Файл отчета в системе Антиплагиат (1222235), страница 12

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Then eqn (1.18) for к = 2 coincides with (1.17).We denote by N the number of secondary delays (in the framework of this model).Lemma 1. For each fixed integer we have(1.19)We now give two corollaries arising from this lemma.Corollary 5. If is a constant value, then.Corollary 6. If are independent identically distributed random variables with the density function , then.Observe that the problems associated with the delays of trains were considered in many papers (see e.g. [2]). However, in these papers theproblem statements and methods differ from ours.1.2 Proof​of Theorem 1In order to prove Theorem 1, we reformulate it in terms of and .Theorem . 1.

If(1.20)then. (1.21)2. Let к Ье a fixed integer, . If(1.22)then(1.23)and hence (1.7).3. If(1.24)then(1.25). (1.26)Lemma 2. Let be a fixed integer.1. Inequality (1.22) entails (1.23).http://dvgups.antiplagiat.ru/ReportPage.aspx?docId=427.12578943&repNumb=123/2508.06.2015Антиплагиат2. Double inequality (1.24) entails equality .Proof. 1. Our proof will be done by mathematical induction.

It is easy to verify the validity of the first statement for к = 2. The inductiveassumption is based on the following logical consequence: iffor some fixed integer thenLet condition (1.22) is valid. Then,(1.27)Because of inductive assumption, we getNext, we obtain by using the first estimate in (1.27) the inequality (1.4), whichinvolves (1.5). Thus,2. Let , then . Otherwise, due to (1.2), (1.3) we have . Now, by using (1.4) and (1.5), we obtain . Because of (1.23) then . This is contrary tothe right hand side of inequality (1.24).

Thereby .Proof of Theorem . 1. Note that . Therefore the condition (1.20)coincides​with (1.2), which entails equality and, consequently, (1.21).2. By Lemma 2 the condition (1.22) involves (1.23) and, consequently, (1.7).Formula (1.25) is a consequence of Lemma 2, namely:Since , then for . This implies (1.26).Thus our theorem is completely proved.1.3 Proof of Theorem 2Lemma 3. The formula (1.11) holds.Proof. We haveObserve that departure time of the first train is . Therefore, according to order of the train departures, if , then and, consequently, .

Otherwiseif , then , i.e. . Hereof we getand thus we obtain the following equalitywhich is equivalent to (1.11).Let is fixed. We consider the random events(1.28)Note that these events are mutually exclusive and .Notice that if, for instance, n = 3, then , and if n = 4, then , i.e. events , in general, depend on n.Lemma 4.

Let. The following formula holds(1.29)Proof. Suppose for clarity that n = 6. We have(1.30)By Theorem 1, item 1, the event entails the following equalitythus,(1.31)The event implies that the inequality is satisfied. Then, from (1.25), we obtainhence(1.32)The​event entails inequality (1.22) as followsAccording to Theorem 1 from this inequality we deduce that . Thus,(1.33)Similarly, we obtain(1.34)Gathering (1.30) – (1.34) we arrive at the equality(1.35)Obviously,(1.36)Next, since , then(1.37)Note that . Consequently,(1.38)From (1.35) – (1.38) we obtain(1.39)Equality (1.39) is equivalent to (1.29). Indeed, given the implication holds. Therefore.Thus,henceThe proof in the case an arbitrary remains unchanged. For n = 3 the proof only simplified.Proof of Theorem 2. The formula (1.11) is proved in Lemma 3 and the formula (1.12) for к = 3 is proved in Lemma 4.The proof of eqn (1.12) for we carry out by analogy with the proof of Lemma 4.We have(1.40)By Theorem 1, item 1, the event entails the following equalityhence(1.41)If the event occurred, then due to (1.26) the following relation holdsfrom which we getLikewise, for all(1.42)From (1.41) and (1.42), bearing in mind thatwe obtain(1.43)Furthermore,​by using (1.25) and relationhttp://dvgups.antiplagiat.ru/ReportPage.aspx?docId=427.12578943&repNumb=124/2508.06.2015Антиплагиатwe find that(1.44)Let us assume that the event occurred.

Then inequality (1.22) holds. Hence according to (1.7) we deduce that . Thus,(1.45)Gathering (1.40), (1.43) – (1.45) we arrive at the equality(1.46)Arguing as at the end of the proof of Lemma 4, we obtainThereforeIt follows that (1.46) entails (1.12).1.4 Proof of corollaries from the theorem 2Proof of Corollary 2. Let .

Taking into account (1.13), we denote(1.47)From Corollary 1 and (1.12) it follows that for(1.48)whereWe have(1.49)and(1.50)Taking into account the continuity of random variable , from (1.48) – (1.50) we obtain (1.14).Proof of Corollary 3. When we replace in (1.14) with T, we obtain eqn (1.15).Proof of Corollary 4. It is well-known, that if and are two independent random variables, and besides has the density function , then for anymeasurable function of two variables и and for each(1.51)Formula (1.51) entails the following form of :(1.52)Formulas ​http://dvgups.antiplagiat.ru/ReportPage.aspx?docId=427.12578943&repNumb=125/25.

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