Theorem 2 (Gnedenko and Kolmogorov (1954), pp. 42-44)

For a sequence of frequency distribution functions $\{ F_n (x)\}$,the relations

 

$$F_n (a_n x + b_n) \Rightarrow F(x),$$ (3)

 

$$F_n (\alpha_n x + \beta_n) \Rightarrow F(x),$$ (4)

as $n \to \infty$, where a_n > 0, $\alpha_n \gt 0$, b_n and $\beta_n$are real constants, and F(x) is a proper distribution function, are satisfied simultaneously if and only if

 

$$\frac{\alpha_n}{a_n} \to 1 \quad \text{and} \quad \frac{b_n - \beta_n}{a_n} \to 0, \qquad \text{as } n \to \infty.$$ (5)

Proof: It will first be proven that (3) and (5) imply (4). Let x₁, x, and x₂ be continuity points of the function F(x), and let x₁ < x < x₂. Then, by (5), if n is sufficiently large,

$$x_1 < \frac{\alpha_n}{a_n} x + \frac{\beta_n - b_n}{a_n} < x_2,$$

$$a_n x_1 + b_n < \alpha_n x + \beta_n < a_n x_2 + b_n.$$

Hence, by the monotonicity of F_n(x),

$$F_n (a_n x_1 + b_n) \leq F_n (\alpha_n x + \beta_n) \leq F_n (a_nx_2 + b_n).$$

In the limit as $n \to \infty$,

$$F(x_1) \leq \begin{array}[t] {c} \underline{\lim} \\ \scrip... ...to \infty\end{array} F_n (\alpha_n x + \beta_n) \leq F (x_2).$$

But, since x is a continuity point of F(x), then as $x_1 \to x$and $x_2 \to x$,

$$F(x) = \lim_{n \to \infty} F_n (\alpha_n x + \beta_n),$$

which is Eq. (4).

It will now be demonstrated that (3) and (4) may imply (5). Set

$$A_n = \frac{\alpha_n}{a_n}, \qquad B_n = \frac{\beta_n - b_n}{a_n}, \quad \text{and} \quad G_n(x) = F_n (a_nx + b_n).$$

Under this change of notation, (3) and (4) become

 

$$G_n (x) \Rightarrow F(X),$$ (6)

 

$$\begin{array} {r@{~}c@{~}l} G_n (A_n x + B_n) & = & \display... ... & = & F_n (\alpha_n x + \beta_n) \Rightarrow F(x).\end{array}$$ (7)

As in the proof of Theorem 1, pick a sequence $n_1 < n_2 < \ldots < n_k < \ldots$ such that as $k \to\infty$,

$$A_{n_k} \to A \quad \text{and} \quad B_{n_k} \to B.$$

By Corollary 1 and Eq. (7), $0 < A < \infty$, and B is finite. From Eq. (6),

$$G_{n_k} (Ax + B) \Rightarrow F(Ax + B).$$

But, by Eq. (7),

$$G_n (Ax + B) \Rightarrow F(x),$$

and hence, for any continuity point of both F(x) and F(Ax + B),

 

F(x) = F(Ax + B). (8)

This equality holds for all but a denumerable, at most, set of values of x.

Suppose $A \neq 1$, then A < 1 or A > 1. Consider first 0 < A < 1. An n-fold application of Eq. (8) gives

$$\begin{array} {r@{~}c@{~}l} F(x) & = & F(Ax + B ) = F(A(Ax+B... ...^2)) = \ldots = F(A^k + B(1+A + \ldots + A^{k-1})).\end{array}$$

Since n is arbitrary and A < 1, $A^n \to 0$, as $n \to \infty$.Hence, as $n \to \infty$,

$$F(x) = F(B(1+A^2 + A^3 + \ldots) ) = F \left( \frac{B}{1-A} \right),$$

or F(x) is constant for all x. As this is impossible for a frequency distribution, the contradiction establishes that A is not less than one.

Now assume that A > 1. Eq. (8) gives

$$F(x) = F(Ax + B) \quad \text{or} \quad F\left( \frac1A x - \frac{B}A \right) = F (x).$$

But this is the same case as considered before, and thus it must be that A = 1. Hence, Eq. (8) becomes

F(x) = F(x+B).

Suppose that $B \neq 0$. The n-fold application of this equation gives

$$F(x) = F(x+B) = F((x+B) + B) = F(x + 2B) = \ldots = F(x+nB),$$

where n is arbitrary. Hence, for every finite x,

$$\begin{array} {r@{~}c@{~}l} F(-\infty) & = & \displaystyle ... ...fty} F(x) = \lim_{Bn \to\infty} F(x+nB) = F(\infty).\end{array}$$

Such a relationship is impossible for a frequency distribution. It follows that B = 0.

It has been shown now that as $k \to\infty$,

$$A_{n_k} \to 1 \quad \text{and} \quad B_{n_k} \to 0.$$

Now, consider the complete sequence $\{ A_n \}$ and $\{B_n\}$.Suppose that there exists a subsequence of the positive integers $\{ n'_k \}$ such that for some $\delta \gt 0$, at least one of the following inequalities is satisfied:

 

$$\lim_{k \to \infty} \left\vert A_{n'_k} - 1 \right\vert \geq... ...lim_{k \to \infty} \left\vert B_{n'_k} \right\vert \geq \delta.$$ (9)

Without loss of generality, the sequence of indices may be chosen so that

$$A_{n'_k} \to A' \quad \text{and} \quad B_{n'_k} \to B', \qquad \text{as } k \to \infty.$$

Exactly the same argument as carried out previously can be used to demonstrate that A' = 1 and B' = 0. But, this contradicts Eq. (9), and thus it must be said that

$$\lim_{n \to \infty} A_n = 1 \quad \text{and} \quad \lim_{n \to \infty} B_n = 0.$$