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.
\[\begin{eqnarray*} \mathbb{E} X &=& \int_0 ^{\infty} x f(x) dx \\ &=& \int_0 ^{\infty} f(x) \Big[ \int_0 ^x 1 dt \Big] dx \\ &=& \int_0 ^{\infty} \int_t ^{\infty} f(x) dx dt \\ &=& \int_0 ^ {\infty} \mathbb{P}(X > t) dt; \end{eqnarray*}\] 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, \[\begin{eqnarray*} \mathbb{E} XY &=& \int_0 ^{\infty} \mathbb{P}(XY >t) dt \\ &=& \int_0 ^{\infty} \int_0 ^{\infty} \mathbb{P}(X > t/y) f_Y(y) dy dt \\ &=& \int_0 ^{\infty} f_Y(y) \Big [ \int_0 ^{\infty} \mathbb{P}(X > t/y) dt \Big]dy \\ &=& \int_0 ^{\infty} y f_Y(y) dy \cdot \int_0 ^{\infty} \mathbb{P}(X >t) dt \\ &=& \mathbb{E} Y (\mathbb{E} X). \end{eqnarray*}\]
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
\[ \mathbb{P}(E(t) \leq y) = F(t+y) - \int_0 ^t [1 - F(t+y -x)] dm(x).\]
Solution.
\[\phi(x) = \mathbb{P}(E(t) > y | X_1 = x)\]
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.
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] 0.688You might guess that this probability is:
pi/(pi + sqrt(2))## [1] 0.68958