Proofs of convergence of random variables

This article is supplemental for “Convergence of random variables” and provides proofs for selected results.

Several results will be established using the portmanteau lemma: A sequence {X_n} converges in distribution to X if and only if any of the following conditions are met:

1. E[f(X_n)] → E[f(X)] for all bounded, continuous functions f;
2. E[f(X_n)] → E[f(X)] for all bounded, Lipschitz functions f;
3. limsup{Pr(X_n ∈ C)} ≤ Pr(X ∈ C) for all closed sets C;

Convergence almost surely implies convergence in probability

$X_n\ \xrightarrow{as}\ X \quad\Rightarrow\quad X_n\ \xrightarrow{p}\ X$

Proof: If {X_n} converges to X almost surely, it means that the set of points {ω: lim X_n(ω) ≠ X(ω)} has measure zero; denote this set O. Now fix ε > 0 and consider a sequence of sets

$A_n = \bigcup_{m\geq n} \left \{ \left |X_m-X \right |>\varepsilon \right\}$

This sequence of sets is decreasing: A_n ⊇ A_n+1 ⊇ ..., and it decreases towards the set

$A_{\infty} = \bigcap_{n \geq 1} A_n.$

For this decreasing sequence of events, their probabilities are also a decreasing sequence, and it decreases towards the Pr(A_∞); we shall show now that this number is equal to zero. Now any point ω in the complement of O is such that lim X_n(ω) = X(ω), which implies that |X_n(ω) − X(ω)| < ε for all n greater than a certain number N. Therefore, for all n ≥ N the point ω will not belong to the set A_n, and consequently it will not belong to A_∞. This means that A_∞ is disjoint with O, or equivalently, A_∞ is a subset of O and therefore Pr(A_∞) = 0.

Finally, consider

$\operatorname{Pr}\left(|X_n-X|>\varepsilon\right) \leq \operatorname{Pr}(A_n) \ \underset{n\to\infty}{\rightarrow} 0,$

which by definition means that X_n converges in probability to X.

Convergence in probability does not imply almost sure convergence in the discrete case

If X_n are independent random variables assuming value one with probability 1/n and zero otherwise, then X_n converges to zero in probability but not almost surely. This can be verified using the Borel–Cantelli lemmas.

Convergence in probability implies convergence in distribution

$X_n\ \xrightarrow{p}\ X \quad\Rightarrow\quad X_n\ \xrightarrow{d}\ X,$

Proof for the case of scalar random variables

Lemma. Let X, Y be random variables, a a real number and ε > 0. Then

$\operatorname{Pr}(Y \leq a) \leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(|Y - X| > \varepsilon).$

(or $\{Y \leq a\} \subset \{X\leq a+\varepsilon\}\cup \{|Y - X| > \varepsilon\}.$)

Proof of lemma:

$\begin{align} \operatorname{Pr}(Y\leq a) &= \operatorname{Pr}(Y\leq a,\ X\leq a+\varepsilon) + \operatorname{Pr}(Y\leq a,\ X>a+\varepsilon) \\ &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(Y-X\leq a-X,\ a-X<-\varepsilon) \\ &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(Y-X<-\varepsilon) \\ &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(Y-X<-\varepsilon) + \operatorname{Pr}(Y-X>\varepsilon)\\ &= \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(|Y-X|>\varepsilon) \end{align}$

Proof of the theorem: Recall that in order to prove convergence in distribution, one must show that the sequence of cumulative distribution functions converges to the F_X at every point where F_X is continuous. Let a be such a point. For every ε > 0, due to the preceding lemma, we have:

$\begin{align} \operatorname{Pr}(X_n\leq a) &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(|X_n-X|>\varepsilon) \\ \operatorname{Pr}(X\leq a-\varepsilon)&\leq \operatorname{Pr}(X_n\leq a) + \operatorname{Pr}(|X_n-X|>\varepsilon) \end{align}$

So, we have

$\operatorname{Pr}(X\leq a-\varepsilon) - \operatorname{Pr} \left (\left |X_n-X \right |>\varepsilon \right ) \leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr} \left (\left |X_n-X \right |>\varepsilon \right ).$

Taking the limit as n → ∞, we obtain:

$F_X(a-\varepsilon) \leq F_X(a+\varepsilon),$

where F_X(a) = Pr(X ≤ a) is the cumulative distribution function of X. This function is continuous at a by assumption, and therefore both F_X(a−ε) and F_X(a+ε) converge to F_X(a) as ε → 0⁺. Taking this limit, we obtain

$\lim_{n\to\infty} \operatorname{Pr}(X_n \leq a) = \operatorname{Pr}(X \leq a),$

which means that {X_n} converges to X in distribution.

Proof for the generic case

We see that |X_n − X| converges in probability to zero, and also X converges to X in distribution trivially. Applying the property proved later on this page we conclude that X_n converges to X in distribution.

Convergence in distribution to a constant implies convergence in probability

$X_n\ \xrightarrow{d}\ c \quad\Rightarrow\quad X_n\ \xrightarrow{p}\ c,$ provided c is a constant.

Proof: Fix ε > 0. Let B_ε(c) be the open ball of radius ε around point c, and B_ε^c(c)its complement. Then

$\operatorname{Pr}\left(|X_n-c|\geq\varepsilon\right) = \operatorname{Pr}\left(X_n\in B_\varepsilon^c(c)\right).$

By the portmanteau lemma (part C), if X_n converges in distribution to c, then the limsup of the latter probability must be less than or equal to Pr(c ∈ B_ε^c(c)),which is obviously equal to zero. Therefore

$\begin{align} \lim_{n\to\infty}\operatorname{Pr}\left( \left |X_n-c \right |\geq\varepsilon\right) &\leq \limsup_{n\to\infty}\operatorname{Pr}\left( \left |X_n-c \right | \geq \varepsilon \right) \\ &= \limsup_{n\to\infty}\operatorname{Pr}\left(X_n\in B_\varepsilon^c(c)\right) \\ &\leq \operatorname{Pr}\left(c\in B_\varepsilon^c(c)\right) = 0 \end{align}$

which by definition means that X_n converges to c in probability.

Convergence in probability to a sequence converging in distribution implies convergence to the same distribution

$|Y_n-X_n|\ \xrightarrow{p}\ 0,\ \ X_n\ \xrightarrow{d}\ X\ \quad\Rightarrow\quad Y_n\ \xrightarrow{d}\ X$

Proof: We will prove this theorem using the portmanteau lemma, part B. As required in that lemma, consider any bounded function f (i.e. |f(x)| ≤ M) which is also Lipschitz:

$\exists K >0, \forall x,y: \quad |f(x)-f(y)|\leq K|x-y|.$

Take some ε > 0 and majorize the expression |E[f(Y_n)] − E[f(X_n)]| as

$\begin{align} \left|\operatorname{E}\left[f(Y_n)\right] - \operatorname{E}\left [f(X_n) \right] \right| &\leq \operatorname{E} \left [\left |f(Y_n) - f(X_n) \right | \right ]\\ &= \operatorname{E}\left[ \left |f(Y_n) - f(X_n) \right |\mathbf{1}_{\left \{|Y_n-X_n|<\varepsilon \right \}} \right] + \operatorname{E}\left[ \left |f(Y_n) - f(X_n) \right |\mathbf{1}_{\left \{|Y_n-X_n|\geq\varepsilon \right \}} \right] \\ &\leq \operatorname{E}\left[K \left |Y_n - X_n \right |\mathbf{1}_{\left \{|Y_n-X_n|<\varepsilon \right \}}\right] + \operatorname{E}\left[2M\mathbf{1}_{\left \{|Y_n-X_n|\geq\varepsilon \right \}}\right] \\ &\leq K \varepsilon \operatorname{Pr} \left (\left |Y_n-X_n \right |<\varepsilon\right) + 2M \operatorname{Pr} \left( \left |Y_n-X_n \right |\geq\varepsilon\right )\\ &\leq K \varepsilon + 2M \operatorname{Pr} \left (\left |Y_n-X_n \right |\geq\varepsilon \right ) \end{align}$

(here 1_{...} denotes the indicator function; the expectation of the indicator function is equal to the probability of corresponding event). Therefore

$\begin{align} \left |\operatorname{E}\left [f(Y_n)\right ] - \operatorname{E}\left [f(X) \right ]\right | &\leq \left|\operatorname{E}\left[ f(Y_n) \right ]-\operatorname{E} \left [f(X_n) \right ] \right| + \left|\operatorname{E}\left [f(X_n) \right ]-\operatorname{E}\left [f(X) \right] \right| \\ &\leq K\varepsilon + 2M \operatorname{Pr}\left (|Y_n-X_n|\geq\varepsilon\right )+ \left |\operatorname{E}\left[ f(X_n) \right]-\operatorname{E} \left [f(X) \right ]\right|. \end{align}$

If we take the limit in this expression as n → ∞, the second term will go to zero since {Y_n−X_n} converges to zero in probability; and the third term will also converge to zero, by the portmanteau lemma and the fact that X_n converges to X in distribution. Thus

$\lim_{n\to\infty} \left|\operatorname{E}\left [f(Y_n) \right] - \operatorname{E}\left [f(X) \right ] \right| \leq K\varepsilon.$

Since ε was arbitrary, we conclude that the limit must in fact be equal to zero, and therefore E[f(Y_n)] → E[f(X)], which again by the portmanteau lemma implies that {Y_n} converges to X in distribution. QED.

Convergence of one sequence in distribution and another to a constant implies joint convergence in distribution

$X_n\ \xrightarrow{d}\ X,\ \ Y_n\ \xrightarrow{d}\ c\ \quad\Rightarrow\quad (X_n,Y_n)\ \xrightarrow{d}\ (X,c)$ provided c is a constant.

Proof: We will prove this statement using the portmanteau lemma, part A.

First we want to show that (X_n, c) converges in distribution to (X, c). By the portmanteau lemma this will be true if we can show that E[f(X_n, c)] → E[f(X, c)] for any bounded continuous function f(x, y). So let f be such arbitrary bounded continuous function. Now consider the function of a single variable g(x) := f(x, c). This will obviously be also bounded and continuous, and therefore by the portmanteau lemma for sequence {X_n} converging in distribution to X, we will have that E[g(X_n)] → E[g(X)]. However the latter expression is equivalent to “E[f(X_n, c)] → E[f(X, c)]”, and therefore we now know that (X_n, c) converges in distribution to (X, c).

Secondly, consider |(X_n, Y_n) − (X_n, c)| = |Y_n − c|. This expression converges in probability to zero because Y_n converges in probability to c. Thus we have demonstrated two facts:

$\begin{cases} \left| (X_n, Y_n) - (X_n,c) \right|\ \xrightarrow{p}\ 0, \\ (X_n,c)\ \xrightarrow{d}\ (X,c). \end{cases}$

By the property proved earlier, these two facts imply that (X_n, Y_n) converge in distribution to (X, c).

Convergence of two sequences in probability implies joint convergence in probability

$X_n\ \xrightarrow{p}\ X,\ \ Y_n\ \xrightarrow{p}\ Y\ \quad\Rightarrow\quad (X_n,Y_n)\ \xrightarrow{p}\ (X,Y)$

Proof:

$\begin{align} \operatorname{Pr}\left(\left|(X_n,Y_n)-(X,Y)\right|\geq\varepsilon\right) &\leq \operatorname{Pr}\left(|X_n-X| + |Y_n-Y|\geq\varepsilon\right) \\ &\leq\operatorname{Pr}\left(|X_n-X|\geq\tfrac{\varepsilon}{2}\right) + \operatorname{Pr}\left(|Y_n-Y|\geq\tfrac{\varepsilon}{2}\right) \end{align}$

Each of the probabilities on the right-hand side converge to zero as n → ∞ by definition of the convergence of {X_n} and {Y_n} in probability to X and Y respectively. Taking the limit we conclude that the left-hand side also converges to zero, and therefore the sequence {(X_n, Y_n)} converges in probability to {(X, Y)}.

See also

• Convergence of random variables