How to prove $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)\cdot ...\cdot P(X_n=x_n)$?

Question

Suppose $n>1$ and $X_1,X_2,...,X_n$ are $n$ independent random variables (say with expotentional distribution) with $X=\sum_{i=1}^n X_n$.

Then $X_1,X_2,...,X_n,X$ are mutually independent random variables.

Thus according to the theorem (https://en.wikipedia.org/wiki/Joint_probability_distribution#Additional_properties) $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)\cdot ...\cdot P(X_n=x_n)\cdot P(X=x)$.

However, the answer was $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)\cdot ...\cdot P(X_n=x_n)$.

How could that be?

Answer 1

Random variable X is dependent on the other random variables.

$\because$ Take an example, Suppose $X_i = i\ \forall i\in\{1,2,...,n\}$.Then we can completely tell whether for a particular n-dimensional vector $x, X=x$ or not.

$\therefore$ The random variable X is not independent to other random variables,however, the result is true for $X_i's$ separately,i.e. $$P(X_1=x_1,X_2=x_2,...,X_n=x_n) = \prod_{i=1}^nP(X_i=x_i)$$

Hope it helps:)