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 $i \in {(\left \{ 1,…,n\right \}}$.
- 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 $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$)
Best Answer
It was claimed that $X_i\sim\mathcal U\{1..365\}$ for all $i\in\{1..n\}$.
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 ...