```{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[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 $\pi$. ```{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)