Random Sample of Bernoulli distribution

Question

Suppose we have a random sample from Bernoulli(p) distribution of size n

let $~~~~Y_1=X_1+X_2$ then $Y_1$~ $b(2,p)$ exact distribution

let $Y_2 = (X_1+X_2)/2$. Is $Y_2$~$b(2,p)$ exact distribution?

Answer 1

We can use a straight transformation to check this.

Let $X$ ~ $(2,p)$, and let $Y = \frac{X}{2}$. Suppose $0 < p < 1$. Then the pmf for $X$ is $${{2}\choose{x}} p^x (1-p)^{2-x}, x = 0, 1, 2$$ and $0$ otherwise.

Since $Y = \frac{X}{2}$, $X = 2Y$. $0 < p < 1$ here as well. Substituting $2y$ in for $x$ and adding the appropriate support gets the pmf for $Y$:

$$ {{2}\choose{2y}} p^{2y} (1-p)^{2-2y}, y = 0, 0.5, 1$$ and $0$ otherwise.

It is easy to see that these pmfs are not the same. Just compare $p_x(1)$ and $p_y(1)$. So $X$ and $Y$ can't have the same distribution, since pmfs uniquely determine the distribution of a discrete random variable.