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

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..

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

• How large is the sample space? In other words, how many copies of $$\frac 1{365^n}$$ are you summing? I should think this would be exactly $365^n$ and hence everything makes perfect sense. Feb 26 at 8:31
• @String all the information given in the question is the information that I have. I have to show it is a valid joint pmf i.e adds up to 1.
– user831952
Feb 26 at 8:32
• en.wikipedia.org/wiki/Birthday_problem
– user831952
Feb 26 at 8:34
• Well, the sample space would be $\left\{1,2,...,365\right\}^n$, which has size $365^n$. Hence you are adding $365^n$ copies of $1/365^n$ together. Feb 26 at 8:36
• OK, to make it very explicit, choose an indexing function $i\mapsto v_i\in\left\{1,2,...,365\right\}^n$ and write it as the sum $$\sum_{i=1}^{365^n}p(v_i)=\frac 1{365^n}\cdot 365^n$$ Feb 26 at 8:40

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