1
$\begingroup$

I need to show that if $X_n \rightarrow X$ and $X_n \rightarrow Y$, then $X\overset{\text{a.s.}}{=}Y$ for convergence in probability, convergence almost surely, as well as for convergence in mean and quadratic mean ($\mathcal L^1$ and $\mathcal L^2$ convergence).

Convergence in Probability:

For any $\epsilon>0$ and for any $n\in\mathbb N$ we have

$$\begin{align} \mathbb P(|X-Y|\geq\epsilon) &\leq\mathbb P(|X-X_n|+|X_n-Y|\geq\epsilon)\\\\ &\leq\mathbb P\left((|X-X_n|\geq\epsilon/2)\cup(|X_n-Y|\geq\epsilon/2)\right)\\\\ &\leq\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2) \end{align}$$

so that

$$\mathbb P(|X-Y|\geq\epsilon)\leq\lim_{n\rightarrow\infty}\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2)=0$$

Since $$\{|X-Y|>0\}=\underbrace{\bigcup_{n=1}^\infty \underbrace{\left\{|X-Y|>\frac{1}{n}\right\}}_{=\emptyset}}_{=\emptyset}=\emptyset$$

we have that $\mathbb P\{|X-Y|>0\}=0$ and so $\mathbb P(X\ne Y)=0$. Hence $\mathbb P(X= Y)=1$ which means that $X\overset{\text{a.s.}}{=}Y$.

Convergence Almost Surely:

Since almost sure convergence implies convergence in probability, the result follows immediately from the last part. However, I'd like to show this without making use of that result. Since $X_n$ converges almost surely to both $X$ and $Y$ then $\mathbb P(\lim_{n\rightarrow\infty}X_n=X)=1$ and $\mathbb P(\lim_{n\rightarrow\infty}X_n=Y)=1$. From here it seems obvious to me that $X\overset{\text{a.s.}}{=}Y$ but I'm not sure how to show this formally.

Convergence in Mean:

$$\begin{align} \mathbb E(|X-Y|) &\leq\mathbb E\left(|X-X_n|+|X_n-Y|\right)\\\\ &=\mathbb E\left(|X-X_n|)+\mathbb E(|X_n-Y|\right) \end{align}$$

so

$$\mathbb E(|X-Y|)\leq\lim_{n\rightarrow\infty}\mathbb E(|X-X_n|)+\mathbb E(|X_n-Y|)=0$$

so $X\overset{\text{a.s.}}{=}Y$

Convergence in Quadratic Mean:

I tried continuing with the same logic but it's not the case that

$$ \mathbb E(|X-Y|^2)\leq\mathbb E\left(|X-X_n|^2+|X_n-Y|^2\right)$$

so I'm not sure how to proceed.

Is my reasoning correct for the first and third? How can I proceed with the other two?

$\endgroup$
1
  • 1
    $\begingroup$ Yes, first and third seems correct. For the second, note that $X_n \to X, X_n \to Y$ almost surely, means you have a set $A$ of full measure, such that for $\omega \in A$ you have $X_n(\omega) \to X(\omega)$ and $X_n(\omega) \to Y(\omega)$ as a sequence of numbers, so easily $|X(\omega)-Y(\omega)| \le |X(\omega)-X_n(\omega)| + |Y(\omega) - X_n(\omega)| \to 0$ which means that you have $X=Y$ on $A$, hence almosts surely. For the fourth: Minkowsky Inequality and goes the same as third one $\endgroup$
    – Presage
    Oct 4, 2020 at 19:58

2 Answers 2

0
$\begingroup$

I think I can answer my own question now for convergence in quadratic mean. By Minkowsky,

$$\begin{align*} \left(\mathbb E\left(|X-Y|^2\right)\right)^{1/2} &=\left(\mathbb E\left(|(X-X_n)+(X_n-Y)|^2\right)\right)^{1/2}\\\\ &\leq \left(\mathbb E\left(|X-X_n|^2\right)\right)^{1/2}+\left(\mathbb E\left(|X_n-Y|^2\right)\right)^{1/2} \end{align*}$$

so

$$\begin{align*} \left(\mathbb E\left(|X-Y|^2\right)\right)^{1/2} &\leq\lim_{n\rightarrow\infty} \left(\mathbb E\left(|X-X_n|^2\right)\right)^{1/2}+\left(\mathbb E\left(|X_n-Y|^2\right)\right)^{1/2}\\\\ &=0 \end{align*}$$

so $X\overset{\text{a.s.}}{=}Y$

$\endgroup$
0
$\begingroup$

It follows from convergence a.s. and from convergence in $L_p$ that there is convergence in probability. So it was enough to condsider convergence in probability.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .