Theorem 13 (Gnedenko (1943), pp. 439-442)

A frequency distribution Fⁿ(x) converges to a limiting frequency distribution of the type $\phi_a (x)$, and n [ 1 - F (x)] converges to a W-function of the type W₁ (a,x) if and only if, for all values of k > 0,

 

$$\lim_{x\to\infty} \frac{1-F(x)}{1 - F (kx)} = k^a, \qquad a \gt 0.$$ (46)

Proof: Suppose first that Eq. (46) is true. It will be demonstrated that $\lim_{n\to\infty} n [1-F (x)]$ is of the type W₁ (a,x). By Eq. (46), F(x) < 1 for all finite x. Assume the contrary. Then, F(x₀) = 1 for some finite x = x₀. But then,

$$\lim_{x\to\infty} \frac{1-F(x)}{1 - F (kx)} = \begin{cases} ... ...\text{if } k = 1, \\ \infty, & \text{if } k \gt 1, \end{cases}$$

which contradicts Eq. (46). It follows that for n sufficiently large, the value of x satisfying

$$1 - F (x) \leq \frac1n$$

will be positive. Let a_n be the smallest value of x (see Theorem 12) satisfying the inequalities

 

$$1 - F (x+0) \leq \frac1n \leq 1 - F (x-0).$$ (47)

In Eq. (46), let x be replaced by a_n x and $k = (1+\epsilon)/x$. Then,

 

$$\lim_{a_n x \to \infty} \frac{1-F (a_n x)}{1-F (a_n (1 + \ep... ...-F (a_n (a+\epsilon))} = \left( \frac{1 + \epsilon}x \right)^a.$$ (48)

Similarly, replace x by a_n x and let $k = (1-\epsilon)/x$. Then,

 

$$\lim_{a_n x \to \infty} \frac{1-F(a_nx)}{1-F (a_n (1-\epsilo... ...{1-F (a_n (1-\epsilon))} = \left( \frac{1-\epsilon}x \right)^a.$$ (49)

The left-hand members of Eq. (48) and (49) are monotonic functions of $\epsilon$, and the right-hand terms are continuous functions of $\epsilon$. Then, by Theorem 10, the convergence in both Eqs. (48) and (49) is uniform. Hence [Rudin (1953), pp. 119-120],

 

$$\begin{array} {r@{~}c@{~}l} \displaystyle \lim_{n\to\infty} ... ...0} \left( \frac{1 + \epsilon}2 \right)^2 = x^{-a}, \end{array}$$ (50)

 

$$\begin{array} {r@{~}c@{~}l} \displaystyle \lim_{n\to\infty} ... ... 0} \left( \frac{1 - \epsilon}2 \right)^2 = x^{-a}.\end{array}$$ (51)

Now, by definition of a_n (see Eq. (47)),

$$1 - F (a_n (1+0)) \leq \frac1n \leq 1 - F (a_n (1-0)).$$

and so

 

$$\frac{1-F (a_nx)}{1 - F (a_n (1-0))} \leq n [ 1 - F (a_n x)] \leq \frac{1 - F (a_n x)}{1 - F (a_n (1+0))}.$$ (52)

In the limit as $n \to \infty$, because of Eqs. (50) and (51), it follows that (52) becomes

$$x^{-a} \leq n [ 1 - F (a_n x) ] \leq x^{-a} \quad \text{or} \quad \lim_{n\to\infty} n [ 1-F (a_nx)] = x^{-a}.$$

Then, by Theorem 7,

$$F^n (a_n x) \Rightarrow e^{-x^{-a}}.$$

The above development was based on the supposition that x > 0, and hence is true for all x > 0. But then,

$$\begin{array} {r@{~}c@{~}l@{\quad}l} \displaystyle \lim_{n\t... ...\to\infty} F^n (a_n x) & = & 0, & \text{for } x < 0,\end{array}$$

since

$$\begin{array} {r@{~}c@{~}l@{\quad}l} x^{-a} & \to & \infty, ... ...0, \\ e^{-x^{-a}} & \to & 0, & \text{as } x \to + 0.\end{array}$$

This proves the sufficiency of Eq. (46).

Now, suppose that Fⁿ(x) converges to a limiting frequency distribution of the type $\phi_a (x)$. By Theorem 7, this is the same as supposing that n [1-F(x)] converges to a W-function of the type W₁ (a,x). Hence, there exists a sequence of constants a_n > 0, b_n, $n = 1,2,3,\ldots$, such that

 

$$\lim_{n\to\infty} n [ 1 - F (a_n x + b_n)] = \begin{cases} x... ...gt 0, \ a \gt 0, \\ \infty, & \text{for } x \leq 0.\end{cases}$$ (53)

For all constants $\beta \gt 1$ and x > 0,

 

$$\lim_{n\to\infty} \beta [ 1 - F (a_n x + b_n)] = \beta x^{a}.$$ (54)

In the subsequent proof, the following convention will be used. Let B be any positive real number. [B] will be taken to represent the largest integer which is less than or equal to B. Furthermore, (B) will be defined as (B) = B-[B]. Using this notation, we can express Eq. (54) as

 

$$\lim_{n\to\infty} \{ [n\beta] + ( n\beta)\} \{ 1 - F (a_n x + b_n ) \} = \beta x^{-a}.$$ (55)

Now, since $n \{ 1 - F (a_n x + b_n)\} \to x^{-a}$ for x > 0, it follows that $\lim_{n\to\infty} \{ 1 - F (a_n x + b_n)\} =0$. Hence, since $(n\beta) < 1$,

$$\lim_{n\to\infty} (n\beta) \{ 1 - F (a_n x + b_n)\} = 0.$$

Eq. (55) becomes

 

$$\lim_{n\to\infty} [n\beta] \{ 1 - F (a_n x + b_n)\} = \beta x^{-a}.$$ (56)

Set $x = z \beta^{1/1}$ in Eq. (56). This yields

 

$$\lim_{n\to \infty} [n\beta] \left\{ 1 - F \left( a_n \beta^{1/1} z + b_n \right) \right\} = z^{-a}.$$ (57)

Since z is a positive real number, the relationship expressed in Eq. (57) is still true if z is replaced with x to give

 

$$\lim_{n\to\infty} [n\beta] \left\{ 1 - F \left( a_n \beta^{1/a} x + b_n \right) \right\} = x^{-a}.$$ (58)

The sequence $\{ [ n\beta]\}$, $n = 1,2,3,\ldots,$ is a subsequence of the positive integers. Hence, by Eq. (53),

 

$$\lim_{n\to\infty} [n\beta] \left\{ 1 - F \left( a_{[n\beta]} x + b_{[n\beta]} \right) \right\} = x^{-a}.$$ (59)

If Eqs. (58) and 59) are simultaneously true, which they are, Theorem 2 requires that as $n \to \infty$,

$$\frac{a_n \beta^{1/a}}{a_{[n\beta]}} \to 1 \quad \text{and} \quad \frac{b_n - b_{[n\beta]}}{a_{[n\beta]}} \to 0.$$

Hence, as $n \to \infty$,

 

$$\frac{a_{[n\beta]}}{a_n} \to \beta^{1/a} \quad \text{and} \quad \frac{b_n-b_{[n\beta]}}{a_{[n\beta]}} \to 0.$$ (60)

A subsequence of $\{ [ n\beta]\}$ will be chosen in the following way. Let n be any fixed integer and define the sequence $\{ n_s\}$ by the relationships

$$\begin{array} {r@{~}c@{~}l} n_1 & = & [n\beta], \\ n_2 & = &... ...\beta], \\ & \cdots & \\ n_s & = & [n_{s-1} \beta].\end{array}$$

Now, by Theorem 2, $a_{[n\beta]}$ and $b_{[n\beta]}$can be defined as

$$a_{[n\beta]} = a_n \beta^{1/a} \quad \text{and} \quad b_{[n\beta]} = b_n$$

without affecting the validity of Eq. (59). Suppose that this change of definition is made. Then,

 

$$\begin{array} {r@{~}c@{~}l@{\quad}l} a_{n_1} & = & a_{[n\bet... ...{1/a} = a_n \beta^{s/a} & \text{and } b_{n_s} = b_n.\end{array}$$ (61)

It follows that

$$\lim_{s\to\infty} \frac{b_{n_s} - 0}{s_{n_s}} = \lim_{s\to\infty} \frac{b_n}{a_n \beta^{s/a}} = 0.$$

Then, by Theorem 2, the constants a'_ns = a_ns and b'_ns = 0 still give the convergence in Eq. (59). This relationship becomes (deleting the primes)

 

$$\lim_{s\to\infty} n_s \{ 1 - F (a_{n_s} x ) \} = x^{-a}.$$ (62)

By Eq. (61), $a_{n_s} \to \infty$ as $s \to \infty$.Hence, for any value of y sufficiently large, there exists an s value such that for any positive x,

$$a_{n_s} x \leq y \leq a_{n_{s+1}} x.$$

Then, by the monotonic properties of F(x),

$$1 - F (a_{n_{s+1}} x ) \leq 1 - F (y) \leq 1 - F (a_{n_s} x),$$

and for k > 0,

$$1 - F (a_{n_{s+1}} kx) \leq 1-F (ky) \leq 1 - F (a_{n_s} kx).$$

From these two inequalities, it results that

 

$$\frac{1-F (a_{n_{s+1}}x)}{1-F (a_{n_s} kx)} \leq \frac{1 - F... ... F (ky)} \leq \frac{1 - F (a_{n_s} x)}{1 - F (a_{n_{s+1}} kx)}.$$ (63)

By definition of n_s,

$$n_{s+1} = [n_s \beta] = n_s \beta - (n_s \beta),$$

where $0 \leq (n_s \beta) \leq 1$. As a consequence,

 

$$\lim_{s\to\infty} \frac{n_{s+1}}{n_s} = \lim_{s\to\infty} \frac{n_s\beta - (n_s \beta)}{n_s} = \beta.$$ (64)

From (62), (63), and (64),

$$\frac{n_{s+1} \{ 1 - F (a_{n_{s+1}} x ) \}}{n_s \{ 1 - F (a_... ...}}{n_{s+1} \{ 1 - F ( a_{n_{s+1}} kx)\}} \ \frac{n_{s+1}}{n_s}$$

or in the limit as $s \to \infty$,

$$\frac{k^a}\beta = \frac{x^{-a}}{(kx)^{-a}} \ \frac1\beta \le... ...)}{1 - F (ky)} \leq \frac{x^{-a}}{(kx)^{-a}} \beta = k^a \beta.$$

But, since $\beta$ was arbitrary, $\beta \gt 1$, it follows that $\beta$can be chosen as close to 1.0 as is wished. Hence,

$$\lim_{y\to\infty} \frac{1-F(y)}{1-F(ky)} = k^a.$$

The change from $s \to \infty$ to $y \to \infty$ is permissible since

$$a_{n_s} x \leq y \leq a_{n_{s+1}} x,$$

and for any fixed x, $y \to \infty$ as $a_{n_s} \to \infty$ with $s \to \infty$.