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.

• 1. If $X_1, X_2, \dots, X_{k-1}$ are independent of $X_k$ then any function of $X_1, X_2, \dots, X_{k-1}$ is independent of $X_k$. And yes using induction is a good idea. 2. Yeah, your direction is correct. Maybe just use some specific values and shows that the probability does not match for $2X_1 + X_2$ and $\text{Pois}(2 \lambda_1 + \lambda_2)$. 3. You are going in the correction direction. Use part 1. – sudeep5221 May 1 '20 at 19:04
• @sudeep5221 Thanks. Do you have an idea for a concrete counter example for 2? – Daniel May 1 '20 at 19:31

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 $$S_{n-1}$$