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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

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  • $\begingroup$ 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. $\endgroup$
    – String
    Feb 26 at 8:31
  • $\begingroup$ @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. $\endgroup$
    – user831952
    Feb 26 at 8:32
  • $\begingroup$ en.wikipedia.org/wiki/Birthday_problem $\endgroup$
    – user831952
    Feb 26 at 8:34
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    $\begingroup$ 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. $\endgroup$
    – String
    Feb 26 at 8:36
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    $\begingroup$ 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 $$\sum_{i=1}^{365^n}p(v_i)=\frac 1{365^n}\cdot 365^n$$ $\endgroup$
    – String
    Feb 26 at 8:40
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Simply

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

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  • $\begingroup$ Okay so it was just that. I overcomplicated it. Thanks! $\endgroup$
    – user831952
    Feb 26 at 8:40
  • $\begingroup$ @mathstudent23: I suspect the relation to the birthday problem (which is not quite as trivial in its full detail) threw you off. $\endgroup$
    – String
    Feb 26 at 8:42
  • $\begingroup$ @String Yes indeed! It is a simplified version that I am dealing with. $\endgroup$
    – user831952
    Feb 26 at 8:45

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