Let $(X_n)_{n\in\mathbb{N}}$, $(X'_n)_{n\in\mathbb{N}}$, $Y$ real random variables such that
$$ X_n\leq Y\leq X'_n \quad\forall n\in\mathbb{N}\ .$$
Suppose that the sequences $X_n$ e $X'_n$ converge to the same limit in distribution (i.e. in law, weakly), that is
$$ X_n\xrightarrow[n\to\infty]{d} X \quad\text{e}\quad X'_n\xrightarrow[n\to\infty]{d} X\ .$$
Then is it true that
$$ Y\ \overset{\,d}=\ X\ ?$$
Probability Theory – Squeeze Theorem for Convergence in Distribution
convergence-divergenceprobability theory
Related Solutions
Weak convergence is equivalent to cdf convergence at all continuity points of the limiting cdf. In this case it is easier to look at the complementary problem: \begin{align*} P(Y^{(n)} > y) &= P\left( \inf_{1 \leq i \leq n}{X_i} > \frac{y}{n} \right)\\ &= P\left( \text{Every $X_i$ is greater than } \frac{y}{n} \right) \\ &= P\left(X_1 > \frac{y}{n} \right)^n \quad \text{by iid-ness} \\ &= \left( 1 - y/n\right)^{n} \quad \text{for $n$ large enough} \end{align*} Which converges to $e^{-y}$ which is exactly what we want.
Assume that (where I conveniently replaced Y with Z) $$\begin{split}X_n-Y_n&\overset p {\rightarrow} 0\\ Y_n&\overset p {\rightarrow} Z\end{split}$$
Fix $\epsilon.$ Notice that $|X_n-Y_n|\le\frac \epsilon 2$ and $|Y_n-Z|\le\frac \epsilon 2$ implies that $|X_n-Z|\le\epsilon$, by the triangle inequality. Reversing the logic, this means that $|X_n-Z|>\epsilon$ implies that $|X_n-Y_n|>\frac \epsilon 2$ (inclusive) or $|Y_n-Z|>\frac \epsilon 2$. In particular, if an event implies that at least one of two other events has occurred, this means that $A\subset B\cup C$, i.e. $P(A)\le P(B\cup C)$.
Now
$$\begin{split}P(|X_n-Z|>\epsilon)&\le P(|X_n-Y_n|>\frac \epsilon 2\cup|Y_n-Z|>\frac \epsilon 2)\text { what we just said}\\ &\le P(|X_n-Y_n|>\frac \epsilon 2)+P(|Y_n-Z|> \frac \epsilon 2)\text { definition of union} \end{split}$$
Take the limit to get $lim_{n\rightarrow\infty}P(|X_n-Z|>\epsilon)\le0$. Since probabilities are positive, it is 0.
Best Answer
Limit in distribution $Z_n\to Z$ for real valued random variables is equivalent to the fact that $\mathbb P(Z_n\leqslant x)\to\mathbb P(Z\leqslant x)$ for every $x$ at which the function $F_Z:z\mapsto\mathbb P(Z\leqslant z)$ is continuous.
In the present case, for every $x$, $\mathbb P(X'_n\leqslant x)\leqslant F_Y(x)\leqslant\mathbb P(X_n\leqslant x)$, and both $\mathbb P(X'_n\leqslant x)$ and $\mathbb P(X_n\leqslant x)$ converge to $F_X(x)$ when $n\to\infty$, for every $x$ at which $F_X$ is continuous. Hence $F_Y=F_X$ at every point of continuity of $F_X$. Both $F_X$ and $F_Y$ are continuous to the right hence $F_X=F_Y$ everywhere, that is, $Y\overset{\,d}=X$.