0
$\begingroup$

If $X_n \to_{d} X$ and $Y_n\to_p 0$ then $X_n+Y_n\to_d X,$ where $X_n\to_d X$ means $X_n$ converges to $X$ in distribution and $Y_n\to_p 0$ means that $Y_n$ converges to $0$ in probability.

In order to show this let $Z_n= X_n+Y_n$ then $$F_{Z_n}(t) = P(X_n+Y_n\leq t)$$ $$= P(X_n+Y_n\leq t\cap |Y_n|\leq \epsilon) + P(X_n+Y_n\leq t\cap |Y_n|> \epsilon)$$ $$ \leq P(X_n\leq t+\epsilon) + P(|Y_n|>\epsilon) \to F_{X}(t+\epsilon).$$

This shows that $\limsup F_{Z_n}(t)\leq F_X(t+\epsilon).$ I could try something similar for finding a lower bound for $\liminf F_{Z_n}(t)$, though I am not sure whether I am on the right path. Any hints will be much appreciated.

$\endgroup$
2
$\begingroup$

You showed $$P(X_n + Y_n \le t) \le P(X_n \le t + \epsilon) + P(|Y_n| > \epsilon).$$ A similar argument yields $$P(X_n \le t - \epsilon) \le P(X_n + Y_n \le t) + P(|Y_n| > \epsilon).$$

The first inequality implies $$\limsup_{n \to \infty} F_{Z_n}(t) \le F_X(t+\epsilon)$$ for any $\epsilon > 0$, while the second implies $$\liminf_{n \to \infty} F_{Z_n}(t) \ge F_X(t-\epsilon)$$ for any $\epsilon > 0$.

Taking $\epsilon \to 0$ yields $F_{Z_n}(t) \to F_X(t)$ for any continuity point $t$ of $F_X$; this is precisely the definition of convergence in distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.