For i.i.d random variables, we can always say \(X_n\overset{d}{\rightarrow}X\) since \[\begin{align}%\label{eq:union-bound} F_{X_n}(x)=F_X(x), \qquad \forall x. \end{align}\]
\(\therefore\)
\[\begin{align}%\label{eq:union-bound} \lim_{n \rightarrow \infty} F_{X_n}(x)=F_X(x), \qquad \forall x. \end{align}\]
For i.i.d random variables we cannot say \(X_n\overset{p}{\rightarrow}X\) since \(\lim_{n\to \infty}P(|X_n-X|\ge\epsilon)>0\). For example, suppose,\(X_n\) is i.i.d \(N(0,1)\), then \((X_n-X)\sim N(0,2)\), then \(|X_n-X|\) can be anything \(\ge 0\).
Another useful relationship for limiting distributions: \(A\Rightarrow B\) means \(A \subseteq B\) means \(\mathbb{P}[A] \leq \mathbb{P}[B]\)
Notes for proving the following theorem
Let \({X_n}\) be a sequence of random variable bounded in probability and let \({Y_n}\) be sequence of random variables that converges to 0 in probability then
\[X_nY_n\overset{P}{\rightarrow}0\] Proof:
Let \(\epsilon>0\), Choose \(B_{\epsilon}>0\) and integer \(N_{\epsilon}\), such that
\[n\ge N_{\epsilon}\Rightarrow P(|X_n|\le B_{\epsilon})\ge 1-\epsilon \tag{1}\] The above inequality is by the definition that \({X_n}\) is bounded by probability.
Then \[\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon) \le \overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|\le B_{\epsilon})+\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|> B_{\epsilon})\tag{2}\]
The above inequality holds is because of
\[A \subset (A\cap B)\cup (A\cap B^{c})\Rightarrow P(A)\leq P(A\cap B) +P(A\cap B^{c})\] Here, we treat \(\{|X_nY_n|\ge \epsilon\}\) as set \(A\), \(\{|X_n|\le B_{\epsilon})\}\) as set \(B\) ,\(\{|X_n|> B_{\epsilon})\}\) as \(B^c\).
From the formula \((2)\), the first part at left hand of \(\ge\), we have:
\[\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|\le B_{\epsilon})\] We can infer, \(|X_nY_n|\ge \epsilon\) and \(|X_n|\le B_{\epsilon}\) \(\Rightarrow \epsilon\le|X_nY_n|\le B_{\epsilon}|Y_n|\Rightarrow |Y_n|\ge \epsilon/B_{\epsilon}\).
Since \(A\Rightarrow B\) means \(P(A)\le P(B)\)
Therefore,
\[\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|\le B_{\epsilon})\le \overline{\lim_{n\to \infty}} P(|Y_n|\ge \epsilon/B_{\epsilon})=0\tag{3}\] \((3)\) equals to zero is because \(Y_n \overset{P}\rightarrow 0\), then by definition of converges in probability, \[\overline{\lim_{n\to \infty}} P(|Y_n-0|\ge \epsilon/B_{\epsilon})=0\], note can can treat \(\epsilon/B_{\epsilon}\) as a new \(\epsilon\).
Now, we look at the second part of formula \((2)\)
\[\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|> B_{\epsilon})\le \overline{\lim_{n\to \infty}}P(|X_n|> B_{\epsilon})=1-\overline{\lim_{n\to \infty}}P(|X_n|\le B_{\epsilon}) \tag{4}\] Here we used \(P(A\cap B)\le P(B)\) and \(P(A)=1-P(A^c)\)
By \((1)\) we have \[\overline{\lim_{n\to \infty}}P(|X_n|\le B_{\epsilon})=1-\epsilon \tag{5}\]
Put \((5)\) back to \((4)\) we get:
\[\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|> B_{\epsilon})\le 1-(1-\epsilon) =\epsilon \tag{6}\] Now, we put \((3)\) and \((6)\) back to \((2)\) we get
\[\begin{align*}\overline{\lim_{n\to \infty}}P(|X_nY_n| \ge \epsilon) &\le \overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|\le B_{\epsilon})+\overline{\lim_{n\to \infty}}P(|X_nY_n|\ge \epsilon,|X_n|> B_{\epsilon})\\&\le 0+\epsilon \end{align*}\]
By the definition of convergence in probability, we have
\[X_nY_n\overset{P}{\rightarrow}0\] then proved the Theorem.
\(\square\)