# Proving $X_n \rightarrow X$ in distribution implies $a+bX_n \rightarrow a+bX$ in distribution by definition

As stated in the title, I am trying to prove that if $$X_n \Rightarrow X$$ in distribution, then $$a+bX_n \Rightarrow a+bX$$ ( where $$a,b\in\mathbb{R}$$) in distribution using the definition as follows: $$X_n$$ converges in distribution to $$X$$ if $$F_{X_n}(x) \Rightarrow F_X(x)$$ for all $$x$$ such that $$F_X(x)$$ is continuous at $$x$$.

I have been able to prove this using characteristic functions, but now I am attempting it using only definitions.

My main idea is to use the definition of $$F_X(x)=\{\omega:X(\omega)\leq x \}$$ for $$\omega\in\Omega$$ (on $$(\Omega,\mathcal{F},P)$$, so then $$F_{X_n}(x) \rightarrow F_X(x) \implies P\{\omega: X_n(\omega)\leq x\} \rightarrow P\{\omega : X(\omega) \leq x\} \Rightarrow P\{\omega: a+bX_n(\omega) \leq a+bx\} \rightarrow P\{\omega: a+bX(\omega) \leq a+bx\}$$,

so then $$a+bX_n\rightarrow a+bX$$ in distribution. However, I am unsure of that last step; and feel like the continuity portion of the definition needs to be addressed.

It is probably worth labelling $$Y_n=a+bX_n$$ and $$Y=a+bX$$ and looking at the sign of $$b$$
• If $$b\gt 0$$ then $$F_{Y_n}(y) = F_{X_n}\left(\frac{y-a}{b}\right) \to F_{X}\left(\frac{y-a}{b}\right) = F_{Y}(y)$$, with the continuity condition carrying across naturally
• If $$b\lt 0$$ then the continuity condition becomes a little more important; essentially you are considering those parts of the distribution where $$\mathbb P\left(X=\frac{y-a}{b}\right) =0$$ and so $$\mathbb P(Y = y)=0$$ and so $$\mathbb P(Y \le y) = \mathbb P(Y \lt y) = 1-\mathbb P(Y \ge y) = 1-\mathbb P(Y \gt y)$$. For such $$y$$s, you have $$F_{Y}(y) = 1-F_{X}\left(\frac{y-a}{b}\right)$$ though not necessarily for $$X_n$$ and $$Y_n$$, and thus $$F_{Y_n}(y) = 1 - F_{X_n}\left(\frac{y-a}{b}\right) + \mathbb P\left(X_n = \frac{y-a}{b}\right) \to 1 - F_{X}\left(\frac{y-a}{b}\right) + 0 = F_{Y}(y)$$
• If $$b=0$$, then $$Y_n=Y=a$$ surely, so trivially you have convergence of all types. In particular $$F_{Y_n}(y)= F_Y(y)$$ as this is $$0$$ when $$y \lt a$$ and $$1$$ when $$y \ge a$$