# 1 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 $$1$$.

## 1.1 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.

g <- function(x){
y=  (x^2) * (sin(x))
y
}
x <- rexp(10000,1)
mean(g(x))
## [1] 0.4828766

# 2 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 $$0.27$$.

## 2.1 Solution

The isFour function checks if a sequence of bits contains a run of four heads.

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.

num=replicate(1000, isFour(rbinom(20,1,0.5))  )
mean(num)
## [1] 0.488

# 3 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 $$\mathbb{E} T$$.

## 3.1 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 $$\mathbb{E} T = e$$.

## 3.2 Solution by R

In R it is easy!

T <-function() {
n=0; s=0
while (s <1) {s <- s + runif(1); n <- n+1}
n
}
z=replicate(1000, T())
mean(z)
## [1] 2.677