Theorem 1 (Gnedenko and Kolmogorov (1954), pp. 40-42)

If the sequence $\{ F_n (x)\}$ of frequency distribution functions converges weakly as $n \to \infty$ to a proper distribution function F(x), then, for any choice of the constants a_n > 0 and b_n, the sequence $\{ F_n (a_n x + b_n)\}$ can converge to a proper distribution only if this is of the same type as F(x).

Proof: Suppose that as $n \to \infty$, $F_n (x) \Rightarrow F(x)$ and $F_n (a_nx + b_n ) \Rightarrow G(x)$, and that both F and G are proper. We wish to prove that there exist a > 0 and b such that

G(x) = F(ax+b).

Choose a sequence of real positive integers $n_1 < n_2 < \ldots < n_k < \ldots$ such that the limits $\lim_{k \to \infty} a_{n_k} = a$ and $\lim_{k \to \infty} b_{n_k} = b$ $(0 \leq a \leq + \infty, - \infty \leq b \leq + \infty)$ exist. For simplicity of notation in the following discussion, we will renumber the sequences

$$a_{n_k} \equiv a_k \quad \text{and} \quad b_{n_k} \equiv b_k \quad (\text{for } k = 1,2,3, \ldots ).$$

Then, the limits become $\lim_{k \to \infty}ak = a$ and $\lim_{k \to \infty} b_k = b.$

We now proceed to prove that $0 < a < + \infty$. Suppose that $a = +\infty$. Let u be the least upper bound of the numbers x for which

$$\begin{array}[t] {c}\overline{\lim} \\ \scriptstyle k \to \infty\end{array} (a_k x + b_k) < + \infty.$$

Choose x and v such that v < x < u. Now

(a_k v + b_k) = (a_k v - a_k x) + (a_k x + b_k)

or

$$\begin{array}[t] {c} \overline{\lim} \\ \scriptstyle k\to\i... ...ine{\lim} \\ \scriptstyle k \to\infty\end{array} (a_k x + b_k),$$

since $a_k \to \infty$ by assumption and v - x < 0; $\overline{\lim} _{k \to \infty} (v-x)a_k = -\infty$. The other limit, $\overline{\lim} _{k \to \infty} (a_kx+b_k) < \infty$, is finite or $-\infty$ since x < u. Thus, $\overline{\lim} _{k \to \infty} (a_k v + b_k) = -\infty$ for all v < u, and so G(v) = 0 for v < u.

If v > u, by definition of u,

$$\begin{array}[t] {c} \overline{\lim} \\ \scriptstyle k\to\infty\end{array} (a_k v + b_k) = + \infty.$$

Hence, G(v) = 1 for v > u. If G(v) < 1 and

$$\begin{array}[t] {c} \underline{\lim} \\ \scriptstyle k \to... ...ine{\lim} \\ \scriptstyle k\to\infty\end{array} (a_k v + b_k),$$

then F_k (a_k v + b_k) would have two limits as $k \to\infty$, and it would be false that $F_k (a_k x + b_k) \Rightarrow G(x)$.

In summary, if we assume that $a = \infty$, then G(v) = 0 for v < u, and G(v) = 1 for $v \geq u$. But then G(x) is improper, which contradicts our hypothesis that F_n (a_nx + b_n) converges to a proper distribution.

It is immediately apparent that b is finite, for if $0 \leq a < \infty$ and $b = \infty$, then $\lim_{k \to \infty} (a_k x + b_k) = +\infty$ and G(x) = 1 for all x. Similarly, if $0 \leq a < \infty$ and $b = -\infty$, then $\lim_{k \to\infty} (a_kx + b_k) = -\infty$ and G(x) = 0. In either case, G(x) is improper, with the discontinuity at $-\infty$ and at $+\infty$, respectively. Hence, b is finite.

Finally, we consider the case where a = 0. Since $\lim_{k\to\infty} b_k = b$, given any $\epsilon \gt 0$, there exists a K₁ such that for k > K₁, $\vert b_k - b\vert < \epsilon/2$ or $b - (\epsilon/2) < b_k < b + (\epsilon/2)$. Then, since a = 0 and $\lim_{k\to\infty} a_k =a$for any finite x, there exists a K₂ such that if k > K₂, then $\vert a_kx\vert < \epsilon/2$. Thus, $-\epsilon/2 < a_k x < \epsilon/2$.Combining the inequalities,

$$b-\epsilon = b - \frac{\epsilon}2 - \frac{\epsilon}2 < a_k x + b_k < b + \frac{\epsilon}2 + \frac{\epsilon}2 = b + \epsilon.$$

Finally, since $\epsilon \gt 0$ is arbitrary, we may choose $\epsilon$so that $b + \epsilon$ and $b - \epsilon$ are continuity points of F(x). Now, for any finite k,

$$F_k (b-\epsilon) \leq F_k (a_k x + b_k) \leq F_k (b+\epsilon).$$

Taking limits as $k \to\infty$,

$$F(b-\epsilon) \leq G(x) \leq F (b+\epsilon).$$

Since x may assume any value,

$$1 = \lim_{x \to\infty} G(x) \leq F(b+\epsilon),$$

$$0 = \lim_{x \to - \infty} G(x) \geq F(b-\epsilon).$$

Then, since $\epsilon$ is arbitrary,

$$F(x) = 1 \text{ for } x \geq b,$$

$$F(x) = 0 \text{ for } x < b.$$

But, this contradicts the hypothesis that F(x) is proper. Hence, it must be that $a \neq 0$.

We have now demonstrated that under the conditions of the hypothesis, the limits of the subsequences $\{ a_k\}$ and $\{ b_k\}$ are such that $0 < a < \infty$ and b is finite.

Now we proceed to prove that G(x)= F(ax+b). Choose x so that F(x) is continuous at the point ax+b and G(x) is continuous at the point x. Then,

 

$$\lim_{k\to \infty} F_k (a_k x + b_k) = G(x), \qquad \text{by definition}.$$ (1)

Now, $\lim_{k\to\infty} (a_kx + b_k) = ax+b$, so that for n sufficiently large and finite x, given any $\epsilon \gt 0$,

$$ax + b - \epsilon \leq a_k x + b_k \leq ax +b+\epsilon,$$

where $\epsilon$ is chosen so that F is continuous at the points $ax+b-\epsilon$ and $ax + b+\epsilon$. It follows that

$$F_k (ax+b-\epsilon) \leq F_k (a_k x + b_k) \leq F_k (ax+b+\epsilon).$$

Taking limits as $k \to\infty$,

$$F(ax+b-\epsilon) \leq \begin{array}[t] {c} \underline{\lim} ... ...\to\infty \end{array} F_k (a_kx + b_k) \leq F (ax+b+\epsilon).$$

Now, since ax+b is a continuity point of F(x), and since $\epsilon$is arbitrary, we obtain

 

$$F(ax+b) = \lim_{k\to\infty} F_k (a_k x + b_k).$$ (2)

Combining Eqs. (1) and (2),

F(ax+b) = G(x),

which proves the theorem.