```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```


# Integration

Approximate the integral 
$$ \int_0 ^{\infty} \sin(x) x^2 e^{-x} dx$$ by appealing the law of large numbers and using R.  Hint:  Consider an i.i.d.\ sequence of exponential random variables all with rate $1$.

## Solution

Let $X_i$ be i.i.d.\ exponential random varaibles with rate $1$.  By the change of variables formula, we have
$$\mathbb{E} g(X_1) = \int_0 ^{\infty} g(x) e^{-x}dx;$$
take $$g(x) = \sin(x) x^2.$$
Since $g(X_i)$ are independent are also i.i.d.\ the law of of large numbers gives
$$ n^{-1}[g(X_1) + \cdots+ g(X_n) ] \to \mathbb{E} g(X_1).$$
This is easily simulated in R.
```{r}
g <- function(x){
y=  (x^2) * (sin(x))
y
}
x <- rexp(10000,1)
mean(g(x))
```







#  Grouping coins

Again, consider a sequence of 20 fair coin flips, as discussed in our first live session.    using $R$, estimate the probability that we will see a run of at least four heads.  Hopefully, we get a number bigger than $0.27$.


## Solution

The *isFour* function checks if a sequence of bits contains a run of four heads.
```{r}
isFour <- function(x){
  ans=0
  for (i in 1:(length(x)+1-4))
    {
  if ( x[i]==1 && x[i+1]==1 && x[i+2]==1 && x[i+3]==1 ){ans<- 1}
  }
  ans
  }
```  
Next, we generate many sequences of fair coin flips, and compute the average number of times a run of four heads occurs.
```{r}
num=replicate(1000, isFour(rbinom(20,1,0.5))  )
mean(num)
```

    
  



#  Pen and paper?

Let $(U_i)_{i \in \mathbb{Z} ^+}$ be a sequence of independent random variables that are uniformly distributed in $[0,1]$.  Let 
$$S_n = X_1 + \cdots + X_n.$$
  Let $$T = \inf\{n \geq 1:  S_n >1\}$$
  so that $T$ is the first time the sum is greater than $1$.  Use R or pen and paper  to compute $\mathbb{E} T$.


## Solution by pen and paper

Let $a \in [0,1]$.  We first show that $$\mathbb{P}(U_1 + \cdots +U_n \leq a ) = \frac{a^n}{n!}.$$
We proceed by induction.  The result is obvious in the case $n=1$.  Assume the result for $n \geq 1$.    Set $Z = U_1 + \cdots +U_n$.  Note that $Z$ is independent of $U_{n+1}$.  
By the induction hypothesis we know the density of $Z$ (at least up to value $a=1$).  Hence we have that 
$$P(Z + U_{n+1} \leq a  ) = \int_0^a\int_0 ^{a-z}  \frac{z^{n-1}}{(n-1)!} du dz,$$
and the inductive step follows from an easily calculation.

Next, we show that for any integer-valued random variable $X \geq 0$, we have
$$\mathbb{E} X = \sum_{n=1} ^{\infty} \mathbb{P}(X \geq n).$$
 Note that 
 $$X = \sum_{i=1} ^{\infty} \mathbf{1}[X \geq i],$$
 so that
 $$
 \begin{eqnarray*}
 \mathbb{E} X &=& \sum_{i=1} ^{\infty} \mathbb{E}  \mathbf{1}[X \geq i] \\
 &=& \sum_{i=1} ^{\infty} \mathbb{P} (X \geq i)
 \end{eqnarray*}
$$
Finally, we note that
$$\{T \geq n\} = \{ U_1 + \cdots +U_{n-1} < 1\}.$$
So, we can now compute $\mathbb{E} T = e$. 	

## Solution by R

In R it is easy!
```{r}
T <-function() {
n=0; s=0
while (s <1) {s <- s + runif(1); n <- n+1}
n
}
z=replicate(1000, T())
mean(z)
```

#   Version: `r format(Sys.time(), '%d %B %Y')`
  * [Rmd Source](https://tsoo-math.github.io/ucl/QHW1-sol.Rmd)

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