ВКР - Математическая модель задержек поездов (1222229), страница 11
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From (1.18) and (1.79) it follows that
(1.84)
where
(1.85)
It is easily seen that
(1.86)
where
We have
(1.87)
(1.88)
The relations (1.85) – (1.88) entail
(1.89)
By using (1.81), we obtain
(1.90)
In addition, similar to (1.79), we deduce
(1.91)
According to (1.80)
Therefore
(1.92)
By combining (1.89) – (1.92), we obtain
(1.93)
Equation (1.73) follows from (1.84) and (1.93).
Let us compute the mean and variance of random variables 
, 
. Note that
, , has a single point of discontinuity 
. Wherein
(1.94)
(1.95)
In order to find 
, , for 
we rewrite (1.72) and (1.73) as follows:
(1.96)
(1.97)
where
Furthermore, taking into account the fact that for any 
and 
for we obtain
(1.98)
Similarly to (1.98) for we have
(1.99)
Consider the integral of the form
where 
, 
, 
and 
are some positive parameters.
It is easy to verify that
It follows that the last two integrals in (1.99) can be rewritten as
(1.100)
(1.101)
Next, due to the property (1.80), from (1.98) – (1.101) we receive
(1.102)
and
(1.103)
For any bounded continuous function 
the following equality holds
(1.104)
Thus, from (1.94), (1.102) and (1.104) we obtain
(1.105)
Consequently,
(1.106)
From (1.105) and (1.106) with consideration the equality
we obtain (1.74) and (1.76).
In turn, using the equality
From (1.95), (1.103) and (1.104) we conclude for 
(1.107)
Consequently,
(1.108)
Thus, equations (1.107) and (1.108) yield equations (1.75) and (1.77) respectively.
Remark 5. Let us show that (1.72) is in some sense a special case of (1.73), although the latter one is shown only for .
It is possible to prove the following assertion: let positive parameters a and b converge to zero and 
. Then, for any bounded continuous on 
function 
the following equality holds
Indeed, for every 
there is 
, that 
when 
. We have
where
Since under these conditions on a and b
then for all sufficiently small a and b the inequalities 
, 
and 
hold. It follows assertion of this Lemma.
Let 
, 
and 
(here we assume that k can take fractional values). Then, according to the statement, given in square brackets expression from (1.73) converges to 
. Last integral in (1.73) converges to 0. Thus, for any t we have 
for .
Remark 6. It can be easily seen that for any t the following equality holds:
.
Graphs of the functions 
from (1.73) for k = 2, 3 with the parameters from (1.66) and 
are shown in Figure 1.6.
Remark 7. It is easy to verify that for 
the formula (1.72) goes over into the formula (1.58) when k = 2, and the formula (1.73) goes over into the formula (1.58) when .
The following calculations are produced as an illustration. We fix the parameters 
, 
and T according to (1.66), as well as we introduce a single designation for the functions (1.72) and (1.73): 
, . For comparison 
with 
from (1.58) Table 1.1 shows the values of these functions for some t and 
.
Figure 1.6 – The plots of and 
in the case when 
and 
are defined in (1.69) with the parameters from (1.66) and
Table 1.1 – Values of the functions 
and 
for some t and 
At the same t and the values of the functions 
and from (1.58) are given in Table 1.2.
Table 1.2 – Values of the functions 
and 
for some t and
Remark 8. By choosing sufficiently small, in the formula (1.73) the integrals over the negative real axis can be made arbitrarily small, and then in the calculations do not take into account. So in the above mentioned example with k = 3, T = 7, , 
the condition (1.70) is certainly satisfied, and estimates of integrals show that
Remark 9. Let the density functions 
and 
are defined by (1.69). Let p – be the maximum allowable probability that at least m of secondary delays occur. What should be the parameters T and in order to inequality 
holds? By Corollary 6 we obtain
(1.109)
where 
, 
. The equation (1.82) entails that
By choosing the parameters T and 
so that the first integral in (1.109) be sufficiently small, we have the condition on these parameters in form of the following inequality
(1.110)
Let 
, 
, , . Calculations show that the condition (1.110) is satisfied for 
. This first integral in (1.109) does not exceed 
.
Remark 10. Consider the behavior of the functions at change the dispersion. For simplicity we take k = 2. Let the parameters , and T are defined by (1.66). Figure 1.7 shows the graphs of in cases and 
.
Figure 1.7 – The plots of with parameters and when
As you might expect, with a decrease of the graph of for becomes steeper.
Now, we fix and consider the behavior of at change . Let the parameters and T are defined by (1.66). Figure 1.8 shows the graphs of in cases and 
. We can see that, in accordance with the formula (1.72), if grows, then the function at each point decreases.
Mark that for 
observed trends persist.
Figure 1.8 – The plots of with parameters and when
Remark 11. It can be easily seen that formulas (1.74) and (1.75), when , transform into the formula (1.59) with 
and , respectively. Likewise, formulas (1.76) and (1.77) transform into the formula (1.60) with and , respectively.
As an illustration the following calculations are produced. Fixing the parameters , and T, according to eqn (1.66), by using the formulas (1.74) – (1.77) at different values of , we obtain with precision to 
(see Tables 1.3 and 1.4)
in the case k = 2:
Table 1.3 – The mean and variance of 
for some
in the case k = 3:
Table 1.4 – The mean and variance of 
for some
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