$\frac{1}{365^n}$ when all of the $x_i \in \{1,2,3,\ldots,365\}$ adds up to 1

Question

Question:

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 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..

Answer:

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 $1$.

Difficulties

I am doing the birthday problem but have to show that $\frac{1}{365^n}$ when all of the $x_i \in \{1,2,3,\ldots,365\}$ adds up to 1. This does not make sense for me as I am adding up powers of a already large denominator over 1. Is there any way to show this

Answer 1

Simply

$$\frac{1}{365^n}\underbrace{\sum_{i=1}^{365}1\dots\sum_{j=1}^{365}1}_{n\text{ times}}=\frac{1}{365^n}\times365^n=1$$