What allows us to divide a random variable into multiple ones?

Question

I can't wrap my head around the solution presented for this problem:

Suppose a trial has a success probability $p$, let $X$ be the random variable for the number of trials it takes to stop at $r$ successes. Calculate $E[X]$.

The solution starts by letting $X=G_1+G_2+...+G_r$, where $G_i$ are geometric random variable; and this is where I do not understand. Intuitively it makes sense because the linguistic translation for both sides is the same, but the book I am reading does not provide a mathematical reasoning.

Can someone please tell me the mathematical structure that underlies the division of a random variable? I am trying to learn concepts from an intuitive point of view, as well as a firm and rigorous mathematical one. For this, the intuition is down, but I need the math.

Thank you very much.

Answer 1

A geometric random variable $G_i$ is the number of Bernoulli trials needed to get one success. Your variable $X$ is the number of trials needed to get $r$ successes, and so if these trials are carried out one after another, you must first have success $1$, success $2$, $\ldots$, success $r$; the number of trials to get success $1$ is $G_1$, the number of trials after the first success up to and including success $2$ is $G_2$, etc. Therefore, the total number of trials $X$ performed to get $r$ successes can be written as $X = G_1 + \cdots + G_r$.

Note that the geometric distributions are the memoryless discrete probability distributions, so that after success $1$, the number of trials until success $2$ is a random variable $G_2$ that is independent of and distributed identically to $G_1$; the $G_i$ are i.i.d. random variables.

Answer 2

We use it because Expectation is Linear.   That is the property that means the expectation of a sum of random variables is the sum of the expectations of the variables.

So if we can express the random variable $X$ as a series of random variables with easier to find expectations, the we can find the expectation of $X$ easier.

$$\begin{align} \mathsf E(X) & = \mathsf E(G_1+G_2+\cdots+G_r) \\ & = \mathsf E(G_1)+\mathsf E(G_2)+\cdots+\mathsf E(G_n) \end{align}$$

Now if $X$ is the count of trials until $r$ successes, and $\{G_1,G_2,\ldots,G_n\}$ is the series of counts of trials until and including each subsequent success (so $G_2$ is the count of trials after the first success until the second, et cetera), then we have our series: $X=G_1+G_2+\cdots+G_n$.

The count of trials until the success number $r$ is the sum of counts of trials between and until each subsequent success, up till success number $r$.

The count of (Bernoulli) trials until a success is a Geometrically Distributed Random Variable (by definition), and the expected value of a Geometrical Distribution of parameter $p$ is known to be $1/p$.

So...

Answer 3

$X(\omega)$ takes on the value $k$ if and only if we reach $r$ successes for the first time after $k$ tries.

$G_i(\omega)$ takes on the value $k_i$ if and only if we reach $1$ successes for the first time after $k_i$ tries

You want to show that $X(\omega) = k \iff G_1(\omega) + G_2(\omega) + \dots + G_r(\omega) = k_1 + k_2 + \dots + k_r = k$

That is, for every atomic event $\omega$ the two random variables $X(\omega)$ and $G_1(\omega) + G_2(\omega) + \dots + G_r(\omega)$ are equal.

Note that this is no different when we write in real analysis $f(x) = g(x) + h(x)$; for every $x$, we have equality between the two sides. Indeed, random variables are functions :)

The proof of why $X(\omega) = G_1(\omega) + \dots + G_r(\omega)$ for every $\omega$ is quite easy.. Can you do it? Just follow the intuition :)