# Are these two distributions independent?

In a probability cheatsheet, there's the following claim:

That $$Z_1$$ and $$Z_2$$ should be independent seems very unintuitive to me. It seems to me that if $$Z_1$$ is high, then $$Z_2$$ is low, and vice-versa. Is there an error in this cheatsheet, and if not, what is wrong with my intuition?

• $Z_1$ will tend to be high in part because $Z$ can be high, in which cases $Z_2$ will simultaneously be high. For more intuition see stats.stackexchange.com/a/261926/919.
– whuber
Oct 19, 2019 at 19:10

Given $$Z$$, $$Z_1$$ and $$Z_2$$ are not independent because $$Z_1+Z_2=Z$$, i.e. there is no conditional independence. But, if we also don't know $$Z$$, which factor enforces a relationship between $$Z_1$$ and $$Z_2$$? e.g. there is no force on $$Z_2$$ being low when $$Z_1$$ is high because we also don't know their sum.
Mathematically (let $$z=z_1+z_2$$), \begin{align}P(Z_1=z_1,Z_2=z_2)&=P(Z_1=z_1,Z=z)\\&=P(Z_1=z_1|Z=z)P(Z=z)\\&={z\choose z_1}p^{z_1}(1-p)^{z-z_1}e^{-\lambda}\lambda^z/z!\\&=\frac{1}{z_1!z_2!}(\lambda p)^{z_1}(\lambda(1-p)^{z_2})e^{-\lambda p}e^{-\lambda(1-p)}\\&=P(Z_1=z_1)P(Z_2=z_2)\end{align}