# If the variance converges to zero, when do we have almost sure convergence

We have that $$\mathbf{E}(X_n)=c$$ where c is a positive constant and $$\lim_{n \rightarrow \infty} \mathtt{Var}(X_n) =0$$. Then $$X_n \rightarrow c \quad \mbox{in probability as} \quad n \rightarrow \infty.$$ If it is also known that $$X_n$$ is bounded, can one show that $$X_n$$ approaches $$c$$ almost surely, and if so why, or is there a counter-example?

No. Consider $$X_n \in \{0,1\}$$, therefore bounded, and the sequence of probabilities $$P_n(X_n = 1) = 1/n$$ (and therefore $$P_n(X_n = 0) = 1 - 1/n$$.) Clearly $$X_n \to 0$$ in probability, but Borel-Cantelli tells us that $$X_n$$ does not converge almost surely to $$0$$, as we expect to see infinitely many $$1$$ values over the sequence.

Edit: I overlooked the constraint of $$\mathrm{E}(X_n) = c$$, but the above example can easily be extended to cover it.

Consider $$X_n \in \{0,1,2\}$$, with the sequence of probabilities $$P_n(X_n = 0) = P_n(X_n = 2) = 1/n$$. $$\mathrm{E}(X_n) = 1 \;\forall\;n$$, so the constraint on the expected value is satisfied. Clearly $$X_n \to 1$$ in probability, but Borel-Cantelli still tells us that we don't get almost sure convergence of $$X_n$$ to $$1$$.

• Thank you for the example @jbowman. Is there also an example where $\mathbf{E}(X_n)$ is a constant for all $n$?
– GCru
Commented Jun 30 at 16:50
• Ah, I overlooked that part of the question :(. But you can just symmetrize the above example; work with $\{0,1,2\}$ with mean $1$, for example. Commented Jun 30 at 17:25

To show $$\{X_n\}$$ converges to $$c$$ almost surely, one can add the extra condition that

$$\sum_{n=1}^\infty \mathrm{Var}(X_n)<\infty \tag{*}\label 1$$

which can be satisfied by, for example, if $$\mathrm{Var}(X_n)=O(2^{-n})$$.

In other words, we need the variance to converge fast enough in order to upgrade convergence in probability to that almost surely. Formally, we have the following proposition, which is a direct result of Borel–Cantelli lemma and Chebyshev's inequality.

Proposition 1. Consider a sequence of random variables $$\{X_n\}$$ satisfying $$EX_n=c$$ and $$\eqref 1$$. Then, we have $$X_n \to \mu$$ almost surely as $$n\to \infty$$.

Proof. Using the Chebyshev's inequality, we have (for any $$\epsilon>0$$),

$$\sum_{n=1}^\infty P(|X_n-c|>\epsilon)\leq \sum_{n=1}^\infty \frac{\mathrm{Var}(X_n)}{\epsilon^2}<\infty$$

Therefore, using the Borel–Cantelli lemma (Thm. 7.5 of here), we conclude $$X_n\to c$$ almost surely.

• +1. But what is "ungrade" intended to mean? "Upgrade," or something else?
– whuber
Commented Jul 3 at 14:01
• I mean "upgrade". Sorry for the typo. Commented Jul 4 at 1:15