6.262 Discrete Stochastic Processes, Problem Set 6 Solutions

ثبت نشده
چکیده

The purpose of this exercise is to show that, for an arbitrary renewal process, N(t), the number of renewals in (0, t], has finite expectation. a) Let the inter-renewal intervals have the distribution FX (x), with, as usual, FX (0) = 0. Using whatever combination of mathematics and common sense is comfortable for you, show that numbers � > 0 and δ > 0 must exist such that FX (δ) ≤ 1 − �. In other words, you are to show that a positive rv must take on some range of of positive values with positive probability. Solution: For any � < 1, we can look at the sequence of events {X ≥ 1/k; k ≥ 1}. The union of these events is the event {X > 0}. Since Pr{X ≤ 0} = 0, Pr{X > 0} = 1. The events {X ≥ 1/k} are nested in k, so that, from (??),

برای دانلود متن کامل این مقاله و بیش از 32 میلیون مقاله دیگر ابتدا ثبت نام کنید

ثبت نام

اگر عضو سایت هستید لطفا وارد حساب کاربری خود شوید

منابع مشابه

6.262 Discrete Stochastic Processes, Problem Set 4 Solutions

The purpose of this problem is to illustrate that for an arrival process with independent but not identically distributed interarrival intervals, X1, X2, . . . ,, the number of arrivals N(t) in the interval (0, t] can be a defective rv. In other words, the ‘counting process’ is not a stochastic process according to our definitions. This illustrates that it is necessary to prove that the countin...

متن کامل

6.262 Discrete Stochastic Processes, Problem Set 5 Solutions

1, and thus in any left eigen­ vector π = π̂(1), π̂(2), · · · , π̂(r), π̂(t) of eigenvalue 1, we have π̂(t) = 0. This means that π̂(i) must be a left eigenvector of [Pi] for each i, 1 ≤ i ≤ r. Since π̂(i) can be given an arbitrary scale factor for each i, we get exactly r independent left� eigenvectors, each a probability vector as desired. We take these eigenvectors to be π(i) = 0, · · · , 0, π̂(i),...

متن کامل

6.262 Discrete Stochastic Processes, Problem Set 12 Solutions

Solution: Since Λ(y) has finitely many (3) possible values, all values of η between any adjacent pair lead to the same threshold test. Thus, for η > 1, Λ(y) > η, leads to the decision ĥ = 0 if and only if (iff) Λ(y) = 1, i.e., iff −1 ≤ y < 0. For η = 1, the rule is the same, Λ(y) > η iff Λ(y) = 1, but here there is a ‘don’t care’ case Λ(y) = 1 where 0 ≤ y ≤ 1 leads to ĥ = 1 simply because of th...

متن کامل

6.262 Discrete Stochastic Processes, 2010 Midterm Exam Solutions

We will give partial credit if you present your thinking in a clear way we can understand (and your thinking is at least partially correct), but otherwise not. If you model a problem using a tool that requires significant computation, it is best to first give the model explicitly and indicate how you will use the results of the computation to determine the final answer. This approach will help ...

متن کامل

6.262 Discrete Stochastic Processes, 2011 Final Exam Solutions

a) Let Z be the time at which the last student finishes. Show that Z has a distribution function FZ (z) given by [1 exp( z] . b) Let T1 be the time at which the first student leaves. Show that the probability density of T1 is given by n e n t . For each i, 2  i  n, let Ti be the interval from the departure of the i 1st student to that of the ith. Show that the density of each Ti is exponentia...

متن کامل

ذخیره در منابع من


  با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید

برای دانلود متن کامل این مقاله و بیش از 32 میلیون مقاله دیگر ابتدا ثبت نام کنید

ثبت نام

اگر عضو سایت هستید لطفا وارد حساب کاربری خود شوید

عنوان ژورنال:

دوره   شماره 

صفحات  -

تاریخ انتشار 2011