Randomly Stopped Sums vs the sum of I.I.D. Random Variables

[CGandC]

I've came across the two following theorems in my studies of Probability Generating Functions:

Theorem 1:
Suppose $X_1, ... , X_n$ are independent random variables, and let $Y = X_1 + ... + X_n$. Then,
$G_Y(s) = \prod_{i=1}^n G_{X_i}(s)$

Theorem 2:
Let $X_1, X_2, ...$ be a sequence of independent and identically distributed random variables with common PGF $G_X$. Let $N$ be a random variable, independent of the $X_i$'s with PGF $G_N$, and let $T_N = X_1 + ... + X_N = \sum_{i=1}^N X_i$. Then the PGF of $T_N$ is:
$G_{T_N}(s) = G_N (G_X(s))$

Question:
I don't understand the difference between these two theorems.
From reading here: https://stats.stackexchange.com/que...topped-sums-vs-the-sum-of-i-i-d-random-variab
I understand that in first theorem $n$ is a number that we know so we know how many $X_i$ will appear in the sum in $Y$.
But in the second theorem $N$ is a random variable so we don't know how many $X_i$ will appear in the sum $Y$.

But I still don't fully understand.

the proof for the first theorem goes as follows:
$G_Y(t) =G_{X_1+X_2+\ldots+X_n}(t)=\mathbb{E}\left[t^{X_1+X_2+\ldots+X_n}\right]=\mathbb{E}\left[\prod_{i=1}^n t^{X_i}\right]=\prod_{i=1}^n \mathbb{E}\left[t^{X_i}\right]=\prod_{i=1}^n G_{X_i}(t)$

Then I tried to prove the second theorem using exactly the same proof as follows:
$G_Y(t) =G_{X_1+X_2+\ldots+X_N}(t)=\mathbb{E}\left[t^{X_1+X_2+\ldots+X_N}\right]=\mathbb{E}\left[\prod_{i=1}^N t^{X_i}\right]=\prod_{i=1}^N \mathbb{E}\left[t^{X_i}\right]=\prod_{i=1}^N G_{X_i}(t)$
this proof is specious, but I don't understand why. I mean, the number of $X_i$ 's that will be multiplied by each other is determined by $N$ ,even if we don't know it, so I don't understand what's the problem.Thanks in advance for any help!

 

[pasmith]

I've came across the two following theorems in my studies of Probability Generating Functions:

Theorem 1:
Suppose $X_1, ... , X_n$ are independent random variables, and let $Y = X_1 + ... + X_n$. Then,
$G_Y(s) = \prod_{i=1}^n G_{X_i}(s)$

Theorem 2:
Let $X_1, X_2, ...$ be a sequence of independent and identically distributed random variables with common PGF $G_X$. Let $N$ be a random variable, independent of the $X_i$'s with PGF $G_N$, and let $T_N = X_1 + ... + X_N = \sum_{i=1}^N X_i$. Then the PGF of $T_N$ is:
$G_{T_N}(s) = G_N (G_X(s))$

Question:
I don't understand the difference between these two theorems.
From reading here: https://stats.stackexchange.com/que...topped-sums-vs-the-sum-of-i-i-d-random-variab
I understand that in first theorem $n$ is a number that we know so we know how many $X_i$ will appear in the sum in $Y$.
But in the second theorem $N$ is a random variable so we don't know how many $X_i$ will appear in the sum $Y$.

But I still don't fully understand.

the proof for the first theorem goes as follows:
$G_Y(t) =G_{X_1+X_2+\ldots+X_n}(t)=\mathbb{E}\left[t^{X_1+X_2+\ldots+X_n}\right]=\mathbb{E}\left[\prod_{i=1}^n t^{X_i}\right]=\prod_{i=1}^n \mathbb{E}\left[t^{X_i}\right]=\prod_{i=1}^n G_{X_i}(t)$

Then I tried to prove the second theorem using exactly the same proof as follows:
$G_Y(t) =G_{X_1+X_2+\ldots+X_N}(t)=\mathbb{E}\left[t^{X_1+X_2+\ldots+X_N}\right]=\mathbb{E}\left[\prod_{i=1}^N t^{X_i}\right]=\prod_{i=1}^N \mathbb{E}\left[t^{X_i}\right]=\prod_{i=1}^N G_{X_i}(t)$
this proof is specious, but I don't understand why. I mean, the number of $X_i$ 's that will be multiplied by each other is determined by $N$ ,even if we don't know it, so I don't understand what's the problem.Thanks in advance for any help!

\prod_{i=1}^N G_{X_i}(t) = (G_{X_1}(t))^N is a random variable: it's a function of N. To find \mathbb{E}(t^Y) you need to remove this dependence on N by using conditional expectation: <br /> \begin{split}<br /> \mathbb{E}(t^{Y}) &amp;= \sum_{n=1}^\infty \mathbb{E}(t^{X_1 + \dots + X_N} | N = n)\mathbb{P}(N = n) \\<br /> &amp;= \sum_{n=1}^\infty \mathbb{E}(t^{X_1 + \dots + X_n})\mathbb{P}(N = n) \end{split}

 

[CGandC]

Ahh! that makes sense, thank you a lot!