Let $X_0$, $X_1$ and $X_2$ be three mutually independent random variables. We define two more random variables $D_1$ and $D_2$ as follows: $$D_1 = X_1 + X_0\\[1ex] D_2 = X_2 + X_0$$ We're interested in arguing (in the most effective and simple way) if
- $D_1$ and $D_2$ are independent
- $D_1|X_0=x_0$ and $D_2|X_0=x_0$ are independent
If $D_1$ and $D_2$ are independent, it must be $\text{Cov}(D_1,D_2)=0$:
$$\text{Cov}(D_1, D_2)=\mathbb{E}[D_1 D_2]-\mathbb{E}[D_1]\mathbb{E}[D_2] =\\ =\mathbb{E}\left[ (X_1+X_0)(X_2+X_0) \right] - \mathbb{E}\left[ X_1+X_0 \right]\mathbb{E}\left[ X_2+X_0 \right]=\\ =\mathbb{E}\left[ X_1 X_2 + X_1 X_0 + X_0 X_2 + X_0^2 \right] - \mathbb{E}\left[ X_1 \right]\mathbb{E}\left[ X_2 \right] - \mathbb{E}\left[ X_1 \right]\mathbb{E}\left[ X_0 \right] - \mathbb{E}\left[ X_0 \right]\mathbb{E}\left[ X_2 \right] - \left(\mathbb{E}\left[ X_0 \right]\right)^2=\\ =\mathbb{E}\left[ X_1 \right]\mathbb{E}\left[ X_2 \right] + \mathbb{E}\left[ X_1 \right]\mathbb{E}\left[ X_0 \right] + \mathbb{E}\left[ X_0 \right]\mathbb{E}\left[ X_2 \right] + \mathbb{E}\left[ X_0^2 \right] - \mathbb{E}\left[ X_1 \right]\mathbb{E}\left[ X_2 \right] - \mathbb{E}\left[ X_1 \right]\mathbb{E}\left[ X_0 \right] - \mathbb{E}\left[ X_0 \right]\mathbb{E}\left[ X_2 \right] - \left(\mathbb{E}\left[ X_0 \right]\right)^2=\\ =\mathbb{E}\left[ X_0^2 \right] - \left(\mathbb{E}\left[ X_0 \right]\right)^2 = \text{Var}(X_0)\neq 0$$
So, if not for the trivial case in which $X_0 = \text{constant}$, $\mathbf{D_1}$ and $\mathbf{D_2}$ are not independent.
2. One could do the same thing of point 1. all over again or just notice that this case is exactly the trivial one in which $X_0 = x_0$, a fixed "constant". So, $\mathbf{D_1|X_0 =x_0}$ and $\mathbf{D_2|X_0 =x_0}$ are independent.
MY QUESTION
Are my proofs correct? Is there a simpler way to prove the same thing or even just argument the same results in a better, more intuitive way?
A BONUS QUESTION
How can I test these theoretical results with the following example? (In order to really "see" what I found)
$X_0$ is the outcome of the fair coin "COIN": $X_0=1$ if Heads, $X_0=0$ if Tails.
$X_1$ is the outcome of the 6-faced fair dice "DICE1": $X_1=\{1,2,3,4,5,6\}$
$X_2$ is the outcome of the 6-faced fair dice "DICE2": $X_2=\{1,2,3,4,5,6\}$
SOMETHING THAT PERPLEXES ME
Intuitively and "constructively", I would say that the event to observe $D_1=k$ is independent from that of observing $D_2 = h$, infact the first is the event of obtaining $X_1 + X_0 = k$ with a throw of DICE1 and a toss of COIN while the second is the event of obtaining $X_2 + X_0 = k$ with a throw of DICE2 and a different toss of COIN: these two "procedures" are totally independent from each other and so should be their probabilities (?)
But this contradicts my analytical results, doesn't it! Why is that?