Exercise 1 (Integrals)


\[\begin{eqnarray*} \mathbb{E} X &=& \int_0 ^{\infty} x f(x) dx \\ &=& \int_0 ^{\infty} f(x) \Big[ \int_0 ^x 1 dt \Big] dx \\ &=& \int_0 ^{\infty} \int_t ^{\infty} f(x) dx dt \\ &=& \int_0 ^ {\infty} \mathbb{P}(X > t) dt; \end{eqnarray*}\] here we note that the interchange in order of integration is permissible since everything is positive and \(\mathbb{E} X < \infty\).

Exercise 2 (Renewal equations) Refering the general theorem on renewal equations, show that if \(m\) be a renewal function with \(F\) as the cumulative distribution for the inter-arrival times, and \[ \phi = H + H*m,\] then \(\phi\) satisfies the renewal-type equation \[ \phi = H + \phi*F.\]

Solution. We will star both sides of the given equation by \(F\) to obtain \[ \phi* F = H*F + H*m*F = H *( F + m*F).\] Since we know from our first renewal equation that \[ m = F + m*F\] we obtain \[ \phi*F = H*m\] from which the desired result follows.

Exercise 3 (Excess life) With the usual notation, let \(E\) be the excess life of a renewal process with renewal function \(m\) and \(F\) for the cumulative distribution of the inter-arrival times.

\[ \mathbb{P}(E(t) \leq y) = F(t+y) - \int_0 ^t [1 - F(t+y -x)] dm(x).\]

\[\lim_{t \to \infty} \mathbb{P}(E(t) \leq y) = \frac{1}{\mu} \int_0 ^y [1-F(x)]dx.\]


\[\phi(x) = \mathbb{P}(E(t) > y | X_1 = x)\]

we have for \(t < x\), we have \(\phi(x)=0\) if \(x-t \leq y\) and \(\phi(x)=1\) if \(x-t >y\). If \(t \geq x = X_1\), then \(E(t)\) has the same law as \(E(t-x)\) and \(\phi(x) = \mathbb{P}(E(t-x))\). Thus

\[\mathbb{P}(E(t) >y) = \mathbb{E}\phi(X_1) = \int_0 ^t \mathbb{P}(E(t-x))dF(x) + \int_{t+y} ^{\infty} 1\cdot dF(x),\] as desired.

\[ \psi(t) = 1- F(t+y) + (\psi*F)(t).\]

Thus by our general theorem on renewal equations, we have

\[ \psi(t) = 1- F(t+y) + \int_0 ^t [1-F(t+y-x)] dm(x)\] and rearranging gives the desired result.

Since \(0 \leq 1- F \leq 1\) is non-increasing and

\[\begin{eqnarray*} \int_0 ^{\infty} k_y(x)dx &=& \int_0 ^{\infty} [1-F(x+y)] dx \\ &=& \mu - \int_0 ^y [1-F(x)] dx \\ &\leq& \mathbb{E}X_1 = \mu < \infty, \end{eqnarray*}\]

the key renewal theorem applies, and gives that

\[\begin{eqnarray*} \lim_{t \to \infty} \int_0 ^t [1-F(t+y-x)] dm(x) &=& \frac{1}{\mu} \int_0 ^{\infty} [1-F(x+y)]dx\\ &=& \frac{1}{\mu} \Big( \mu - \int_0 ^y [1-F(x)] dx \Big). \end{eqnarray*}\]

Now, remember, mind your surroundings, and we still owe the universe a one minus, so the desired conclusion follows.

Exercise 4 (Random tiles) I have two types of tiles, one of length \(\pi\) and another of length \(\sqrt{2}\). Suppose that I tile the half line \([0, \infty)\), via the following procedure, I pick one of two types of tiles with equal probability, then I can place it, starting at the origin. I continue this procedure indefinitly, and independently.


tile <- function(t){
  types = rbinom(1,1,0.5)
  tlength = sqrt(2) * (1-types[1]) +  pi*types[1]
  while(t > tlength){
    types = c(types, rbinom(1,1,0.5))
     tlength <- tlength +  sqrt(2) * (1-types[length(types)]) +  pi*types[length(types)]

Now for some large \(t\), we sample from our function a large number of times, and find the average number of times that \(t\) land in a tile of length \(\pi\).

y = replicate(1000, tile(111))

## [1] 0.688

You might guess that this probability is:

pi/(pi + sqrt(2))
## [1] 0.68958