convergence modes

Question

There are many results on which convergence mode implies the other. For example almost sure convergence implies convergence in probability, implies convergence in Distribution. Now I make some thoughts on this and came up with following question, which I don't know they are true or not.

Let $X_{n},Y_{n}$ be random variables with $\mathbb{P}_{X_{n}} = \mathbb{P}_{Y_{n}}$ for all n. Now I want to prove or disprove the following Claims:

1. If $X_{n}$ $\rightarrow$ $0$ in probability does this implies $Y_{n} \rightarrow 0$ in probability?
2. If $X_{n} \rightarrow 0$ almost surely implies $Y_{n} \rightarrow 0$ almost surely?
3. If $X_{n} \rightarrow 0$ in $\mathcal{L}^{1}$ implies $Y_{n} \rightarrow 0$ in $\mathcal{L}^{1}$?

What are you thinking?

Answer 1

1. Yes, that's true. Since $X_n$ and $Y_n$ have, by assumption, the same distribution, we know that $$\mathbb{P}(X_n \in B) = \mathbb{P}(Y_n \in B)$$ for any measurable set $B$. In particular, we can choose $B := (-\infty,-\epsilon) \cup (\epsilon,\infty)$ for fixed $\epsilon>0$ to find that $$\mathbb{P}(|X_n|>\epsilon) = \mathbb{P}(|Y_n|>\epsilon).$$ If $X_n \to 0$ in probability, then the left-hand side converges to $0$ as $n \to \infty$, and so does the right-hand side. Hence, $Y_n \to 0$ in probability.

2. No, this is, in general, wrong. Consider the sequence of random variables $(Y_n)_{n \in \mathbb{N}}$ on the probability space $((0,1],\mathcal{B}((0,1]))$ (endowed with Lebesgue measure $\lambda$) defined by $$\begin{align*} Y_1(\omega) &:= 1_{\big(\frac{1}{2},1 \big]}(\omega) \\ Y_2(\omega) &:= 1_{\big(0, \frac{1}{2}\big]}(\omega) \\ Y_3(\omega) &:= 1_{\big(\frac{3}{4},1 \big]}(\omega) \\ Y_4(\omega) &:= 1_{\big(\frac{1}{2},\frac{3}{4} \big]}(\omega)\\ &\vdots \end{align*}$$ Then $Y_n$ does not convergence almost surely to $0$ (since for any $\omega \in (0,1]$ and $N \in \mathbb{N}$ there exist $m,n \geq N$ such that $Y_n(\omega)=1$ and $Y_m(\omega)=0$). On the other hand, we can define $$\begin{align*} X_1(\omega) &:= 1_{\big(0,\frac{1}{2} \big]}(\omega) \\ X_2(\omega) &:= 1_{\big(0, \frac{1}{2}\big]}(\omega) \\ X_3(\omega) &:= 1_{\big(0,\frac{1}{4} \big]}(\omega) \\ X_4(\omega) &:= 1_{\big(0,\frac{1}{4} \big]}(\omega)\\ &\vdots \end{align*}$$ For each fixed $n \in \mathbb{N}$, $X_n$ and $Y_n$ have the same distribution, and it follows easily that $X_n \to 0$ almost surely.

3. Yes, that's true. If we denote by $\mathbb{P}_{X_n}$ the distribution of $X_n$, then $$\mathbb{E}(f(X_n)) = \int_{\mathbb{R}} f(y) \, d\mathbb{P}_{X_n}(y)$$ holds for any measurable function $f: \mathbb{R} \to \mathbb{R}$ such that $f(X_n) \in L^1(\mathbb{P})$. Since $X_n$ and $Y_n$ have the same distribution, this implies $$\mathbb{E}(f(X_n)) = \mathbb{E}(f(Y_n))$$ for any such $f$. If $X_n \to 0$ in $L^1$, then we can choose $f(x)=|x|$ to conclude that $$\mathbb{E}(|Y_n|) = \mathbb{E}(|X_n|) \to 0$$ i.e. $Y_n \to 0$ in $L^1$.