Let $0 $$
\sum_{d=0}^{n-1}\binom{n-1}{d} p^d (1-p)^{n-1-d}=(n-1)p
$$ Proof: By the binomial formula $$
(p+q)^{n-1}=\sum_{d=0}^{n-1}\binom{n-1}{d}p^d q^{n-1-d}
$$ Differentiating both sides with respect to $p$ and replacing $q=1-p$ we get
$$
(n-1)(p+1-p)^{n-2}=\frac{1}{p}\sum_{d=0}^{n-1}\binom{n-1}{d}p^d (1-p)^{n-1-d}
$$ and the result follows. Question: let's consider the example with $n=3$, $p=0.3$. Then
$$
\sum_{d=0}^{n-1}\binom{n-1}{d} p^d (1-p)^{n-1-d}= 0.7^2+(2*0.3*0.7)+(0.3^2)=1
$$
and
$$
(n-1)p=2*0.3=0.6
$$ What am I doing wrong?
Application of binomial formula
1
$\begingroup$
combinatorics
permutations
combinations
1 Answers
1
You differentiated wrong. If you differentiate $p^d$ with respect to $p$ you should get $d\; p^{d-1}$, not just $p^{d-1}$. The binomial theorem should give you $$ \sum_{d=0}^{n-1} {n-1 \choose d} p^d (1-p)^{n-1-d} = 1$$ not $(n-1)p$. Perhaps you meant
$$ \sum_{d=0}^{n-1} {n-1 \choose d} d \; p^d (1-p)^{n-1-d} = (n-1) p $$