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.


1 Answer 1


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 $y \ge a$


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