How do I find the distribution of the following iid variables?

$X_i\sim Ber(p)$ , $n>m$

$$ P(\sum_{i=1}^m X_i | \sum_{i=1}^n X_i ) = \frac {P(\sum_{i=1}^m X_i , \sum_{i=1}^n X_i )}{P(\sum_{i=1}^n X_i )}$$

  • $\begingroup$ It might be clearer if you wrote $P\left(\sum\limits_{i=1}^m X_i =a \mid \sum\limits_{i=1}^n X_i=b \right)$. This is not equal to $\dfrac{P\left(\sum\limits_{i=1}^m X_i =a\right)}{ P\left(\sum\limits_{i=1}^n X_i=b \right)}$, and obviously not if $a > b$ $\endgroup$
    – Henry
    Jun 19, 2020 at 13:19
  • $\begingroup$ I understand. Thank you! $\endgroup$
    – MC1325
    Jun 19, 2020 at 13:25

1 Answer 1



  • In effect, you will have $\sum\limits_{i=1}^n X_i$ ones out of $n$ with the rest zeros, and you want to know how many of these ones are in the first $m$

  • The conditional probability will not depend on $p$ and will end up being a hypergeometric distribution

  • If you do not take that shortcut, you could investigate $\dfrac{P\left(\sum\limits_{i=1}^m X_i =a\right)P\left(\sum\limits_{i=m+1}^n X_i =b-a\right)}{ P\left(\sum\limits_{i=1}^n X_i=b \right)}$


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