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.

  • $\begingroup$ 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. $\endgroup$ – sudeep5221 May 1 '20 at 19:04
  • $\begingroup$ @sudeep5221 Thanks. Do you have an idea for a concrete counter example for 2? $\endgroup$ – 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 $S_{n-1}$

  • $\begingroup$ Thanks. I managed to do 1 and 3, but I don't have a counter example for 2 $\endgroup$ – Daniel May 1 '20 at 19:30

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