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Consider $n$ integer-valued and independent random variables $e_1, e_2, \dots, e_n$ with known distribution functions $m_1, m_2, \dots, m_n$. Let's denote with $E^{1..n}$ the random variable given by the sum of $e_1, e_2, \dots e_n$, that is $E^{1..n} = \displaystyle\sum_i{e_i}$. The PDF of $E^{1..n}$, $m^{1..n}$, is given by the convolution of $m_1, m_2, \dots, m_n$.

I would like to calculate $P(E^{1..n} = 1 \,| \,e_1 = 1)$.

My intuition tells that that this is equal to $P(E^{2..n} = 0) P(e_1=1)$ but I just can't figure out the math behind it.

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This would be a better place for your question: math.stackexchange.com –  Dr_Sam Feb 26 '13 at 7:24
    
But the answer is no: $P(E^{1..n} = 1 | e_1 = 1)$ = $P(E^{2..n} + e_1 = 1 | e_1 = 1)$ = $P(E^{2..n} = 0 | e_1 = 1)$ = $P(E^{2..n} = 0)$ because of the independance. –  Dr_Sam Feb 26 '13 at 8:06
    
Yes, you're right, thanks! –  user92382 Feb 26 '13 at 9:27
    
@Dr_Sam: Please post your comment as an answer, so the OP can accept it. –  Paul Feb 27 '13 at 0:23
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1 Answer

up vote 1 down vote accepted

As said in my comments of the question, a better place for asking this kind of question would be on the math part of stackexchange.

However, we can find

$P(E^{1..n}=1|e_1=1) = P(E^{2..n}+e_1=1|e_1=1) = P(E^{2..n}=0|e_1=1) = P(E^{2..n}=0)$,

the last equality being the result of the independance of the $e_i$.

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