Theorem 4 (Gnedenko (1943), pp. 431-434)

Let a_n > 0 and b_n $(n=1,2,3,\ldots)$ be any finite constants. If, as $n \to \infty$, $G_{1,n} (a_n x + b_n) \Rightarrow \phi (x)$($\phi(x)$ a proper frequency distribution), then $\phi(x)$ belongs to one of the three following types:

$$\begin{array} {r@{~}c@{~}l} \text{(a)} \quad \phi_a (x) & = ... ...c)} \quad \phantom{_a}\Lambda (x) & = & e^{-e^{-x}},\end{array}$$

where a is a positive constant.

Proof: By hypothesis, there exist finite constants a_n > 0 and b_n ($n = 1,2,3,\ldots$) such that as $n \to \infty$,$G_{1,n}(a_n x + b_n) = F^n (a_n x + b_n) \Rightarrow \phi (x)$($\phi(x)$ proper). Now, any subsequence of Fⁿ (a_n x + b_n) also converges weakly to $\phi(x)$. In particular, for any positive integer k,

 

$$F^{nk} (a_{nk} + b_{nk} ) = \left[ F^n (a_{nk} x + b_{nk} ) \right]^k \Rightarrow \phi (x).$$ (13)

Then, $F^n(a_{nk} x + b_{nk} ) \Rightarrow [\phi (x)]^{1/k}$ as $n \to \infty$, and so Fⁿ(a_nk + b_nk) converges to a proper limit. But by Theorem 1, the limit of F^N (a_nk + b_nk) must then be of the same type as $\phi(x)$. Hence, there exist constants $\alpha_k \gt 0$ and $\beta_k$ for every $k = 1,2,\ldots$, such that

 

$$F^n (a_{nk} + b_{nk} ) \Rightarrow \phi (\alpha_k x + \beta_k).$$ (14)

Combining Eqs. (13) and (14) as $n \to \infty$,

 

$$\phi^k (\alpha_k x + \beta_k ) = \phi (x)$$ (15)

for every x such that x and $\alpha_k x + \beta_k$ are continuity points of $\phi$. Since $\phi(x)$ is a proper frequency distribution and so is monotonic increasing, it follows that such points are everywhere dense on the x axis. As will be shown however, Eq. (15) must hold for all values of x. For any x, choose a sequence $\{ \epsilon_n \}$, $\epsilon_n \gt 0$,$n = 1,2,3,\ldots$, such that $\lim_{n \to \infty} \epsilon_n = 0$,and such that $x + \epsilon_n$ and $\alpha_k (x + \epsilon_n) + \beta_k$ are continuity points of $\phi$. For every n then,

$$Q (x,\epsilon_n) = \phi^k (\alpha_k (x + \epsilon_n) + \beta_k ) - \phi (x + \epsilon_n) = 0.$$

But by Definition 8, a frequency distribution is right-continuous, and so $Q(x,\epsilon_n)$ is right-continuous since it is a continuous function of $\phi(x)$. But then,

$$\phi^k (\alpha_k x + \beta_k) - \phi(x) = \lim_{n\to\infty} Q (x,\epsilon_n) = 0.$$

It follows that $\phi(x)$ must satisfy Eq. (15) for every x and for all natural numbers k.

The following three cases will be considered separately:

$$\begin{array} {rl} \alpha_k < 1, \quad & \text{for at least ... ..._k = 1, \quad & \text{for at least one integer $k$.}\end{array}$$

It will be shown that the $\alpha_k$, $k = 1,2,3, \ldots$ for any specific $\phi(x)$, must all belong to the same case, and that each case leads to a particular one of the three limiting types for $\phi(x)$.

Limiting Type (1): $\alpha_k < 1$ for at least one integer, say s. If

$$x \geq \frac{\beta_s}{1 - \alpha_s},$$

then

$$x(1 + \alpha_s) \geq \beta_s \quad \text{and} \quad x \geq \alpha_s x + \beta_s.$$

Since $\phi(x)$ is monotonic increasing,

$$\phi(x) \geq \phi (\alpha_s x + \beta_s).$$

Combining this result with Eq. (15) gives

$$\phi^s (x) \geq \phi^s (\alpha_s x + \beta_s) = \phi (x).$$

This conclusion can only hold if $\phi (x) = 1$ or $\phi(x) = 0$ for $x \geq \beta_s/(1 - \alpha_s)$. Of the two possibilities, only $\phi (x) = 1$ is consistent with $\phi(x)$ being a frequency distribution. Hence,

 

$$\phi(x) = 1, \qquad \text{for } x \geq \frac{\beta_s}{1 - \alpha_s}.$$ (16)

It will now be demonstrated that $\phi(x) < 1$ for $x < \beta_s/(1-\alpha_s)$. Suppose the contrary. Then, there exists $x_0 < \beta_s/(1-\alpha_s)$ such that $\phi(x_0) = 1$. Let x₁ be any value of x such that x₁ < x₀. Then, $x_1 < x_0 < \beta_s/(1-\alpha_s)$. Since $\alpha_s < 1$, it follows that there exists an N such that for n > N,

$$\left( \frac{\beta_s}{1-\alpha_s} - x_1 \right) \alpha^n_s < \frac{\beta_s}{1-\alpha_s} - x_0.$$

But then,

$$\begin{array} {r@{~}c@{~}l} \displaystyle x_0 < \frac{\beta... ...ha_s + \alpha^2_s + \ldots + \alpha^{n-1}_s \right).\end{array}$$

By the monotonicity of $\phi(x)$, the definition of x₀, and the condition that $\phi(x) \leq 1$, it follows that

$$1 = \phi(x_0) \leq \phi \left(\alpha^n_s x_1 + \beta_s \left... ... \alpha_s + \alpha^2_s + \ldots + \alpha^{n-1}_s \right)\right)$$

or

 

$$\phi \left( \alpha^n_s x_1 + \beta_s \left( 1 + \alpha_s + \ldots + \alpha^{n-1}_s \right)\right) = 1.$$ (17)

If Eq. (15) is used in conjunction with Eq. (17), it is found that

$$\begin{array} {r@{~}c@{~}l} 1 & = & \phi^{s^n} \left( \alpha... ...ght)\right]^{s^{n-2}} \\ & = & \ldots = \phi (x_1).\end{array}$$

But, since x₁ was any value of x less than x₀, it follows that $\phi (x) = 1$ for every x < x₀. This contradicts the hypothesis that $\phi(x)$ is proper and together with Eq. (16) establishes that

 

$$\begin{array} {rl} \phi(x) < 1, \qquad & \displaystyle \tex... ...yle \text{for } x \geq \frac{\beta_s}{1 - \alpha_s}.\end{array}$$ (18)

It will now be demonstrated that if $\alpha_k < 1$ for at least one integer k, then $\alpha_k < 1$ for every value of k. Suppose the contrary. Then, there exist positive integers r and s such that $\alpha_s < 1$ and $\alpha_r \geq 1$. Consider $\alpha_r = 1$ first. Then, Eq. (15) becomes

 

$$\begin{array} {r@{~}c@{~}l} \phi^r(x + \beta_r) & = & \phi(x), \\ \phi^r(x) & = & \phi (x - \beta_r).\end{array}$$ (19)

In Eq. (19) set $x = \beta_s/(1-\alpha_s)$ to give, together with Eq. (18),

 

$$\begin{array} {r@{~}c@{~}l} \displaystyle \phi^r \left( \fra... ...phi^r \left( \frac{\beta_s}{1-\alpha_s} \right) = 1.\end{array}$$ (20)

If $\beta_r \neq 0$, define x₀ as

$$x_0 = \min \left( \frac{\beta_s}{1-\alpha_s} + \beta_r, \frac{\beta_s}{1-\alpha_s} - \beta_r \right).$$

Then, by Eq. (20), $\phi(x_0) = 1$ and $x_0 < \beta_s/(1-\alpha_s)$. By the reasoning following Eq. (16), this situation leads to a contradiction. If $\beta_r = 0$, then $\phi^r (x) = \phi (x)$. Combining this with the conditions already imposed on $\phi(x)$ by Eq. (18), it must be that

$$\begin{array} {rl} \phi(x) = 1, \qquad & \displaystyle \text... ...ystyle \text{for } x < \frac{\beta_s}{1 - \alpha_s}.\end{array}$$

This contradicts the hypothesis that $\phi(x)$ is a proper frequency distribution. Thus, it has been proven that $\alpha_r \neq 1$.

Suppose now that $\alpha_r \gt 1$. If $x \leq \beta_s/(1-\alpha_r)$,then

$$(1-\alpha_r)x \geq \beta_r \quad \text{and} \quad x \geq \alpha_r x + \beta_r.$$

Since $\phi(x)$ is monotonic increasing,

$$\phi(x) \geq \phi (\alpha_r x + \beta_r).$$

Combining this result with Eq. (15),

$$\phi (x) = \phi^r (\alpha_r x + \beta_r) \leq \phi^r (x).$$

Hence, $\phi (x) = 1$ or $\phi(x) = 0$ for $x \leq \beta_r/(1-\alpha_r)$. If $\beta_r/(1-\alpha_r) = \beta_s/(1-\alpha_s)$, then $\phi(x) = 0$ for $x \leq \beta_s/(1-\alpha_s)$ and $\phi (x) = 1$ for $x \gt \beta_s/(1-\alpha_s)$. But then $\phi(x)$ is improper, which contradicts the hypothesis of the theorem. If $\beta_r/(1-\alpha_r) \gt \beta_s/(1-\alpha_s)$, the contradiction is immediate since for $\beta_s/(1-\alpha_s) < x < \beta_r/(1-\alpha_r)$, $\phi (x) = 1$, and for $x \leq \beta_s/(1-\alpha_s) < \beta_r/(1-\alpha_r)$, $\phi(x) = 0$ (see Eq. (18)). Hence, once again $\phi(x)$ would be improper and thus contradict the hypothesis. The only possibility left is that

$$\frac{\beta_r}{1-\alpha_r} < \frac{\beta_s}{1-\alpha_s}.$$

It will be demonstrated that this also results in $\phi(x)$ being improper. Let x₁ be any x value such that

$$x_1 < \frac{\beta_r}{1 - \alpha_r} < \frac{\beta_s}{1-\alpha_s}.$$

Since $\phi(x_1) = 1$ or $\phi (x_1) = 0$ if $x_1 < \beta_r/(1-\alpha_r)$, and since by Eq. (18) $\phi (x_1) < 1$ if $x_1 < \beta_s/(1-\alpha_s)$, it follows that $\phi (x_1) = 0$. Choose any $\epsilon \gt 0$. Then, since $\alpha_s < 1$, there exists N such that if n > N,

$$\left( \frac{\beta_s}{1-\alpha_s} - x_1 \right) \alpha^n_s < \epsilon.$$

Then,

$$- \epsilon < - \left( \frac{\beta_s}{1-\alpha_s} - x_1 \right) \alpha^n_s < \epsilon,$$

$$\begin{array} {r@{~}l} \displaystyle \frac{\beta_s}{1 - \al... ... \equiv z < \frac{\beta_s}{1 - \alpha_s} + \epsilon,\end{array}$$

where z is introduced as a definition. Then,

$$\frac{\beta_s}{1 - \alpha_s} - \epsilon < z < \frac{\beta_s}{1 - \alpha_s} + \epsilon,$$

$$\left\vert \frac{\beta_s}{1-\alpha_s} - z \right\vert < \epsilon,$$

and z can be made to approach arbitrarily close to $\beta_s/(1-\alpha_s)$. But by the definition of z, together with Eq. (15),

$$\begin{array} {r@{~}c@{~}l} \phi^{s^n} (z) & = & \left[ \phi... ...ght) \right]^{s^{n-2}} \\ & = & \ldots = \phi(x_1).\end{array}$$

Since x₁ was defined so that $\phi (x_1) = 0$,

$$\phi^{s^n} (z) = 0 \quad \text{and} \quad \phi (z) = 0.$$

But, z is arbitrarily close to $\beta_s/(1-\alpha_s)$, and hence this situation would lead to $\phi(x) = 0$ for $x < \beta_s/(1-\alpha_s)$, and $\phi (x) = 1$ for $x \geq \beta_s/(1 - \alpha_s)$. This contradicts the hypothesis that $\phi(x)$ is proper.

By the enumeration of all possible cases, it has now been proven that if $\alpha_k < 1$ for at least one integer k, then $\alpha_k < 1$for all integers k (k positive). Then, the argument leading to Eq. (18) could have been applied to any integer k, and for any two integers r and s,

$$\frac{\beta_s}{1-\alpha_s} = \frac{\beta_r}{1-\alpha_r}.$$

For otherwise, by Eq. (18), there would be values of x for which $\phi (x) = 1$ and $\phi(x) < 1$ simultaneously. Define

$$\begin{array} {r@{~}c@{~}l} y & = & \displaystyle x - \frac{... ...phi \left( y + \frac{\beta_k}{1 - \alpha_k} \right).\end{array}$$

In terms of this new variable, which was obtained by a simple translation, Eq. (15) becomes

$$\begin{array} {r@{~}c@{~}l} \bar{\phi}^k (\alpha_k,y) & = & ... ...rac{\beta_k}{1 - \alpha_k} \right) = \bar{\phi} (y).\end{array}$$

The function $\bar{\phi}(y)$, which satisfies

$$\bar{\phi}^k (\alpha_k y) = \bar{\phi}(y) \qquad (\alpha_k < 1),$$

was determined in Theorem 3 as

$$\psi_a (y) = \begin{cases} e^{-(-y)^a}, & \text{for } x \leq 0, \\ 1, & \text{for } x \gt 0. \end{cases}$$

But, y and x are related by a simple translation, and it results that $\phi(x)$ is of the same type as $\psi_a(x)$.

Limiting Type (2): Now, consider the case where $\alpha_k \gt 1$for at least one integer s. Then, for $x_1 \leq \beta_s/(1-\alpha_s)$,

$$\begin{array} {r@{~}c@{~}l} x_1 (1-\alpha_s) & \geq & \beta_... ... \phi (x_1) & \geq & \phi (\alpha_s x_1 + \beta_s ).\end{array}$$

Combining this result with Eq. (15),

$$\phi (x_1) = \phi^s (\alpha_sx_1 + \beta_s ) \leq \phi^s (x_1).$$

Hence, since $\phi (x_1) < 1$ and $\phi(x)$ is a frequency distribution,

$$\phi(x) = 0, \qquad \text{for } x < \frac{\beta_s}{1-\alpha_s}.$$

Now, assume that $x_0 \gt \beta_s/(1-\alpha_s)$ and $\phi (x_0) = 0$. Let x₂ be any x such that $x_2 \gt x_0 \gt \beta_s/(1-\alpha_s)$. Then, since $\alpha_s \gt 1$, there exists an N such that if n > N,

$$\frac1{\alpha^n_s} \left( x_2 - \frac{\beta_s}{1-\alpha_s} \right) < x_0 - \frac{\beta_s}{1-\alpha_s},$$

 

$$x_0 \gt \frac{x_2}{\alpha^n_s} - \frac{\beta_s (1-\alpha^n_s... ...a_s} + \frac1{\alpha^2_s} + \ldots + \frac1{\alpha^n_s}\right),$$ (21)

where z is defined as shown.

Equation (15) can be restated as

$$\phi^s (x) = \phi \left( \frac{x}{\alpha_s} - \frac{\beta_s}{\alpha_s} \right).$$

If this form is applied n times,

 

$$\begin{array} {r@{~}c@{~}l} \phi^{s^n} & = & \displaystyle [... ...^2_s} + \ldots + \frac1{\alpha^n_s} \right) \right).\end{array}$$ (22)

Combining Eqs. (21) and (22),

$$\begin{array} {r@{~}c@{~}l} \phi^{s^n} (x_2) & = & \phi (z) \leq \phi (x_0) = 0, \\ \phi (x_2) & = & 0.\end{array}$$

But then, $\phi(x) = 0$ for every x > x₀, which contradicts $\phi(x)$ being a frequency distribution. Hence, $\phi (x) \gt 0$ for $x \gt \beta_s/(1-\alpha_s)$.

Finally, suppose that $\phi(x_0) = 1$ for some $x_0 \gt \beta/(1-\alpha_s)$. Let x₃ be any value of x such that

 

$$x_0 \gt x_3 \gt \frac{\beta_s}{1-\alpha_s}.$$ (23)

As shown in Eq. (21), with an interchange of the variables in correspondence with inequalities (23),

$$x_3 \gt \frac{x_0}{\alpha^n_s} - \beta_s \left( \frac1{\alph... ...+ \frac1{\alpha^2_s} + \ldots + \frac1{\alpha^n_s} \right) = z.$$

Hence, by Eq. (22),

$$1 = \phi^{s^n} (x_0) = \phi (z) \leq \phi (x_3).$$

By definition of x₃, this result leads to $\phi (x) = 1$ for $x \gt \beta_s/(1-\alpha_s)$, and $\phi(x) = 0$ for $x \leq \beta_s/(1-\alpha_s)$, which contradicts that $\phi(x)$ is proper. In summary, it has now been demonstrated that

 

$$\begin{array} {rl} 0 < \phi (x) < 1, \qquad & \displaystyle ... ...tyle \text{for } x \leq \frac{\beta_s}{1-\alpha_s}.\end{array}$$ (24)

Suppose that there exists an integer r such that $\alpha_r < 1$although $\alpha_s \gt 1$. By the argument given in Case (a), if $\alpha_r < 1$, then $\alpha_k < 1$ for all k. and so $\alpha_s < 1$. This contradiction shows that for all k, $\alpha_k \geq 1$.Now, suppose that $\alpha_s \gt 1$ and $\alpha_r = 1$. Then, by the reasoning preceding Eq. (20), together with Eq. (20),

$$\begin{array} {r@{~}c@{~}l} \displaystyle \phi^r \left( \fra... ...phi^r \left( \frac{\beta_s}{1-\alpha_s} \right) = 0.\end{array}$$

If $\beta_r \neq 0$, set $x_0 = \max \left[ \beta_s/(1-\alpha_s) + \beta_r, \beta_s/(1-\alpha_s) - \beta_r \right]$. Then,

$$x_0 \gt \frac{\beta_s}{1-\alpha_s} \quad \text{and} \quad \phi (x_0) = 0.$$

But, in the discussion preceding Eq. (21), this situation was shown to lead to a contradiction. Hence, $\beta_r \neq 0$ is eliminated as a possibility if $\alpha_r = 1$. If $\beta_r = 0$, then

$$\phi^r (x) = \phi (x),$$

and $\phi (x) = 1$ or $\phi(x) = 0$ for all x. This contradicts that $\phi(x)$ is a proper frequency distribution. Taken together, all these contradictions require that $\alpha_r \gt 1$. Hence, $\alpha_k \gt 1$ for all k.

By Eq. (24), together with the preceding result that $\alpha_k \gt 1$ for all k, it results that

$$\frac{\beta_s}{1 - \alpha_s} = \frac{\beta_s}{1 - \alpha_r}$$

for all pairs of integers r and s. Set

$$\begin{array} {r@{~}c@{~}l} y & = & \displaystyle x - \frac{... ...phi \left( y + \frac{\beta_k}{1 - \alpha_k} \right).\end{array}$$

Then, Eq. (15) becomes

$$\begin{array} {r@{~}c@{~}l} \bar{\phi}^k (\alpha_k y) & = & ... ...\frac{\beta_k}{1-\alpha_k} \right) = \bar{\phi} (y).\end{array}$$

As shown in Theorem 3, the solution of this equation, with $\alpha_k \gt 1$, is of the same type as

$$\phi_a (x) = \begin{cases} 0, & \text{for } x \leq 0, \\ e^{-x^{-a}}, & \text{for } x \gt 0. \end{cases}$$

Limiting Type (3): Lastly, consider the case where $\alpha_k = 1$ for at least one integer s. By the arguments given in the discussions of Cases (a) and (b), $\alpha_k = 1$ for all k. It shall be demonstrated that $0 < \phi (x) < 1$ for all finite x. If $\alpha_k = 1$, Eq. (15) becomes

 

$$\phi^k(x + \beta_k) = \phi(x).$$ (25)

If $\beta_k = 0$, $\phi^k(x) = \phi(x)$, and $\phi (x) = 1$ or $\phi(x) = 0$. This contradicts that $\phi(x)$ is proper. Hence, $\beta_k \neq 0$.

The n-fold application of Eq. (25) gives

 

$$\phi (x) = \phi^k (x + \beta_k) = \phi^{k^2} (x + 2 \beta_k) = \ldots = \phi^{k^n} (x + n \beta_k),$$ (26)

 

$$\phi^{k^n} (x) = \phi (x - n \beta_k).$$ (27)

Suppose that there exists a finite x₀ such that $\phi (x_0) = 0$.Let x₁ be any value of x such that x₁ > x₀. Choose n so that

$$\begin{array} {r@{~}c@{~}l} (x_1 - x_0) & < & n \vert\beta_k\vert, \\ x - n \vert\beta_k\vert & < & x_0.\end{array}$$

If $\beta_k \gt 0$, then by Eq. (27) and the definition of x₀,

$$\phi^{k^n} (x_1) = \phi (x_1 - n \beta_k) \leq \phi (x_0) = 0$$

or $\phi(x) = 0$ for all x > x₀. This contradicts that $\phi(x)$ is a frequency distribution and establishes that for $\beta_k \gt 0$, $\phi (x) \gt 0$ for all x.

Now, $\beta_k$ cannot be negative, for if $\beta_k < 0$, then $x + \beta_k < x$ and $\phi (x + \beta_k) \leq \phi (x)$. Then, Eq. (25) yields

$$\phi (x) = \phi^k (x + \beta_k) \leq \phi^k (x).$$

But then, $\phi(x) = 0$ or $\phi (x) = 1$ for all x, which contradicts that $\phi(x)$ is proper.

Now, suppose that there exists a finite number of x₀ such that $\phi(x_0) = 1$. Let x₂ be any value of x such that x₂ < x₀. Choose n large enough so that

$$\begin{array} {r@{~}c@{~}l} (x_0 - x_2) & < & n \beta_k, \\ x_0 & < & x_2 + n \beta_k,\end{array}$$

and, by Eq. (17),

$$\phi (x_2) = \phi^{k^n} (x_2 + n \beta_k ) \geq \phi^{k^n} (x_0) = 1$$

or $\phi (x_2) = 1$. But then, $\phi (x) = 1$ for all x < x₀, which is a contradiction. In summary, it has been demonstrated that

 

$$0 < \phi (x) < 1, \qquad \text{for all finite $x$}.$$ (28)

Set z = e^x, $c_k = e^{\beta_k} \gt 1$, and $\bar{\phi}(z) = \left\{ \begin{smallmatrix} \phi (\log z), & \text{for } z \gt 0, \\ 0, & \text{for }, z < 0.\end{smallmatrix}\right.$ In view of the limitations placed on $\phi(x)$ by Eq. (28), these transformations are permissible. Equation (25) becomes

$$\bar{\phi}^k (c_k z) = \phi^k (\log c_k + \log z) = \phi^k (x +\beta_k) = \phi (x) = \phi (\log z) = \bar{\phi}(z).$$

By Theorem 3, the solution of

$$\bar{\phi}^k (c_kz) = \bar{\phi}(z) \qquad (c_k < 1)$$

is of the same type as

$$\phi_a (z) = \begin{cases} 0, & \text{for } z < 0, \\ e^{-z^{-a}}, & \text{for } z \geq 0. \end{cases}$$

This transforms to a frequency distribution type of

$$\Lambda (x) = e^{-e^{-x}}.$$