����������� ������ 51 (1999), 29–37

UDK 519.213

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EXTREME VALUES OF THE SEQUENCES OF INDEPENDENT RANDOM VARIABLES WITH MIXED DISTRIBUTIONS

Pavle Mladenović

Abstract. In this paper we consider some examples of the sequences of independent random variables with the same mixed distribution. In these cases we determine the type of extreme value distribution and the normalizing constants.

1. Introduction

Let (X_{n}) be a sequence of independent random variables with the common distribution function F. If for some constants a_{n} > 0 and b_{n}

P \{ max_{1⩽j⩽n} X_{j} ⩽ \frac{x}{a_{n}} + b_{n}\} →_{d} G(x),  (1.1)

where G is non-degenerated distribution function, then the function G belongs to one of three classes of the maximum stable distributions, and the functions in these classes have the following forms (maybe after linear transformation of the argument):

Type I. G_{1}(x) = exp(−e^{−x}), −∞ < x < +∞;

Type II. G_{2}(x) = \{ \begin{matrix}0, & if x ⩽ 0, \\ exp(−x^{−α}), & if x > 0, \end{matrix} for some α > 0;

Type III. G_{3}(x) = \{ \begin{matrix}exp(−(−x)^{α}), & if x ⩽ 0, \\ 1, & if x > 0. \end{matrix} for some α > 0.

These three types of distributions are called the extreme values distributions. If for some distribution functions F and G the relation (1.1) holds true, then we say that the common distribution function F of the random variables X_{1}, X_{2}, X_{3}, … belongs to the domain of attraction of the function G. We shall use the notation M_{n} = max{X_{1},…, X_{n}}. The constants a_{n} > 0 and b_{n} from the relation (1.1) are called the normalizing constants. Note that for a_{n} > 0, the inequality M_{n} ⩽ x/a_{n} + b_{n} is equivalent to a_{n}(M_{n} − b_{n}) ⩽ x.

AMS Subject Classification: 60G70.

Keywords and phrases: mixed distributions; extreme value distributions; normal, Cauchy, uniform and truncated exponential distributions.

2. Necessary and sufficient conditions for convergence to the extreme value distributions

If for some distribution function F there exists a domain of attraction, then it is determined by the asymptotic behaviour of the tail 1 − F (x), as x → +∞. The following useful theorem 1.5.1 from [1] holds true:

THEOREM 1. [1] Let (X_{n}) be a sequence of independent random variables with the common distribution function F (x), −∞ < x < +∞, (u_{n}) a sequence of real numbers, 0 ⩽ τ ⩽ +∞ and M_{n} = max{X_{1}, X_{2},…, X_{n}}. Then, the equality

lim_{n→∞} P {M_{n} ⩽ u_{n}} = e^{−τ} ,

holds true if and only if lim_{n→∞} n(1 − F (u_{n})) = τ.

Necessary and sufficient conditions for the function F to belong to some domain of attraction can also be formulated as in the following theorem 1.6.2 from [1]:

THEOREM 2. [1] Let (X_{n}) be a sequence of independent random variables with the common distribution function F, and x_{F} = sup{x| F (x) < 1}. Necessary and sufficient conditions for the function F to belong to the domain of attraction of possible types are given by:

Type I. There exists a strictly positive function g(t) defined on the set (−∞, x_{F}),

such that for every real number x the equality lim_{t↑x_{F}} \frac{1 − F (t + xg(t))}{1 − F (t)} = e^{−x} holds true.

Type II. x_{F} = +∞ and lim_{t→∞} \frac{1 − F (tx)}{1 − F (t)} = x^{−α}, for some α > 0 and all x > 0.

Type III. x_{F} < +∞ and lim_{h↓0} \frac{1 − F (x_{F} − hx)}{1 − F (x_{F} − h)} = x^{α}, for some α > 0 and all x > 0.

3. Mixed distributions

Let X_{1} and X_{2} be random variables with distribution functions F_{1}(x) and F_{2}(x), respectively, and

X = \{ \begin{matrix}X_{1} & with probability p, \\ X_{2} & with probability q, \end{matrix}

where p + q = 1. The distribution function of the random variable X is given by

F (x) = P {X ⩽ x} = pP {X_{1} ⩽ x} + qP {X_{2} ⩽ x}
= pF_{1}(x) + qF_{2}(x).

The distribution of probability determined by the distribution function F is called the mixture of the distributions determined by the functions F_{1} and F_{2}.

We shall consider some examples of sequences of independent random variables with common mixed distribution. In these cases we are going to determine the type of extreme value distribution and the normalizing constants.

3.1. Mixture of normal distributions. Let (X_{n}) be a sequence of independent random variables with normal N (0, 1) distribution and M_{n} = max{X_{1},…, X_{n}}. As is well known, the limiting distribution of the maximum M_{n} is given by

P {a_{n}(M_{n} − b_{n}) ⩽ x} → e^{−e^{−x}}, n → ∞,  (3.1)

where the normalizing constants a_{n} and b_{n} are

a_{n} = \sqrt{2 ln n}, b_{n} = \sqrt{2 ln n} − \frac{1}{2} \frac{ln ln n + ln 4π}{\sqrt{2 ln n}} .  (3.2)

Corollary. Let (Y_{n}) be a sequence of independent random variables with normal N (m, σ^{2}) distribution. Then, we have X_{n} = (Y_{n} − m)/σ ∈ N (0, 1). If M_{n} = max{X_{1},…, X_{n}} and M̃_{n} = max{Y_{1},…, Y_{n}}, then

P \{a_{n}(M_{n} − b_{n}) ⩽ x\} = P \{a_{n}(\frac{M̃_{n} − m}{σ} − b_{n}) ⩽ x\}
= P {ã_{n}(M̃_{n} − b̃_{n}) ⩽ x} → exp(−e^{−x}), n → ∞,

where the constants ã_{n} and b̃_{n} are given by:

ã_{n} = \frac{a_{n}}{σ} = \frac{\sqrt{2 ln n}}{σ},
b̃_{n} = m + σb_{n} = m + σ\sqrt{2 ln n} − σ \frac{ln ln n + ln 4π}{2\sqrt{2 ln n}} .

THEOREM 3. Let (Z_{n}) be a sequence of independent random variables such that

Z_{n} ∈ \{ \begin{matrix}N (m_{1}, σ_{1}^{2}), & with probability p, \\ N (m_{2}, σ_{2}^{2}), & with probability q, \end{matrix} for all n,

where p + q = 1. Let us denote M_{n}^{∗} = max{Z_{1},…, Z_{n}}. If

(a) σ_{1} > σ_{2}, m_{1}, m_{2} ∈ R or (b) σ_{1} = σ_{2} and m_{1} > m_{2}, then for every real number x the equality

lim_{n→∞} P {a_{n}^{∗}(M_{n}^{∗} − b_{n}^{∗}) ⩽ x} = e^{−e^{−x}}  (3.3)

holds true, where the constants a_{n}^{∗} and b_{n}^{∗} are given by

a_{n}^{∗} = \frac{\sqrt{2 ln n}}{σ_{1}}, b_{n}^{∗} = m_{1} + σ_{1}\sqrt{2 ln n} − \frac{σ_{1}}{2\sqrt{2 ln n}}(ln ln n + ln \frac{4π}{p^{2}} ).  (3.4)

REMARK. Since M_{n}^{∗} in distribution is the same as M̃_{np}, then all normalizing constants with a_{n}^{∗}/ã_{np} → 1 and a_{n}^{∗}(b̃_{np} − b_{n}^{∗}) → 0 will work. The maximum M_{n}^{∗} asymptotically comes from F_{1}.

Proof of Theorem 3. Let φ(x) = (2π)^{−1/2}e^{−x^{2}/2} and Φ(x) = ∫_{−∞}^{x} φ(t) dt. We shall use the following asymptotic relation

1 − Φ(x) ∼ x^{−1}φ(x), x → ∞.  (3.5)

If X_{i} ∈ N (m_{i}, σ_{i}^{2}), i = 1, 2, then distribution function of the random variable X_{i} can be represented in the form

F_{i}(x) = P \{X_{i} ⩽ x\} = P \{\frac{X_{1} − m_{1}}{σ_{1}} ⩽ \frac{x − m_{1}}{σ_{1}} \} = Φ(\frac{x − m_{1}}{σ_{1}} ), i = 1, 2.

Distribution function of the random variable Z_{n} has the form F (x) = pF_{1}(x) + qF_{2}(x). Using this representation of F (x) we obtain

1 − F (t) = p[1 − Φ(\frac{t − m_{1}}{σ_{1}} )] + q[1 − Φ(\frac{t − m_{2}}{σ_{2}} )]

∼ \frac{pσ_{1}}{t − m_{1}} φ(\frac{t − m_{1}}{σ_{1}} ) + \frac{qσ_{2}}{t − m_{2}} φ(\frac{t − m_{2}}{σ_{2}} )

= \frac{1}{\sqrt{2π}} \{\frac{pσ_{1}}{t − m_{1}} exp[−\frac{1}{2}(\frac{t − m_{1}}{σ_{1}} )^{2}] + \frac{qσ_{2}}{t − m_{2}} exp[−\frac{1}{2}(\frac{t − m_{2}}{σ_{2}} )^{2}]\}

= \frac{1}{\sqrt{2π}} \frac{pσ_{1}}{t − m_{1}} exp[−\frac{1}{2}(\frac{t − m_{1}}{σ_{1}} )^{2}](1 + o(1)), t → ∞;

\frac{1 − F (t + xg(t))}{1 − F (t)} = exp\{−\frac{(t − m_{1} + xg(t))^{2} − (t − m_{1})^{2}}{2σ_{1}^{2}} \}\frac{(t − m_{1})(1 + o(1))}{t − m_{1} + xg(t)}

= exp(−\frac{xg(t)(t − m_{1})}{σ_{1}^{2}}) exp(−\frac{x^{2}g^{2}(t)}{2σ_{1}^{2}})\frac{(t − m_{1})(1 + o(1))}{t − m_{1} + xg(t)}.

For g(t) = σ_{1}/(t − m_{1}), we get

\frac{1 − F (t + xg(t))}{1 − F (t)} = e^{−x} exp(−\frac{x^{2}σ_{1}^{2}}{2(t − m_{1})^{2}} )\frac{1}{1 + xσ_{1}^{2}(t − m_{1})^{−2}} (1 + o(1))
→ e^{−x}, as t → +∞.

Using theorem 1 we conclude that the distribution function F (x) belongs to the domain of attraction of type I, i.e. there exist constants a_{n}^{∗} and b_{n}^{∗}, such that the following equality holds true:

lim_{n→∞} P \{M_{n}^{∗} ⩽ \frac{x}{a_{n}^{∗}} + b_{n}^{∗}\} = e^{−e^{−x}}.

The constants a_{n}^{∗} and b_{n}^{∗} can be determined as follows: let us first determine the constant u_{n}, such that 1 − F (u_{n}) ∼ \frac{1}{n}e^{−x} as n → ∞ i.e.

1 − pF_{1}(u_{n}) − qF_{2}(u_{n}) ∼ \frac{1}{n}e^{−x}, n → ∞.

Using the equalities F_{1}(u_{n}) = Φ((u_{n} − m_{1})/σ_{1}) and F_{2}(u_{n}) = Φ((u_{n} − m_{2})/σ_{2}) and asymptotic relation (3.5), we obtain

1 − pF_{1}(u_{n}) − qF_{2}(u_{n}) = p[1 − Φ(\frac{u_{n} − m_{1}}{σ_{1}} )] + q[1 − Φ(\frac{u_{n} − m_{2}}{σ_{2}} )]

∼ \frac{pσ_{1}}{u_{n} − m_{1}} φ(\frac{u_{n} − m_{1}}{σ_{1}} ) + \frac{qσ_{2}}{u_{n} − m_{2}} φ(\frac{u_{n} − m_{2}}{σ_{2}} )

∼ \frac{1}{n}e^{−x}, as n → ∞.

Let us denote: v_{n} = (u_{n} − m_{1})/σ_{1} and w_{n} = (u_{n} − m_{2})/σ_{2}. For large values of n the inequality v_{n} < w_{n} holds true, and

p\frac{φ(v_{n})}{v_{n}} + q \frac{φ(w_{n})}{w_{n}} = \frac{1}{\sqrt{2π}} (\frac{p}{v_{n}} e^{−v_{n}^{2}/2} + \frac{q}{w_{n}} e^{−w_{n}^{2}/2})

= \frac{1}{\sqrt{2π}} \frac{p}{v_{n}} e^{−v_{n}^{2}/2}(1 + \frac{v_{n}}{p} \frac{q}{w_{n}} e^{−(w_{n}^{2}−v_{n}^{2})/2}).

Let Δ_{n} = w_{n}^{2} − v_{n}^{2} = (u_{n} − m_{1})^{2}/σ_{1}^{2} − (u_{n} − m_{2})^{2}/σ_{2}^{2}. If σ_{1} > σ_{2} > 0, then

Δ_{n} = (\frac{1}{σ_{2}^{2}} − \frac{1}{σ_{1}^{2}} )u_{n}^{2} + Au_{n} + B → +∞, as n → ∞.

If σ_{1} = σ_{2} = σ and m_{1} > m_{2}, then Δ_{n} = [2u_{n}(m_{1} − m_{2}) + m_{2}^{2} − m_{1}^{2}]/σ^{2} → +∞, as n → ∞. In both cases we have the following asymptotic equality

p\frac{φ(v_{n})}{v_{n}} + q \frac{φ(w_{n})}{w_{n}} = \frac{1}{\sqrt{2π}} \frac{p}{v_{n}} e^{−v_{n}^{2}/2}(1 + o(1)), as n → ∞.

Hence, the constant u_{n} should be determined from the conditions u_{n} = σ_{1}v_{n} + m_{1} and

p\frac{φ(v_{n})}{v_{n}} ∼ \frac{1}{n}e^{−x}, as n → ∞.  (3.6)

Asymptotic relation (3.6) can be transformed in the following way:

\frac{1}{pn}e^{−x} · \frac{v_{n}}{φ(v_{n})} → 1,

− ln n − ln p − x + ln v_{n} − ln φ(v_{n}) → 0,
− ln n − ln p − x + ln v_{n} + \frac{1}{2} ln 2π + \frac{v_{n}^{2}}{2} → 0.  (3.7)

It follows from (3.7) that v_{n}^{2}/(2 ln n) → 1 as n → ∞, and

ln v_{n} = \frac{1}{2}(ln 2 + ln ln n) + o(1).  (3.8)

The relation (3.7) can also be writen in the form

\frac{v_{n}^{2}}{2} = x + ln n + ln p − \frac{1}{2} ln 2π − ln v_{n} + o(1).  (3.9)

Now let us substitute the value of ln v_{n} from (3.8) into (3.9). We obtain

\frac{v_{n}^{2}}{2} = x + ln n + ln p − \frac{1}{2} ln 2π − \frac{1}{2} ln 2 − \frac{1}{2} ln ln n + o(1)
= x + ln n + ln p − \frac{1}{2} ln 4π − \frac{1}{2} ln ln n + o(1),
v_{n}^{2} = 2 ln n\{1 + \frac{x + ln p − \frac{1}{2} ln 4π − \frac{1}{2} ln ln n}{ln n} + o(\frac{1}{ln n})\}.

Note that ln p − \frac{1}{2} ln 4π = −\frac{1}{2} ln \frac{4π}{p^{2}} . Using the formula \sqrt{1 + x} = 1 + \frac{1}{2}x + o(x) as x → 0, we get

v_{n} = \sqrt{2 ln n}\{1 + \frac{x + ln p − \frac{1}{2} ln 4π − \frac{1}{2} ln ln n}{2 ln n} + o(\frac{1}{ln n})\}

= \sqrt{2 ln n}\{1 + \frac{1}{2 ln n}(x − \frac{1}{2} ln \frac{4π}{p^{2}} − \frac{1}{2} ln ln n) + o(\frac{1}{ln n})\}.

Since u_{n} = m_{1} + σ_{1}v_{n}, it follows that

u_{n} = \frac{σ_{1}x}{\sqrt{2 ln n}} + m_{1} + σ_{1}\sqrt{2 ln n} − \frac{σ_{1}}{2\sqrt{2 ln n}}(ln ln n + ln \frac{4π}{p^{2}} ) + o(\frac{1}{\sqrt{ln n}}).

On the other hand u_{n} ∼ \frac{x}{a_{n}^{∗}} + b_{n}^{∗}, as n → ∞. Consequently it is easy to obtain the

normalizing constants a_{n}^{∗} and b_{n}^{∗} in the form (3.4).

3.2. Mixture of Cauchy distributions. Let (X_{n}) be a sequence of independent random variables with the Cauchy distribution K(1, 0), determined by the distribution function

F (x) = \frac{1}{2} + \frac{1}{π} arctg x.

Let us denote M_{n} = max{X_{1},…, X_{n}}. For x > 0 we have

\frac{1 − F (tx)}{1 − F (t)} = \frac{\frac{π}{2} − arctg(tx)}{\frac{π}{2} − arctg t} → \frac{1}{x}, t → ∞.  (3.10)

Indeed, on substituting arctg t = π/2 − ϑ, we obtain t = tg(π/2 − ϑ) = ctg ϑ,

lim_{t→∞}(\frac{π}{2} − arctg t) · t = lim_{ϑ→0} ϑ ctg ϑ = 1.

Similarly, lim_{t→∞}(\frac{π}{2} − arctg(tx))tx = 1, and (3.10) follows easily. In this case, the

distribution function F belongs to the domain of attraction of the function G_{2}(x), and we have the type II of limiting distribution. The normalizing constants are a_{n} = 1/γ_{n} and b_{n} = 0, where the constant γ_{n} can be determined from the equality

1 − F (γ_{n}) = \frac{1}{2} − \frac{1}{π} arctg γ_{n} = \frac{1}{n}.

Hence, γ_{n} = tg(\frac{π}{2} − \frac{π}{n}) = ctg \frac{π}{n}. For every x > 0 we have:

lim_{n→∞} P \{M_{n} · tg \frac{π}{n} ⩽ x\} = e^{−x^{−1}}.

Corollary. Let (Y_{n}) be a sequence of independent random variables with the Cauchy distribution K(λ, 0), which is determined by the distribution function

F (x) = \frac{1}{2} + \frac{1}{π} arctg \frac{x}{λ}. Let us denote M̃_{n} = max\{Y_{1},…, Y_{n}\} and Y_{n}/λ = X_{n}.

Then, M̃_{n} = λM_{n}. It is easy to see that the random variables Y_{n}/λ has K(1, 0) distribution. Since the inequality M_{n} tg \frac{π}{n} ⩽ x is equivalent to \frac{1}{λ} M̃_{n} tg \frac{π}{n} ⩽ x, it follows that the normalizing constants in this case are given by ã_{n} = \frac{1}{λ} tg \frac{π}{n} and b̃_{n} = 0. For these values of normalizing constants and every x > 0 we have

lim_{n→∞} P \{\frac{1}{λ}M̃_{n} · tg \frac{π}{n} ⩽ x\} = e^{−x^{−1}}.

THEOREM 4. Let K(λ_{i}, 0) be the class of random variables with the distribution function F_{i}(x) = \frac{1}{2} + \frac{1}{π} arctg \frac{x}{λ_{i}}, i = 1, 2. Let (Z_{n}) be a sequence of independent random variables such that for every n,

Z_{n} ∈ \{ \begin{matrix}K(λ_{1}, 0), & with probability p, \\ K(λ_{2}, 0), & with probability q, \end{matrix}

where p + q = 1, and M_{n}^{∗} = max{Z_{1},…, Z_{n}}. Then, for all x > 0 we have

lim_{n→∞} P \{\frac{π}{n(pλ_{1} + qλ_{2})}M_{n}^{∗} ⩽ x\} = e^{−x^{−1}}.  (3.11)

REMARK. For Cauchy variables with different scale parameters, M_{n}^{∗} will come from either of the two parent distributions, which explains why the extremal distribution is also a mixture. More specifically, for pure Cauchy variables, πM_{n}/nλ → U with F_{U}(x) = e^{−x^{−1}}. In a sample of size n from the mixture of two Cauchy distributions one observes approximately np and nq variables of the two types, and hence M_{n}^{∗} ≈ max(M_{np}^{(1)}, M_{nq}^{(2)}) in distribution. Since F_{U} is a max-stable distribution, this explains the result.

Proof of Theorem 4. The distribution function of the random variable Z_{n} is

F (x) = \frac{1}{2} + \frac{p}{π} arctg \frac{x}{λ_{1}} + \frac{q}{π} arctg \frac{x}{λ_{2}} .

It is easy to prove that lim_{t→∞} t(\frac{π}{2} − arctg(at)) = \frac{1}{a} for a > 0. Consequently, for every x > 0 we obtain the following relations:

\frac{1 − F (tx)}{1 − F (t)} = \frac{\frac{1}{2} − \frac{p}{π} arctg \frac{tx}{λ_{1}} − \frac{q}{π} arctg \frac{tx}{λ_{2}}}{\frac{1}{2} − \frac{p}{π} arctg \frac{t}{λ_{1}} − \frac{q}{π} arctg \frac{t}{λ_{2}}}

= \frac{pt(\frac{π}{2} − arctg \frac{tx}{λ_{1}} ) + qt(\frac{π}{2} − arctg \frac{tx}{λ_{2}} )}{pt(\frac{π}{2} − arctg \frac{t}{λ_{1}} ) + qt(\frac{π}{2} − arctg \frac{t}{λ_{2}} )}

→ \frac{pλ_{1}/x + qλ_{2}/x}{pλ_{1} + qλ_{2}} = \frac{1}{x}, as t → ∞.

It follows from theorem 2 that the function F belongs to the domain of attraction of the function G_{2}, i.e. in this case we have the type II of extreme value distribution. The normalizing constants are a_{n}^{∗} = 1/γ_{n} and b_{n}^{∗} = 0, where 1−F (γ_{n}) = 1/n. This equation can be transformed equivalently as follows:

\frac{1}{2} − \frac{p}{π} arctg \frac{γ_{n}}{λ_{1}} − \frac{q}{π} arctg \frac{γ_{n}}{λ_{2}} = \frac{1}{n},

γ_{n}p(\frac{π}{2} − arctg \frac{γ_{n}}{λ_{1}} ) + γ_{n}q(\frac{π}{2} − arctg \frac{γ_{n}}{λ_{2}} ) = \frac{π}{n}γ_{n}.

From the last equality we obtain pλ_{1} + qλ_{2} − \frac{π}{n}γ_{n} → 0, as γ_{n} → ∞. Consequently,

we have γ_{n} ∼ \frac{n(pλ_{1} + qλ_{2})}{π} and a_{n}^{∗} = \frac{1}{γ_{n}} ∼ \frac{π}{n(pλ_{1} + qλ_{2})}. Now it is easy to derive

the equality (3.11).

3.3. Mixture of uniform and truncated ecponential distribution. Let (X_{n}) be a sequence of independent random variables with the uniform U[0, c] distribution. The distribution function of the random variable X_{n} is given by F_{1}(x) = x/c, 0 ⩽ x ⩽ c. For n ⩾ τ > 0 it follows from the equality 1 − F_{1}(u_{n}) = τ/n that u_{n} = c(1 − τ/n). Using theorem 1 we obtain P {M_{n} ⩽ c(1 − τ/n)} → e^{−τ} as n → ∞. For x < 0 and τ = −x we obtain P \{\frac{n}{c}(M_{n} − c) ⩽ x\} → e^{x}, as n → ∞. Hence, in this case we have the extreme value distribution of type III, where α = 1. The normalizing constants are a_{n} = \frac{n}{c} and b_{n} = c.

Let (Y_{n}) be a sequence of independent random variables with the same truncated exponential distribution E(λ, c). This distribution is determined by the distribution function F_{2} which is given by F_{2}(x) = 0 for x ⩽ 0, F_{2}(x) = 1 for x ⩾ c and

F_{2}(x) = \frac{1 − e^{−λx}}{1 − e^{−λc}} , if 0 ⩽ x ⩽ c.

For x < 0 and τ = −x, it follows from theorem 1 that

lim_{n→∞} P \{M_{n} ⩽ c + \frac{x(e^{λc} − 1)}{λn} + o(\frac{1}{n})\} = e^{x}.

In this case we also obtain the limiting distribution of type III. The normalizing constants are given by ã_{n} = λn(e^{λc} − 1)^{−1} and b_{n} = c.

THEOREM 5. Let U[0, c] be the class of random variables with the uniform distribution on the interval [0, c] and E(λ, c) the class of random variables with the truncated exponential distribution. Let (Z_{n}) be a sequence of independent random variables such that for every n,

Z_{n} ∈ \{ \begin{matrix}U[0, c], & with probability p, \\ E(λ, c), & with probability q, \end{matrix}

where p + q = 1. If M_{n}^{∗} = max{Z_{1},…, Z_{n}}, then for every x < 0 the following equality holds true:

lim_{n→∞} P \{M_{n}^{∗} ⩽ c + \frac{x}{n}(\frac{p}{c} + \frac{λq}{e^{λc} − 1})^{−1}\} = e^{x}.  (3.12)

Proof. The distribution function of the random variable Z_{n} is given by

F (x) = \frac{px}{c} + q \frac{1 − e^{−λx}}{1 − e^{−λc}} , if 0 ⩽ x ⩽ c.

It is easy to prove that for every x > 0 the equality

lim_{h↓0} \frac{1 − F (c − hx)}{1 − F (c − x)} = x,

holds true. Using theorem 2 we conclude that the function F belongs to the domain of attraction of the function G_{3}. The normalizing constants are b_{n}^{∗} = c and a_{n}^{∗} = n/k, where k should be determined. Using theorem 1 we obtain that for x < 0 the condition P {M_{n} ⩽ c + xk/n} → e^{x} as n → ∞, can be writen in the form:

1 − F (c + \frac{kx}{n}) ∼ −\frac{x}{n}, n → ∞,

1 − \frac{p}{c}(c + \frac{kx}{n}) − \frac{q(1 − e^{−λc}e^{−λkx/n})}{1 − e^{−λc}} ∼ −\frac{x}{n}, n → ∞,

1 − p − \frac{pkx}{cn} − \frac{q}{1 − e^{−λc}} \{1 − e^{−λc} + e^{−λc} \frac{λkx}{n} + o(\frac{1}{n})\} ∼ −\frac{x}{n}, n → ∞.

Now, we obtain that −\frac{pkx}{cn} − \frac{e^{−λc}qλkx}{n(1 − e^{−λc})} ∼ −\frac{x}{n} as n → ∞. Consequently, the

constant k is given by k = (\frac{p}{c} + \frac{λq}{e^{λc} − 1})^{−1}, and the equality (3.12) holds true.

Acknowledgment. I would like to thank two anonymous referees for helpful comments.

REFERENCES

[1] Leadbetter, M.R., Lindgren, G. and Rootzén, H., Extrems and Related Properties of Random Sequences and Processes, Springer-Verlag, New York–Heidelberg–Berlin, 1986.

[2] Mladenović, P., Mixed distributions and extreme values, Petrović, R; Radojević, D. (eds), SYM–OP–IS ′98 (1998), 435–438.

(received 14.10.1998) Mathematical Faculty, Belgrade University, Studentski trg 16, 11000 Belgrade, Yugoslavia