Stochastics Applied to Communications Engineering I'm struggling with a communications math problem.
A stream of data is transmitted. For the first transmitted packet, the packet error rate (PER) is $0.1$, but for every following transmission, the PER will be less. I have to figure out the necessary PER for an average number of transmission attempts of $1.105$ . This includes the successful one. 
My intuition is the following: 
$N = 0.9 * 1 + 0.1*(1-PER) * 2 + 0.1*PER*(1-PER)*3 + 0.1*PER*(1-PER) *4 \cdots $
$N =$ average number of transmission attempts
This, I think, will yield the correct result. But apparently, and I just can't figure out why, this is equivalent to
$N = 1 + 0.1 + 0.1*PER + 0.1*PER*PER + \cdots = 1 + 0.1(1+p+p^2 + \cdots)$
Can anyone please explain this to me? What's the idea behind it? My stochastics is not the best.
 A: I've figured out your question. Let's start with the meaning of the first formula and let's tidy up your notation a little bit. What you're trying to do is to transmit a packet without error. And you keep trying until it is transmitted errorless. Since the packet error rate or $\text{PER}$ is $0.1$, you need a few tries. The probability of succeeding after $k$ tries is
$$P[N=k]=(1-\text{PER})\cdot\text{PER}^{k-1}$$
Indeed, if you succeed in $k$ tries, that means you have failed $k-1$ times to transmit the package and finally succeeded the $k$th time. Hence this formula for the probability. This is known as a geometric distribution in statistics and probability theory.
Then, we want to know the average number of tries
$$E[N]=\sum_{k=1}^{\infty} k\cdot P[N=k] = \sum_{k=1}^{\infty} k \cdot (1-\text{PER})\cdot\text{PER}^{k-1} \\ = 1\cdot (1-\text{PER}) + 2 \cdot (1-\text{PER}) \cdot \text{PER} + 3 \cdot (1-\text{PER}) \cdot \text{PER}^2 + \ldots$$
This is exactly your first formula, derived from the definition of average or expected number of tries for the geometric distribution.
But let's look at it from another viewpoint. Imagine step by step, we look at what the average can be. Well, you have to do at least one try, so
$$E[N]=1+\ldots$$
But that try might fail, with probability $\text{PER}$, so with probability $\text{PER}$ you need to make another try, so on average $\text{PER}$ new tries extra
$$E[N]=1+\text{PER}+\ldots$$
But that new try might also fail, thus you need to make a new try again with probability $\text{PER}$, but since the previous step was already on average $\text{PER}$ you now add $\text{PER}^2$
$$E[N]=1+\text{PER}+\text{PER}^2+\ldots$$
Etc... This will get you your second formula. This can be however condensed in a neat way:
Since your first try will either succeed, either fail, in the case you succeed, you'll make $1$ try in total and that's it. But in the case you fail, you make $1$ try and then you're still nowhere, you have to keep trying. But since the errors are independent of each other (or at least they are assumed to be), at that point, on average the number of extra tries you'll have to do is exactly the same as if you hadn't already done a try. Thus, we can resume this as follows:
$$E[N] = (1-\text{PER}) \cdot 1 + \text{PER} \cdot (1+E[N])$$
and you can work this out to be
$$E[N] = 1+ \text{PER} \cdot E[N]$$
If you keep susbtituting back the formula into itself, you get your second formula. If however you just rearrange the equation to take out $E[N]$ in function of $\text{PER}$ you get
$$E[N] = \frac{1}{1-\text{PER}}$$
And this is probably the most compact and convenient of all the formulas. Maybe you'll remember this from Taylor series courses.
