{r setup, include=FALSE} knitr::opts_chunk$set(echo = TRUE)  {exercise, name="Integrals"} * Let$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, name="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, name="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](https://www.youtube.com/watch?v=Uk280jVuH1w), and we still owe the universe a *one minus*, so the desired conclusion follows.  {exercise, name="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-$0$.  {r} tile <- function(t){ types = rbinom(1,1,0.5) tlength = sqrt(2) * (1-types) + pi*types 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\$. {r} y = replicate(1000, tile(111)) mean(y)  You might guess that this probability is: {r} pi/(pi + sqrt(2)) 
* Version: r format(Sys.time(), '%d %B %Y') * [Rmd Source](https://tsoo-math.github.io/ucl/QHW6.1-sols.Rmd)