Previously I asked the following question, that does not generally hold:
The following question is very similar to this reference.
Let $(X_{1},X_{2},X_{3})$ be a random vector with continuous independent variable entries. We denote by $X_{1}^{1}, X_{1}^{2}$ two variables that are i.i.d. as $X_{1}$
Claim: $(X_{1}^{1}+ X_{1}^{2},X_{2}, X_{3})$ are again independent.
I would required this for a sum of $n$ i.i.d. variables, which would follow by induction.
Proposed proof:
Let $Z := X_{1}^{1}+ X_{1}^{2}$ and $Y := (X_{2}, X_{3})$.
In similar fashion to here we repeat the application of the convolution formula
\begin{equation} f_{z, y}(z, y) =f_{z \mid y}(z \mid y) \cdot f_{y}(y) =f_{y}(y) \cdot \int_{-\infty}^{+\infty} \int_0^{z-x_1^2} f_z\left(x_1^1, x_1^2 \mid y\right) d x_1^1 d x_1^2 = f_{y}(y) \cdot \int_{-\infty}^{+\infty} f_z\left(z-x_1^2, x_1^2|y\right) d x_1^2 \end{equation}
Here comes a step, where one requires that $X_{1}^{1}|Y$ is independent from $X_{1}^{2}| Y$. I am not certain whether this is actually satisfied
\begin{equation} f_{y}(y) \cdot \int_{-\infty}^{+\infty} f_z\left(z-x_1^2, x_1^2|y\right) d x_1^2 = f_{y}(y) \cdot \int_{-\infty}^{+\infty} f_{x_{1}^{1}} (z-x_1^2|y) f_{x_{1}^{1}} (x_1^2|y) d x_1^2 =f_{y}(y) \cdot \int_{-\infty}^{+\infty} f_{x_{1}^{1}} (z-x_1^2) f_{x_{1}^{1}} (x_1^2) d x_1^2 = f_{y} (y) f_{z} (z) \end{equation}
Are $X_{1}^{1}|Y$ and $X_{1}^{2}|Y$ conditionally independent ?
Where can I make use of the i.i.d. property ?
The answer was that this statement does not generally hold.
I cite the answer: This is not true. Let $X_1 \sim N(0,1)$, $X_2 \sim N(1,1)$, and $X_3 \sim N(-1,1)$ all be independent, so $(X_1, X_2, X_3)$ satisfies the assumptions of the problem. Observe that $X_1^1 := X_2 - 1$ and $X_1^2 := X_3 + 1$ are both $N(0,1)$ random variables and independent, so are i.i.d. as $X_1$. However, $(X_1^1 + X_1^2, X_2, X_3) = (X_2 + X_3, X_2, X_3)$ are not independent.
My question is:
- What condition would ensure for it to work ?
- When does it work ? Examples of distributions would help.