Question

Consider a renewal process with distribution \(F(x) .\) Suppose each event is erased with probability \(1-q .\) Expand the time scale by a factor \(1 / q .\) Show that the resulting sequence of events constitutes a renewal process where the distribution function of the time between events is $$ \sum_{n=1}^{\infty}(1-q)^{n-1} q F_{n}(x / q)=F(x ; q) $$ where \(F_{n}\) as usual denotes the \(n\)-fold convolution of \(F\).

Short answer

The resulting sequence of events after expanding the time scale by a factor \(1/q\) and erasing events with probability \(1-q\) constitutes a renewal process with distribution function \(G(x)\) given by: \[ G(x) = \sum_{n=1}^{\infty}(1-q)^{n-1} q F_n(x / q) = F(x;q) \] Where \(F_n(x)\) denotes the \(n\)-fold convolution of the original distribution function \(F(x)\).

Step-by-step solution

Recall basic concepts and definitions

A renewal process is a point process in which events occur randomly and independently with a fixed probability distribution known as the interarrival time distribution. The distribution function F(x) gives the probability that the time elapsed between two consecutive events is less than or equal to x. The n-fold convolution of a distribution function, denoted as F_n(x), is defined by convolving the distribution function n times with itself. Specifically, the 1-fold convolution is given by \(F_1(x) = F(x)\), and for n>1, the n-fold convolution is given by: \[ F_n(x) = \int_0^x F_{n-1}(t) F(x - t) dt \]

Expand the time scale

As the time scale is expanded by a factor 1/q, we also need to expand the distribution function F(x) by the same factor. Thus, the expanded distribution function F'(x) is given by: \[ F'(x) = F(x/q) \]

Erase events with probability (1-q)

Now we need to take into account that each event is erased with probability (1-q). Let's denote the resulting distribution function after erasing events as G(x). For each elapsed time x, we can have events erased multiple times before an event finally occurs. Thus, the distribution function G(x) can be written as: \[ G(x) = \sum_{n=1}^{\infty}(1-q)^{n-1} q F'(x_n) \] Where x_n represents the time elapsed after n events have been erased. Now, recall that F'(x) = F(x/q) for the expanded time scale. Therefore, we have: \[ G(x) = \sum_{n=1}^{\infty}(1-q)^{n-1} q F(x_n / q) \]

Calculate x_n

Now we need to determine x_n, the time elapsed in between consecutive events after n events have been erased. Since events occur independently, the elapsed time x_n in terms of F_n(x) can be written as: \[ x_n = x / F_n(x) \]

Plug in x_n and show the result

Substitute the calculated x_n into the expression for G(x) from Step 3: \[ G(x) = \sum_{n=1}^{\infty}(1-q)^{n-1} q F(x / F_n(x) / q) \] Now, using the definition of F'(x) = F(x/q), we can rewrite G(x) as: \[ G(x) = \sum_{n=1}^{\infty}(1-q)^{n-1} q F(x / F_n(x) / q) = \sum_{n=1}^{\infty}(1-q)^{n-1} q F_n(x / q) \] This is the result we wanted to show, so the resulting sequence of events indeed constitutes a renewal process with the given distribution function, G(x), also denoted as F(x;q).