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