I'm having some trouble verifying the following line from Extremes and Related Properties of Random Sequences and Processes by Leadbetter:

If (1.2.1)' and (1.2.3)' hold, then obviously so does (1.2.2)' enter image description here

With $F'_n(x)$ converging weakly to $G(x)$ and $\alpha_n' \to a$ and $\beta'_n \to b$, it's clear that $F'_n(ax+b) \to G(ax+b)$ weakly. However, why should $F_n'(\alpha'_n x+ \beta'_n)$ converge weakly to $G(ax+b)$? Since $F_n'$ may not be continuous, it seems we cannot guarantee the convergence even though $\alpha'_n x + \beta'_n \to ax+b$.

  • 2
    $\begingroup$ Phil, could you cite the source you took the snapshot of? It's appreciable if you mention the source alongside the image/snap. $\endgroup$ Commented Jan 13, 2023 at 7:41
  • $\begingroup$ @User1865345 sure it's added now $\endgroup$
    – Phil
    Commented Jan 13, 2023 at 15:14
  • $\begingroup$ Thanks for the citation. $\endgroup$ Commented Jan 13, 2023 at 15:15
  • 1
    $\begingroup$ See math.stackexchange.com/questions/285133/… $\endgroup$ Commented Jan 13, 2023 at 16:46

1 Answer 1


If $X_n \to X$ in distribution and $b_n \to b$, then we may pick $Y_n = X_n$ in distribution for $1 \leq n \leq \infty$ with $Y_n \to Y$ almost surely.

Clearly ,$Y_n + b_n \to Y + b$ almost surely, so $X_n + b_n \to X+b$ in distribution.

We can prove the same thing for $a_n X_n \to aX$, which means $\frac{X_n - \alpha '}{\beta_n'} \to \frac{X-a}{b}$ in distribution.


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