Homework 6.1

Exercise 1 (Integrals)

Let $X \geq 0$ be a continuous random variable with finite first moment. Prove that $\mathbb{E} X = \int_0 ^{\infty} \mathbb{P}(X >t) dt = \int_0 ^{\infty}[ 1- F_X(t)]dt$ Hint: use a double integral.
Let $X$ and $Y$ be nonnegative independent continuous random variables. Prove that for $t >0$ , we have $\mathbb{P}(XY > t) = \int_0 ^{\infty} \mathbb{P}(X >\tfrac{t}{y}) f_Y(y) dy,$ where $f_Y$ is the probability density function for $Y$ .
Using the previous results prove that $\mathbb{E}( X Y) = (\mathbb{E} X )(\mathbb{E} Y),$ assuming all the expectations are finite.

Solution.

We have

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

Since $X$ and $Y$ are independent, we have $\begin{eqnarray*} \mathbb{P}(XY > t) &=& \int_0 ^{\infty} \int_{t/y} ^{\infty} f_X(x) f_Y(y) dxdy \\ &=& \int_0 ^{\infty} \mathbb{P}(X > t/y) f_Y(y) dy, \end{eqnarray*}$ as desired.
By the previous results, $\begin{eqnarray*} \mathbb{E} XY &=& \int_0 ^{\infty} \mathbb{P}(XY >t) dt \\ &=& \int_0 ^{\infty} \int_0 ^{\infty} \mathbb{P}(X > t/y) f_Y(y) dy dt \\ &=& \int_0 ^{\infty} f_Y(y) \Big [ \int_0 ^{\infty} \mathbb{P}(X > t/y) dt \Big]dy \\ &=& \int_0 ^{\infty} y f_Y(y) dy \cdot \int_0 ^{\infty} \mathbb{P}(X >t) dt \\ &=& \mathbb{E} Y (\mathbb{E} X). \end{eqnarray*}$

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.

By conditioning on the first arrival, show that $\mathbb{P}(E(t) >y) = \int_0 ^t \mathbb{P}(E (t-x) >y)dF(x) + \int_{t+y} ^{\infty} dF(x)$
Apply the general theorem on renewal equations to obtain that

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

Assuming the inter-arrivals are non-lattice type, apply the key
renewal theorem to obtain that

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

Solution.

Let $X_1, X_2, \ldots$ be the inter-arrival times. It is not difficult to see that with

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

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.

Thus with $\psi(t) = \mathbb{P}(E(t) >y)$ , we have

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

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.

For each fixed $y$ , we will apply the key renewal theorem on the function $k_y(x) = 1- F(x+y).$

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.

Suppose that I pick a large $t$ , is it equally likely that it would be covered the tile types?
Run a simulation to estimate the probability that $t$ is covered by tile of length $\pi$ .

Solution.

If we pick some large $t$ , from our experience with size-biasing, we know it is more likely that we will land in the longer tile of length $\pi$
The following code, excutes the tiling procedure up to time $t$ and returns the type of tile. We till label the $\pi$ tile as type- $1$ and the $\sqrt{2}$ tile as type- $0$ .

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)]
  }
  
  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))

mean(y)

## [1] 0.688

You might guess that this probability is:

pi/(pi + sqrt(2))

## [1] 0.68958

Version: 06 December 2020
Rmd Source