Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that if the sequence $X_1, X_2,\ldots$ converges to a constant $\theta$ in probability then it converges to $\theta$ in distribution

Can't seem to figure it out.... Any assistance would be greatly appreciated!

share|cite|improve this question
It's worth noting that the converse is also true. – Nate Eldredge Oct 31 '11 at 12:58

A characterization of convergence in distribution of $(X_n)$ to $\theta$ is the fact that $\mathrm E(h(X_n))\to h(\theta)$ for every continuous bounded function $h$. Once you start with this, the road is clear: decompose the difference $\mathrm E(h(X_n)-h(\theta))$ into a part where $X_n$ is close to $\theta$ and a part where $X_n$ is not.

More precisely, assume that $|h(x)|\leqslant M$ for every $x$ and choose a positive $\varepsilon$. Since $h$ is continuous at $\theta$, there exists $\delta$ such that $|x-\theta|\leqslant \delta$ implies $|h(x)-h(\theta)|\leqslant\varepsilon$, hence $$ |\mathrm E(h(X_n))-h(\theta)|\leqslant \mathrm E(|h(X_n)-h(\theta)|)\leqslant2M\mathrm P(|X_n-\theta|\geqslant\delta)+\varepsilon. $$ When $n\to\infty$, $\mathrm P(|X_n-\theta|\geqslant\delta)\to0$ hence $\limsup|\mathrm E(h(X_n))-h(\theta)|\leqslant\varepsilon$. This holds for every positive $\varepsilon$ hence $\mathrm E(h(X_n))\to h(\theta)$.

Nota: This result (convergence in probability implies convergence in distribution) holds for every limit in probability, not necessarily deterministic. As @Nate mentioned in a comment, the reverse implication (convergence in distribution implies convergence in probability) is true as well, but only when the limit is almost surely constant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.