I'm trying solving this problem of probability, but but after a certain moment I was unable to progress.
In this exercise, I have two sequences $\{X_n\}$ and $\{Y_n\}$ of random variables such that $\dfrac{Y_n-E(Y_n)}{\sqrt{\mathrm{Var}(Y_n)}}\to \mathcal{N}(0,1)$ in distribution and:
$$ \lim_{n\to\infty} \dfrac{E[(Y_n-X_n)^2]}{\mathrm{Var}(Y_n)}=0$$
And he asks to show that $\dfrac{X_n-E(X_n)}{\sqrt{\mathrm{Var}(X_n)}}\to \mathcal{N}(0,1)$ in distribution.
I already showed that:
$$ \lim_{n\to \infty} \dfrac{E(Y_n)-E(X_n)}{\sqrt{\mathrm{Var}(Y_n)}}=0$$
and that $\dfrac{Y_n-X_n}{\sqrt{\mathrm{Var}(Y_n)}}\to 0$ in probability. If I can show that:
$$
\lim_{n\to\infty} \dfrac{\mathrm{Var}(X_n)}{\mathrm{Var(Y_n)}}= 1,
$$
then I can complete the result. Can someone help me show this latest convergence? My teacher gave a tip to use the triangular inequality, but even so I was not able to show that this limit is equal to 1.
Best Answer
We can write $$\left|X_n - EX_n\right| \le \left|X_n - Y_n\right| + \left|Y_n - EY_n\right| + \left|EX_n - EY_n\right|.$$ Now, square both sides, and then take expectations. For the other side, swap the role of $X_n$ and $Y_n$ in the first inequality and follow the same idea.
More hint: if $E(A_n^2) \to 0$ and $E(B_n^2)$ is bounded sequence, then an application of Cauchy-Schwarz inequality yields that $E|A_n B_n| \to 0.$