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.


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.


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