Theorem 15 (Gnedenko (1943), pp. 446-448)

In order for n [ 1 - F (x)] to converge to a W-function of the type W₃(x) (or, what is the same, Fⁿ(x)) to converge to a frequency distribution of the type $\Lambda(x)$, as $n \to \infty$), it is necessary and sufficient that

 

$$\lim_{n\to\infty} n [1-F (a_n x + b_n)] = e^{-x}$$ (88)

for all values of x, where a_n and b_n are given by the following conditions:

(a)

b_n is the greatest lower bound for the values of x satisfying

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

(b)

a_n is the greatest lower bound for the values of x satisfying

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

Proof: If Eq. (88) is satisfied by the a_n and b_n as defined, then Fⁿ(x) converges to a frequency distribution of the type $\Lambda(x)$ by Theorem 7. Also, under these conditions, Eq. (88) becomes W₃ (x). Hence, this demonstrates sufficiency. It remains to prove the necessity.

Assume that n [ 1 - F (x)] converges to a W-function of the type W₃ (x). Then, there exists $\alpha_n \gt 0$ and $\beta_n$, $n = 1,2,3,\ldots$, such that for all x,

 

$$\lim_{n\to\infty} n [ 1 -F (\alpha_n x + \beta_n)] = e^{-x}.$$ (89)

It will be proved that the $\alpha_n$ and $\beta_n$ may be replaced by a_n and b_n without affecting the convergence of Eq. (89). For convenience, define w_n (x) to be

$$w_n (x) = n [1 - F (\alpha_n x + \beta_n )].$$

Then, Eq. (84) gives

$$\lim_{n\to\infty} w_n (x) = e^{-x}.$$

Hence, for any $\epsilon \gt 0$:

(A)

There exists N₁ such that for n > N₁,

$$\left\vert w_n (\epsilon) - e^{-\epsilon} \right\vert < (1-e... ...} \right) = 1 - e^{-\epsilon} - \frac1{2e} (1 - e^{-\epsilon})$$

or

 

$$\begin{array} {c} \displaystyle w_n (\epsilon) -e^{-\epsilo... ..._n (\epsilon) + \frac1{2e} (1 - e^{-\epsilon}) < 1. \end{array}$$ (90)

(B)

There exists N₂ such that for n > N₂,

$$\left\vert w_n (-\epsilon) - e^{\epsilon} \right\vert < (e^\epsilon - 1 ) \left( 1 - \frac1{2e^{1+\epsilon}} \right)$$

or

 

$$\begin{array} {c} \displaystyle w_n (-\epsilon) \gt e^\epsil... ...w_n(-\epsilon) - \frac1{2e} (1-e^{-\epsilon}) \gt 1.\end{array}$$ (91)

(C)

There exists N₃ such that for n > N₃,

$$\left\vert w_n (1+\epsilon) - e^{-(1+\epsilon)}\right\vert < \frac1{2e} (1 - e^{-\epsilon})$$

or

 

$$\begin{array} {c} \displaystyle w_n (1 + \epsilon) < e^{-1-... ...epsilon) + \frac1{2e} (1 - e^{-\epsilon}) < \frac1e.\end{array}$$ (92)

(D)

There exists N₄ such that for n > N₄,

$$\left\vert w_n ( 1 - \epsilon) - e^{-(1-\epsilon)} \right\ve... ... \frac1e \left(1 - \frac1{2e^\epsilon} \right) (e^\epsilon - 1)$$

or

 

$$\begin{array} {c} \displaystyle w_n ( 1 - \epsilon) \gt e^{-... ...epsilon) - \frac1{2e} (1-e^{-\epsilon}) \gt \frac1e.\end{array}$$ (93)

Now, combining Eqs. (90), (91), (92), and (93), if $n \gt \max$ (N₁, N₂, N₃,N₄),

 

$$\begin{array} {r@{~}c@{~}l} w_n (\epsilon) + \eta < & 1 & < ... ...n) + \eta < & \frac1e & < w_n (1 - \epsilon) - \eta,\end{array}$$ (94)

where $\eta = (1/2e) (1-e^{-\epsilon})$. By definition of W_n (x), this is the same as

$$\begin{array} {r@{~}c@{~}l} n [ 1 - F (\alpha_n \epsilon + \... ... n [ 1 - F (\alpha_n (1-\epsilon) + \beta_n)] - \eta\end{array}$$

or

 

$$\begin{array} {r@{~}c@{~}l} F (\alpha_n \epsilon + \beta_n) ... ... (\alpha_n (1-\epsilon) + \beta_n) + \frac{\eta}{n}.\end{array}$$ (95)

From Eq. (95) and the definition of a_n and b_n it follows that

 

$$- \alpha_n \epsilon + \beta_n \leq b_n \leq \alpha_n + \beta_n,$$ (96)

 

$$\alpha_n (1 - \epsilon) + \beta_n \leq a_n + b_n \leq \alpha_n (1+\epsilon) + \beta_n.$$ (97)

From Eq. (96),

 

$$\left\vert \frac{b_n - \beta_n}{\alpha_n} \right\vert < \epsilon.$$ (98)

From Eq. (97),

$$\begin{array} {r@{~}c@{~}l} \alpha_n ( 1 - \epsilon) & \leq ... ...} + \frac{b_n - \beta_n}{\alpha_n} \leq 1 + \epsilon\end{array}$$

or

 

$$\left\vert \frac{a_n}{\alpha_n} + \frac{b_n - \beta_n}{\alpha_n} - 1 \right\vert < \epsilon.$$ (99)

Then by Eqs. (98) and (99),

 

$$\begin{array} {r@{~}c@{~}l} \displaystyle \left\vert \frac{a... ...} \right\vert \leq \epsilon + \epsilon = 2 \epsilon.\end{array}$$ (100)

It follows from Eqs. (98) and (100) that a_n and b_n defined in Theorem 2 are such that

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

This proves the necessity of the conditions.