Let \(X=(X_i)_{i=1}^{\infty}\) be an iid sequence of random variables. Let \(N\) be a random nonnegative integer that is independent of \(X\). Suppose \(X_1\) and \(N\) have finite expectations. Prove that \[ \mathbb{E} \big( \sum_{i=1} ^N X_i \big) = \mathbb{E}N \mathbb{E} X_1\]

Observe that

\[\sum_{i=1} ^N X_i = \sum_{n=0} ^{\infty}\sum_{i=1} ^n X_i\mathbf{1}[N=n].\]

Note that we take as convention that when \(n=0\), we have the empty sum, which is zero. Since \(N\) and the iid sequence \(X\) are independent, we have that

\[\mathbb{E}(X_i \mathbf{1}[N=n] ) = \mathbb{E}(X_i) \mathbb{E}(\mathbf{1}[N=n] ) = \mathbb{E}(X_i) \mathbb{P}(N=n).\] Thus taking expectations on both sides (and bring the expectation operator over an infinite sum) we have

\[ \mathbb{E} \big( \sum_{i=1} ^N X_i \big) = \sum_{n=0} ^{\infty}\sum_{i=1} ^n \mathbb{E}(X_i) \mathbb{P}(N=n) = \mathbb{E}(X_1)\sum_{n=0}n\mathbb{P}(N=n) = \mathbb{E}(X_1) \mathbb{E}(N),\]

as desired.

How would you do Exercise 4 analytically?

What would you do if there is more than two tiles, and the tiles did not occur with equal probability?

Use simulations to test your formula.

- Consider two tile types: a and b, of lengths \(a\) and \(b\), respectively. We can consider this an alternating renewal process: the working time is the time it remains in an a-tile, before switching to a b-tile, having service time, the time it remains in the b-tile. In order to compute the working time, we note that we spend time \(a\) on each time we get an a-tile; thus by the Baby Wald result the working time is given by \(aN_a\), where \(N_a\) is geometric with parameter \(p=\tfrac{1}{2}\). Similarly, the service time is given by \(bN_b\), and again \(N_b\) is just geometric with parameter \(p=\tfrac{1}{2}\). Applying the theorem on alternating renewal processes we have

\[\lim_{t \to \infty} \mathbb{P}(\text{that we are on an a-tile at time } t) = \frac{\mathbb{E}( aN_a) }{\mathbb{E}( aN_a) + \mathbb{E}( bN_b) } = \frac{a}{a+b}\]

- Consider the case of three tiles \(a\), \(b\), and \(c\) that occur with probabilities \(p_a + p_b + p_c =1\). Again, we can treat this as an alternating renewal process. The working time is \(aN_a\), where \(N_a\) is now a geometric \(1-p_a\). The service time is a bit more complicated, as remaining on a b-tile or a c-tile counts as service; the number of such tiles is again geometric \(1-p_b-p_c = p_a\), but we really do need to use the Baby Wald, as the tiles could be or length \(b\) or \(c\), with probabilites \(\tfrac{p_b}{p_b+p_c}\) and \(\tfrac{p_c}{p_b+p_c}\) respectively, giving an expecting service time of

\[\big( b \cdot \tfrac{p_b}{p_b+p_c} + c \cdot \tfrac{p_c}{p_b+p_c} \big) \cdot \tfrac{1}{p_a}.\]

Thus

\[\lim_{t \to \infty} \mathbb{P}(\text{that we are on an a-tile at time } t) = \frac{\tfrac{a}{1-p_a}} { \tfrac{a}{1-p_a} + (b \cdot \tfrac{p_b}{p_b+p_c} + c \cdot \tfrac{p_c}{p_b+p_c} \big) \cdot \tfrac{1}{p_a} }.\]

- We test this formula in Python, with \(a=1, b=\sqrt{2}, c=\pi\) and \(p_a = 2/9\), \(p_b=3/9\), and \(p_c = 4/9\). We check the tile type at \(t=300\), for a total of \(1000\) independent times.

```
import numpy as np
pa= 2/9
pb= 3/9
pc= 4/9
a=1
b=np.sqrt(2)
c=np.pi
def type():
x=a
u=np.random.uniform()
if (u > pa and u < pa+pb):
x=b
if (u > pa+pb):
x=c
return x
def tile(t):
ctile = type()
tlength = ctile
while(t > tlength):
ctile = type()
tlength = tlength + ctile
return ctile
y = [tile(300) for _ in range(1000) ]
freq = np.unique(y,return_counts = True)
print(freq)
```

`## (array([1. , 1.41421356, 3.14159265]), array([125, 228, 647], dtype=int64))`

```
tile1 = (1/1000)*freq[1][0]
tile1theory = (a/(1-pa)) / ( a/(1-pa) + (b*(pb/(pb+pc)) +c *(pc/(pb + pc)))*(1/pa) )
print(tile1 - tile1theory )
```

`## 0.018667988819657297`

Let \(\Pi\) be a Poisson point process on \([0, \infty)\). Pick a (large) number, say \(x=\sqrt{2} + 100\). Find the smallest interval \((A,B)\) such that \(x \in (A,B)\), and \(A\) and \(B\) are points of \(\Pi\), thus \(B\) is the next arrival after \(A\).

Find the distribution of \(S=B - A\).

Is \(S\) exponentially distributed? Explain.

Do simulations to confirm your findings.

Consider a Poisson process of intensity \(\lambda\)

- We already know from the memoryless property that \(B-x\) is still exponentially distributed with rate \(\lambda\). It is not hard to argue from reversibility that \(x-A\) is also exponentially distributed with rate \(\lambda\).

One can also think about generating a two-sided Poisson process on \((-\infty, -\infty)\), this can be accomplished by putting together two independent Poisson processes on \((0, \infty)\) glued at \(0\), where one is flipped.

We see that \(S\) has actually the sum of two independent exponentials.

We can also see this in the following Python code, where we take \(\lambda\)=1:

```
# Poisson arrivals just past the end time
def pois(end):
T= np.array([np.random.exponential()])
while(T[-1]<end):
T = np.append(T, T[-1] + np.random.exponential())
return T
def find(times,x):
n=0
while(x > times[n]):
n = n+1
return (times[n] - times[n-1])
print(find(pois(200), 100+np.sqrt(2) ))
```

`## 1.2188956938953623`

```
y = [find(pois(200), 100+np.sqrt(2) ) for _ in range(5000) ]
print(np.mean(y))
```

`## 2.0200540996767087`

```
import matplotlib.pyplot as plt
supress=plt.hist(y, bins = 100, density=True, label='Proability Histogram')
t=np.linspace(0,10,num=1000)
plt.plot(t,t*np.exp(-1*t), label='Exact Density')
plt.legend(loc='upper left')
plt.show()
```

See these notes

- Version: 29 November 2022
- Rmd Source