I'm currently trying to figure out a proof given in a elementary probability lecture, involving a trick using the conditional probability.
$\mathbf {Assumptions}:$
a) $N$ is a random variable with support $\mathbb{N}_0$.
b) $X_1, X_2,...$ are i.i.d. random variables.
c) $N, X_1, X_2,...$ are independend.
$\mathbf {Problem}:$ I'm citing the proof in the lecture notes. The goal is to rewrite $$P(X_1 + ...+X_N \leq t) = \sum_{n=0}^{\infty}P(X_1+...+X_N \leq t, N = n)=\sum_{n=0}^{\infty}P(X_1+...+X_N \leq t| N = n) P(N=n)$$ $$=\sum_{n=0}^{\infty}P(X_1+...+X_n \leq t) P(N=n)$$
I understand why the first and the second equation holds, but I'm a bit confused about the third equation, i.e. mainly the step $P(X_1+...+X_N \leq t| N = n) = P(X_1+...+X_n \leq t)$ is what confuses me the most. Intuitively it totally makes sense to me but I'm having problems proving this statement rigorously.
$\mathbf {My Approach}:$ I tried to rewrite $$P(X_1+...+X_N \leq t| N = n) = \frac{P(X_1+...+X_N \leq t, N = n)}{P(N=n)} =\frac{P(X_1+...+X_n \leq t, N = n)}{P(N=n)} = \frac{P(X_1+...+X_n \leq t) P(N = n)}{P(N=n)} = P(X_1+...+X_n \leq t)$$ In the second equation I used that $\{\omega \in \Omega: X_1+...+X_N \leq t, N = n\} = \{\omega \in \Omega: X_1+...+X_n \leq t, N = n\}$ (Which holds since, we use the intersection with $N=n$). The first equation is just the definition of the conditional probability and the third equation I used the independence of $N, X_1, X_2,...$ Is this proof correct, or am I missing something?
$\mathbf {Question}:$
a) Is this proof above correct, or am I missing something?
b) The second step in the proof cited in the section "Problem" is not really needed, or am I wrong?
c) The equation $P(X_1+...+X_N \leq t| N = n) = P(X_1 + ... + X_n \leq t)$ does only hold if $N, X_1, X_2,...$ are independent right? Is there a way to rewrite $P(X_1 + ... + X_N \leq t)$ if the independence does not hold?
Thanks in advance! Sorry if this question is kind of dumb...just wanted to be sure.