independence between one R.V. and its sum with an independent R.V.

Question

If I have two independent, poisson-distributed random variables $X$ and $Y$, and their sum, $ {X+Y}=Z $, is $X$ independent of $Z$? I know that $X$ would be independent of any function of numerous individual random variables of which it is independent, but I'm running into a problem applying this reasoning to a function that involves X itself.

Answer 1

Given that $Z=0$, then for sure $X=0$, since both $X$ and $Y$ are nonnegative. So knowledge of the value of $Z$ changes the distribution of $X$. This is one way to see that $X$ and $Z$ are not independent.

Answer 2

Let $X \sim Pois(3)$ and independently, let $Y \sim Pois(5).$ Then it is easy to see that $Z = X + Y \sim Pois(8).$ (One way is to use moment generating functions.) So $E(Z) = 8$ and $Var(Z) = 8.$

Also, $Cor(X, Y) = 0,$ by independence. But $Cor(X, Z) \approx 0.61$ (illustrated by simulation below), and so $X$ and $Z$ cannot be independent. You can start the exact evaluation of $Cor(X, Z)$ by finding $Cov(X, Z) = Cov(X, X + Y) = Var(X) + Cov(X,Y) = Var(X) = 3.$

Some of these results are approximated in the simulation below using R statistical software. With a million iterations most results will have 2 or 3 place accuracy.

 m = 10^6;  x = rpois(m, 3);  y = rpois(m, 5)
 z = x + y;  mean(z);  var(z)
 ## 7.997419      # aprx E(Z) = 8
 ## 8.002252      # aprx Var(Z) = 8
 cor(x, y);  cor(x, z)
 ## -0.001629232  # aprx Cor(X, Y) = 0
 ## 0.6113947     # aprx Cor(X, Z), not 0
 cov(x, z)
 ## 2.997578      # aprx Cov(X, Z) = 3

Below is a scatterplot of $(X,Z)$ with integer values slightly 'jittered' to prevent massive overplotting. It shows a clear pattern of association.

Finally notice that the plot shows, $P(X = 6) > 0$ and $P(Z = 3) > 0,$ but $P(X = 6, Z = 3) = 0.$ This alone is enough to to show lack of independence, and is obvious because neither $X$ nor $Y$ can be negative.

Answer 3

One simple proof that they are dependent involves showing that the conditional distribution of Z given X is not the same as the unconditional distribution. That is,

$P(Z \leq z | X=x) = P(x + Y \leq z) = P(Y \leq z-x) \neq P(Z \leq z)$.