I have $n$ independent random variables: $X_1, ... , X_n$, and $X_i ∼ \operatorname{Pois}(\lambda_i)$ for $i=1, ... , n$.
Defined a new random variable $S_n=\displaystyle\sum_{i=1}^n X_i$
I have the following question:
1) Prove that $S_n ∼ \operatorname{Pois}(\displaystyle\sum_{i=1}^n \lambda_i)$, for any $n\ge1$
2) Prove/give counter example for the following statement: $(2X_1+X_2)∼ \operatorname{Pois}(2\lambda_1+\lambda_2)$
3) Find the distribution of $(X_1\mid S_n=s)$ (The distribution of $X_1$ given that $S_n=s$).
For 1), I thought to use induction (because I know that it's true for one or two independent variables), but I don't know how to show why $X_k$ and $\displaystyle\sum_{i=1}^{k-1} X_i$ are independent.
For 2), I think it's wrong, because it's a sum of dependent variables $(X_1 + X_1+X_2)$, but I don't have any counter example.
And for 3), I started it, but got stuck and didn't get a known distribution:
$$P(X_1=x_1\mid S_n=s)=\frac{P(X_1=x_1, S_n=s)}{P(S_n=s)}=\frac{P\Bigl(X_1=x_1, \displaystyle\sum_{i=2}^{n} X_i=s-x_1\Bigr)}{P(S_n=s)}=\frac{P(X_1=x_1)\,P\Bigl(\displaystyle\sum_{i=2}^{n} X_i=s-x_1\Bigr)}{P(S_n=s)}$$
And then I used the formula for the poisson distribution.
For question 1, using induction is straightforward. Since you already know that $X_{12} = X_1 + X_2$ follow a Poisson distribution with the parameter $\lambda_{12}\lambda_1 + \lambda_2$, now when we have new independent variable, say, $X_3$ added to $X_{12}$, by induction, the parameter becomes $\lambda_3 + \lambda_{12}$ which is nothing but $\lambda_3 + \lambda_1 + \lambda_2$.
For question 2, we can intuitively say the result does not hold. This is because we are summing random variables which are not independent i.e $X_1, X_1$ and $X_2$ are NOT independent. So, result from question 1, is not expected to be valid.
For question 3, as pointed out in the comments, use the result from question 1 to compute the probability of $S_n$ and $S_{n-1}$