With $X_1,...X_4 \stackrel{i.i.d}{\sim} Ber(p)$ find $\mathbb{P}(X_1=x_1,...,X_4=x_4|X_1+...+X_4=2)$

Question

My attempt: $\mathbb{P}(X_1,...,X_4|X_1+...+X_4=2)=\frac{\mathbb{P}(X_1,...,X_4,X_1+...+X_4=2)}{\mathbb{P}(X_1+...+X_4=2)}$

$\mathbb{P}(X_1+...+X_4)\sim Binom(4,p)\Leftrightarrow \mathbb{P}(X_1+...+X_4=2)= {4 \choose 2}p^2(1-p)^2$

And I dont know how to find the following. $\mathbb{P}(X_1=x_1,...,X_4=x_4,X_1+...+X_4=2)$

edit:

By drawing an event table of $X_1,...,X_4$ against $\sum X_i$ we see that for all the ${4 \choose 2}$ cases of two $\ 0s$ and two $1s$, each case have a probability of $p^2(1-p)^2$. Which is $\frac{1}{{4 \choose 2}}\mathbb{P}(\sum X_i=2)$.

Is this the answer for $\mathbb{P}(X_1=x_1,...,X_4=x_4|X_1+...+X_4=2)$?

Answer 1

Your work is essentially correct, but there is a missing case (specifically the case $x_1+x_2+x_3+x_4 \ne 2$).

$$P(X_1=x_1, \ldots, X_4 = x_4 \mid X_1 + \cdots + X_4 = 2) = \begin{cases} 1/\binom{4}{2} & x_1 + \cdots + x_4 = 2 \\ 0 & \text{otherwise}. \end{cases}$$

Answer 2

This is the classical example to introduce the definition of Sufficient statistics.

They asked you to prove that, Given $S=\Sigma_i X_i=s$, the conditional distribution given $S=s$ does not depend anymore on the parameter $p$.

As shown, the conditional distribution is uniform over the 6 possible cases

1. $1 1 0 0 $
2. $1 0 1 0$
3. $1 0 0 1$
4. $0 1 1 0$
5. $0 1 0 1$
6. $0 0 1 1 $

... and zero elsewhere.

In other words, all the information needed to do inference about $p$ are included in the statistic $S=\Sigma_i X_i$