Birthday $pmf$ and joint $pmf$ for $n$ people 
Let a discrete r.v be denoted $X_1,..,X_n$ denote the birthdays of $n$ people in a room. Assume that $X_1,..,X_n$ are mutually independent and that $X_i$ is a distribution such that $X_i \sim U(\left \{ 1,2,...,365 \right \})$ for all $i \in {(\left \{ 1,...,n\right \}}$.


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*Find and write out the pmf $p_i: \mathbb{R} \rightarrow \mathbb{R}$ for the r.v $X_i$.

*Find the joint pmf $p:\mathbb{R}^n \rightarrow \mathbb{R}$ for the vector $(X_1,..,X_n)$ and show that it is a valid joint pmf.

My try
First: The pmf would just be the pmf of a uniformal distribution. That would mean that $p_i(x)={x_i}^{-1}$ for the $x=1,...,n$. (Now the notation is weird for me as it is for the $i$'th $X_i$). (I know that $p_i: \mathbb{R} \rightarrow \mathbb{R}$)
Second: I know that for it to be a valid pmf the total probability has to sum up to $1$. The joint pmf should be denoted as $p_{X_1,...,X_n}(x_1,...,X_n):=P(X_1=x_1,...,X_n=x_n)$ but no idea on how to write this out from my results. Wouldnt I just have to replace $x_i$ with $1/365$? (I know that $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$)
 A: It was claimed that $X_i\sim\mathcal U\{1..365\}$ for all $i\in\{1..n\}$.

The probability mass function $p(u)$ for a uniform discrete distribution of $\mathcal U\{a{.}{.}b\}$, is $$p(u)=(b-a+1)^{-1}\mathbf 1_{u\in\{a..b\}}$$

So for any $i\in\{1..n\}$, the marginal probability mass function $p_i(x)$ is...

For the second, the keywords are "mutually independent".    There's something about the joint probability mass function of mutually independent random variables ...
A: The first question is easier than you think: each day is equally likely and there are $365$ possibilities so $p_i(x)=\frac1{365}$ for $x \in \{1,2,3,\ldots,365\}$ and $0$ otherwise
Since the birthdays are independent you have $p(\mathbf{x}) $ $=p(x_1,x_2,\ldots,x_n) $ $= \mathbb P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n) $ $= \mathbb P(X_1=x_1)\mathbb P(X_2=x_2)\cdots\mathbb P(X_n=x_n) $ $= p_1(x_1)p_1(x_2)\cdots p_n(x_n)$.  This will be $\frac{1}{365^n}$ when all of the $x_i \in \{1,2,3,\ldots,365\}$ i.e. when $\mathbf{x}\in \{1,2,3,\ldots,365\}^n$, and $0$ otherwise.  You can easily check that it adds up to $1$.
