Suppose that random variables $\{X_n, n\geq 1\}$ and $\{Y_n, n\geq 1\}$ are all defined on a common probability space and that $X_n\Rightarrow X$ and $Y_n\Rightarrow c$, with $c$ a constant (here $\Rightarrow$ means convergence in distribution). Then $X_n + Y_n \Rightarrow X + c$.
Exercise: Prove the above statement. You may use the following outline:
a) An integer $n_\epsilon$ exists, such that $P(X_n\leq x)$ and $P(X\leq x)$ differ less than $\epsilon$ and also $P(\left|Y_n-c\right|\leq \epsilon)\geq 1-\epsilon$.
b) Note the inclusion of events $$\{X_n\leq x\}\cap \{Y_n\leq c + \epsilon\}\subset\{X_n + Y_n\leq x + c\}\subset \{X_n\leq x + \epsilon\}\cup \{Y_n\leq c - \epsilon\}$$
c) Use the elementary inequalities $P(A\cap B)\geq P(A) - P(B^c)$ and $P(A\cup B)\leq P(A) + P(B)$ to derive a 'squeeze' on $P(X_n + Y_n\leq x +c)$.
d) Letting $\epsilon \to 0$ then finishes it.
What I've tried:
There exists an integer such that $P(X_n\leq x)$ and $P(X\leq x)$ differ less than $\epsilon$. Yes, this is true, as $X_n\Rightarrow X$, so we know that $F_n^X(x)\to F^X(x)$ in probability. Furthermore, $F_n^Y\to c$ in probability, so it's also true that $P(\left|Y_n -c\right|\leq \epsilon)\geq 1-\epsilon$ (because this is equivalent to $P(\left|Y_n - c\right|>\epsilon)\to 0$). Using b), I think I should look at $P(P(X_n\leq x) - P(X\leq x))\cdot P(\left|Y_n - c\right|\leq \epsilon) \leq P(\left|F_n^X(x)-F^X(x)\right| + \left|Y_n - c\right|\leq \epsilon)$. Since I don't know how to break up this function I don't really know what to do next. I think that one of the reasons I don't know how to proceed is that I don't know how to handle the absolute value signs.
Question: Could anyone give me a hint so that I can proceed in the right direction? Is my thinking up and until this part correct?
Thanks!
