Assume we have two independent random variables $X_1$, $X_2$ with values in the set $Z_2 = {0,1}$. Prove that if $X_2$ has a uniform distribution then $X_1 \oplus_2 X_2$ has also the uniform distribution. (Used in "coin tossing by phone").
I know from a theorem that this is true and that the uniform probability distribution will be $P(y=0)=1/2$ and $P(y=0)=1/2$. But I don't know how to prove it. Any idea?
Thanks for your help.
Let $\Pr(X_1=0)$ be $p$, so $\Pr(X_1=1)=1-p$.
The event $X_1\oplus X_2=0$ can occur in two possible disjoint ways: (i) $X_1=0$ and $X_2=0$ or (ii) $X_1=1$ and $X_2=1$.
The probability of (i) is $(p)(1/2)$, and the probability of (ii) is $(1-p)(1/2)$.
If we add them, we get: $$\Pr(X_1\oplus X_2=0)=(p)(1/2)+(1-p)(1/2)=1/2.$$
Since the only possible values of $X_1\oplus X_2$ are $0$ and $1$, it follows that $\Pr(X_1\oplus X_2)=1/2$.