarrow2. $$
Worked Example: Rare Spikes Defeat Expectation Convergence
On the same probability space, let $$ Y_n(\omega)=n\mathbf 1_{(0,1/n)}(\omega). $$ For every \(\omega>0\), \(Y_n(\omega)\) is eventually zero, and \(Y_n(0)=0\). Hence \(Y_n\to0\) almost surely. Nevertheless, $$ \mathbb E Y_n=n\,\mathbb P(0<\omega<1/n)=1 $$ for every \(n\), so the expectations do not converge to \(\mathbb E0=0\).
The failure is precisely a failure of uniform integrability. If \(K>0\), choose an integer \(n>K\). On \((0,1/n)\), \(Y_n=n>K\), and therefore $$ \mathbb E\!\left[|Y_n|\mathbf 1_{\{|Y_n|>K\}}\right] =n\cdot\frac1n=1. $$ The supremum over \(n\) is at least \(1\) for every \(K\), so the required tail limit is not zero. The spike becomes taller as it becomes rarer, leaving its expected contribution unchanged.
Worked Example: An Integrable Envelope Controls a Shrinking Tail
Define \(Z(\omega)=\omega^{-1/2}\) for \(\omega\in(0,1]\), and set \(Z(0)=0\). Then \(Z\) is integrable, since $$ \mathbb E Z=\int_0^1\omega^{-1/2}\,d\omega=2. $$ Let \(W_n=Z\mathbf 1_{(0,1/n)}\). For every \(\omega>0\), \(W_n(\omega)=0\) eventually, and \(W_n(0)=0\); hence \(W_n\to0\) almost surely.
Since \(|W_n|\le Z\), for every \(K>0\), $$ \mathbb E\!\left[|W_n|\mathbf 1_{\{|W_n|>K\}}\right] \le \mathbb E\!\left[Z\mathbf 1_{\{Z>K\}}\right]. $$ The right-hand side tends to zero as \(K\to\infty\), by integrability of \(Z\). Thus \((W_n)\) is uniformly integrable. Direct integration gives $$ \mathbb E W_n=\int_0^{1/n}\omega^{-1/2}\,d\omega =\frac{2}{\sqrt n}\longrightarrow0, $$ in agreement with Vitali’s theorem.
A Practical Decision Process
When a problem asks whether expectations converge, it helps to separate the logical steps rather than infer an answer directly from the mode of convergence.
Check whether \(X_n\to X\) in probability. Almost-sure convergence is sufficient because it implies convergence in probability.
Verify uniform integrability, for example through a uniform higher-moment bound or an integrable envelope.
Conclude that \(\mathbb E|X_n-X|\to0\), and then use \(|\mathbb E X_n-\mathbb E X|\le\mathbb E|X_n-X|\).
A common pitfall is to treat convergence in probability as if it controlled the size of the errors. It controls the probabilities of errors exceeding each fixed threshold, but the rare-spike example shows that the magnitudes on those events can still grow enough to prevent expectation convergence. Conversely, uniform integrability without convergence to a particular limit does not by itself imply that expectations converge. Both ingredients in Vitali’s theorem have a distinct role.
Check Your Understanding
For each question, identify which part of the convergence-and-tail criterion is relevant.
- Why does almost-sure convergence suffice for the convergence-in-probability hypothesis of Vitali’s theorem?
- What uniform tail quantity must tend to zero for a sequence to be uniformly integrable?
- If \(X_n\to X\) in probability and \((X_n)\) is uniformly integrable, what stronger form of convergence does Vitali’s theorem give?
- For \(Y_n=n\mathbf 1_{(0,1/n)}\), compute the expectation and explain why the sequence is not uniformly integrable.
- How can a uniform bound on \(\mathbb E|X_n|^p\), for some \(p>1\), control the tail expectations?