Let's say $\{X_1,X_2,X_3,...\}$ is a set of positive IID random variable has distribution on a general distribution F
The summation of these random variable is a renewal sequence
$$S_n=X_1+X_2+X_3+...+X_n$$
Let N(t) define as a counting process that $ N(t)=\sum_{n=1}^{\infty}\mathbf{1}_{(0,t]}(S_n)$.
We would like to find the probability that $P[N(t) \text{ is odd}]$
My reasoning for this problem is define a new $N'(t)=\sum_{n=1}^{\infty} \mathbf{1}_{(0,t]}(S_{2n-1})$ but I have no idea how to derive the probability out of $N'(t)$ since the distribution of $X$ is general. My guess is that $P(N'(t))$ will be some kind of renewal type equation.
This is a question that I think is a general version for the splitting event of Poisson process discussed before.
Update 1
I tried to borrow the idea from Poisson process splitting problem. The interarrival time for the new counting process $N'(t)$ is $X_i+X_{i+1}$ which is the convolution of the distribution $F(t)*F(t)=F^{2*}$. If we know about the new interarrival distribution, can we derive the probability out of it?
Update 2
I think I come up a solution but I would appriciate if others could check my answer.
For a renewal process $$N(t) \ge n \Leftrightarrow S_n \leq t$$ (I do confuse about this relation but it appears in many stochastic process textbooks)
Therefore
$$ \begin{align} P[N(t) = n] &= P[N(t)\ge n]-P[N(t) \ge n+1] \\ &=P[S_n \le t]-P[S_{n+1} \le t] \end{align} $$
Thus
$$ \begin{align} P(N(t) \text{ is odd}) &= \sum_{n=1}^{\infty}\{P(S_{2n-1}\le t) - P(S_{2n}\le t)\} \\ &=\sum_{n=1}^{\infty}(F^{(2n-1)*}-F^{(2n)*}) \end{align} $$
Here, $F^{n*}$ is n-fold convolution for the distribution F (from sum of random r.v)