Properties of Convergence in Probability

Theorem

Let

• $X_n,Y_n,A_n$ : Sequences of random sample
• $X_n\stackrel{P}{\to }X$
• $Y_n\stackrel{P}{\to }Y$
• $A_n\stackrel{P}{\to }a$
• $B_n\stackrel{P}{\to }b$
• $a\in \mathbb{R}$

Then

1. $X_n+Y_n\stackrel{P}{\to }X+Y$
2. $X_n\stackrel{P}{\to }X⟹a{X}_n\stackrel{P}{\to }aX$
3. $g(X_n)\stackrel{P}{\to }g(X)$¹
4. $X_n{Y}_n\stackrel{P}{\to }XY$
5. $g(A_n)\stackrel{P}{\to }g(a)$
6. $A_n/a\stackrel{P}{\to }1$
7. $\sqrt{A_n}\stackrel{P}{\to }\sqrt{c}$
8. $A_n{B}_n\stackrel{P}{\to }ab$
9. $b\ne 0⟹A_n/B_n\stackrel{P}{\to }a/b$

Proofs

1.

$X_n+Y_n\stackrel{P}{\to }X+Y$

Let $ϵ>0$ be given.

Notice that if $|(X_n+Y_n)-(X+Y)|\ge ϵ$, then by triangle inequality,

$$ϵ\le |(X_n+Y_n|=|(X_n-X)+(Y_n-Y)|\le |X_n-X|+|Y_n-Y|$$

Therefore, the event $\{|(X_n+Y_n)-(X-Y)|\ge ϵ\}$ is contained in the event $\{|X_n-X|+|Y_n-Y|\ge ϵ\}$.

Since the probability set function is monotone with respect to set containment, then

$$P[|(X_n+Y_n)-(X+Y)\ge ϵ]\le P[|X_n-X|+|Y_n-Y|\ge ϵ]$$

If $|X_n-X|+|Y_n-Y|\ge ϵ$, then at least one of the followign must hold:

• $|X_n-X|\ge ϵ/2$, or
• $|Y_n-Y|\ge ϵ/2$, or

Therefore,

$$P[|X_n-X|+|Y_n-Y|\ge ϵ]\le P[|X_n-X|\ge ϵ/2]+P[|Y_n-Y|\ge ϵ/2]$$

By hypothesis, $X_n\stackrel{P}{\to }X$ and $Y_n\stackrel{P}{\to }Y$, so both terms on the right converge to $0$ as $n\to \infty$.

Because $P[|X_n-X|+|Y_n-Y|\ge ϵ]=0$ as $n\to \infty$, by definition of convergence in probability, we have that $X_n+Y_n\stackrel{P}{\to }X+Y$.

2.

$X_n\stackrel{P}{\to }X⟹a{X}_n\stackrel{P}{\to }aX$

Suppose $a\ne 0$. Let $ϵ>0$. Then,

$$\begin{array}{rl}P[|a{X}_n-aX|\ge ϵ]& =P[|a||X_n-X|\ge ϵ]\\ & =P[|X_n-X|\ge ϵ/|a|]\end{array}$$

Notice that the last term converges to $0$. Thus, $P[|a{X}_n-aX|\ge ϵ]$ also converges to $0$.

3.

$g(X_n)\stackrel{P}{\to }g(X)$

4.

$X_n{Y}_n\stackrel{P}{\to }XY$

By 1., 2. and 3., we have

$$\begin{array}{rl}{X}_n{Y}_n& =\frac{1}{2}{X}_n^2+\frac{1}{2}{Y}_n^2-\frac{1}{2}(X_n-Y_n{)}^2\\ & \stackrel{P}{\to }\frac{1}{2}{X}^2+\frac{1}{2}{Y}^2-\frac{1}{2}(X-Y{)}^2\end{array}$$

5.

$g(A_n)\stackrel{P}{\to }g(a)$

6.

$A_n/a\stackrel{P}{\to }1$

7.

$\sqrt{A_n}\stackrel{P}{\to }\sqrt{c}$

8.

$A_n{B}_n\stackrel{P}{\to }ab$

9.

$b\ne 0⟹A_n/B_n\stackrel{P}{\to }a/b$

Footnotes

1. Page 104 of Tucker (1967) ↩