Proofs of convergence of random variables

${\displaystyle X_{n}\ {\overset {\mathrm {as} }{\rightarrow }}\ X\quad \Rightarrow \quad X_{n}\ {\overset {p}{\rightarrow }}\ X}$

Proof: If ${\displaystyle \{X_{n}\}}$ converges to ${\displaystyle X}$ almost surely, it means that the set of points ${\displaystyle O=\{\omega \mid \lim X_{n}(\omega )\neq X(\omega )\}}$ has measure zero. Now fix ${\displaystyle \varepsilon >0}$ and consider a sequence of sets

${\displaystyle A_{n}=\bigcup _{m\geq n}\left\{\left|X_{m}-X\right|>\varepsilon \right\}}$

This sequence of sets is decreasing (${\displaystyle A_{n}\supseteq A_{n+1}\supseteq \ldots }$) towards the set

${\displaystyle A_{\infty }=\bigcap _{n\geq 1}A_{n}.}$

The probabilities of this sequence are also decreasing, so ${\displaystyle \lim \operatorname {Pr} (A_{n})=\operatorname {Pr} (A_{\infty })}$; we shall show now that this number is equal to zero. Now for any point ${\displaystyle \omega }$ outside of ${\displaystyle O}$ we have ${\displaystyle \lim X_{n}(\omega )=X(\omega )}$, which implies that ${\displaystyle \left|X_{n}(\omega )-X(\omega )\right|<\varepsilon }$ for all ${\displaystyle n\geq N}$ for some ${\displaystyle N}$. In particular, for such ${\displaystyle n}$ the point ${\displaystyle \omega }$ will not lie in ${\displaystyle A_{n}}$, and hence won't lie in ${\displaystyle A_{\infty }}$. Therefore, ${\displaystyle A_{\infty }\subseteq O}$ and so ${\displaystyle \operatorname {Pr} (A_{\infty })=0}$.

Finally, by continuity from above,

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

which by definition means that ${\displaystyle X_{n}}$ converges in probability to ${\displaystyle X}$.

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.

${\displaystyle X_{n}\ {\xrightarrow {p}}\ X\quad \Rightarrow \quad X_{n}\ {\xrightarrow {d}}\ X,}$

${\displaystyle X_{n}\ {\xrightarrow {d}}\ c\quad \Rightarrow \quad X_{n}\ {\xrightarrow {p}}\ c,}$ 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

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

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,

${\displaystyle {\begin{aligned}\lim _{n\to \infty }\operatorname {Pr} \left(\left|X_{n}-c\right|\geq \varepsilon \right)&\leq \limsup _{n\to \infty }\operatorname {Pr} \left(\left|X_{n}-c\right|\geq \varepsilon \right)\\&=\limsup _{n\to \infty }\operatorname {Pr} \left(X_{n}\in B_{\varepsilon }(c)^{c}\right)\\&\leq \operatorname {Pr} \left(c\in B_{\varepsilon }(c)^{c}\right)=0\end{aligned}}}$

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

${\displaystyle |Y_{n}-X_{n}|\ {\xrightarrow {p}}\ 0,\ \ X_{n}\ {\xrightarrow {d}}\ X\ \quad \Rightarrow \quad Y_{n}\ {\xrightarrow {d}}\ X}$

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:

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

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

${\displaystyle {\begin{aligned}\left|\operatorname {E} \left[f(Y_{n})\right]-\operatorname {E} \left[f(X_{n})\right]\right|&\leq \operatorname {E} \left[\left|f(Y_{n})-f(X_{n})\right|\right]\\&=\operatorname {E} \left[\left|f(Y_{n})-f(X_{n})\right|\mathbf {1} _{\left\{|Y_{n}-X_{n}|<\varepsilon \right\}}\right]+\operatorname {E} \left[\left|f(Y_{n})-f(X_{n})\right|\mathbf {1} _{\left\{|Y_{n}-X_{n}|\geq \varepsilon \right\}}\right]\\&\leq \operatorname {E} \left[K\left|Y_{n}-X_{n}\right|\mathbf {1} _{\left\{|Y_{n}-X_{n}|<\varepsilon \right\}}\right]+\operatorname {E} \left[2M\mathbf {1} _{\left\{|Y_{n}-X_{n}|\geq \varepsilon \right\}}\right]\\&\leq K\varepsilon \operatorname {Pr} \left(\left|Y_{n}-X_{n}\right|<\varepsilon \right)+2M\operatorname {Pr} \left(\left|Y_{n}-X_{n}\right|\geq \varepsilon \right)\\&\leq K\varepsilon +2M\operatorname {Pr} \left(\left|Y_{n}-X_{n}\right|\geq \varepsilon \right)\end{aligned}}}$

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

${\displaystyle {\begin{aligned}\left|\operatorname {E} \left[f(Y_{n})\right]-\operatorname {E} \left[f(X)\right]\right|&\leq \left|\operatorname {E} \left[f(Y_{n})\right]-\operatorname {E} \left[f(X_{n})\right]\right|+\left|\operatorname {E} \left[f(X_{n})\right]-\operatorname {E} \left[f(X)\right]\right|\\&\leq K\varepsilon +2M\operatorname {Pr} \left(|Y_{n}-X_{n}|\geq \varepsilon \right)+\left|\operatorname {E} \left[f(X_{n})\right]-\operatorname {E} \left[f(X)\right]\right|.\end{aligned}}}$

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

${\displaystyle \lim _{n\to \infty }\left|\operatorname {E} \left[f(Y_{n})\right]-\operatorname {E} \left[f(X)\right]\right|\leq K\varepsilon .}$

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.

${\displaystyle X_{n}\ {\xrightarrow {d}}\ X,\ \ Y_{n}\ {\xrightarrow {p}}\ c\ \quad \Rightarrow \quad (X_{n},Y_{n})\ {\xrightarrow {d}}\ (X,c)}$ 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:

${\displaystyle {\begin{cases}\left|(X_{n},Y_{n})-(X_{n},c)\right|\ {\xrightarrow {p}}\ 0,\\(X_{n},c)\ {\xrightarrow {d}}\ (X,c).\end{cases}}}$

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

${\displaystyle X_{n}\ {\xrightarrow {p}}\ X,\ \ Y_{n}\ {\xrightarrow {p}}\ Y\ \quad \Rightarrow \quad (X_{n},Y_{n})\ {\xrightarrow {p}}\ (X,Y)}$

Proof:

${\displaystyle {\begin{aligned}\operatorname {Pr} \left(\left|(X_{n},Y_{n})-(X,Y)\right|\geq \varepsilon \right)&\leq \operatorname {Pr} \left(|X_{n}-X|+|Y_{n}-Y|\geq \varepsilon \right)\\&\leq \operatorname {Pr} \left(|X_{n}-X|\geq \varepsilon /2\right)+\operatorname {Pr} \left(|Y_{n}-Y|\geq \varepsilon /2\right)\end{aligned}}}$

where the last step follows by the pigeonhole principle and the sub-additivity of the probability measure. 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)}.

• Convergence of random variables