I’d like to ask wether my proof of the following proposition is correct. Thanks..
Proposition:
(Prop.1):
$\forall p\in (0,\frac{1}{2}),\forall n\in\mathbb{N},\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }\gt p^{i+1}(1-p)^{n-(i +1)}$
(Prop.2):
$\forall p\in \{\frac{1}{2}\},\forall n\in\mathbb{N},\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }= p^{i+1}(1-p)^{n-(i +1)}$
(Prop.3): $\forall p\in (\frac{1}{2},1),\forall n\in\mathbb{N},\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }\lt p^{i+1}(1-p)^{n-(i +1)}$
Proof:
(Prop.1):
Let $p\in (0,\frac{1}{2})$ and $n\in\mathbb{N}$, We’ll prove that $\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }>p^{i+1}(1-p)^{n-(i +1)}$:
Since $n\in\mathbb{N}$, We have $1\leq n$, And thus $0\leq n-1$, And we conclude that $0\in\{0,...,n-1\}$, And thus we’ll prove it first for the case $i=0\in\{0,...,n-1\}$:
For $i=0$ we have:
(Eq.1) $p^i(1-p)^{n-i }=p^0(1-p)^{n-0}=(1-p)^n$
And:
(Eq.2) $p^{i+1}(1-p)^{n-(i +1)}=p^{0+1}(1-p)^{n-(0+1)}=p(1-p)^{n-1}$
Now since $p\lt \frac{1}{2}$, We get that $2p\lt 1$, And so (Ineq.1) $p\lt 1-p$, But since $0<p$, We get by (Ineq.1) that it is also the case that $0\lt 1-p$, And thus $0\lt (1-p)^{n-1}$ (because the exponential function is always positive),
Now by multiplying (Ineq.1) by $(1-p)^{n-1}>0$, We conclude that $p(1-p)^{n-1}\lt (1-p)(1-p)^{n-1}=(1-p)^n$, But this implies by (Eq.1) and (Eq.2) that $p^{i}(1-p)^{n-i }\gt p^{i+1}(1-p)^{n-(i +1)}$ as was to be shown.
Now since $n\in\mathbb{N}$ we get that there are two cases (case 1) $n=1$ or (case 2) $2\leq n$:
(case 1): If $n=1$, Then we get $n-1=0$, And thus $\{0,...,n-1\}=\{0\}$,
But since we’ve already shown that for $i=0$, We have $p^{i}(1-p)^{n-i }\gt p^{i+1}(1-p)^{n-(i +1)}$, We conclude that $\forall i\in\{0\}, p^i(1-p)^{n-i }>p^{i+1}(1-p)^{n-(i +1)}$, And thus $\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }>p^{i+1}(1-p)^{n-(i +1)}$.
(case 2): If $2\leq n$, Then we have $0\leq n-2$, And thus $0\in\{0,...,n-2\}$.
Now we’ll prove by terminating induction on $i$ that $\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }>p^{i+1}(1-p)^{n-(i +1)}$:
(Note: We use terminating induction since $i\in\{0,...,n-1\}$ and thus we have to terminate the induction at some point)
(base case): For $i=0\in\{0,...,n-2\}$, We have to show that $p^i(1-p)^{n-i }>p^{i+1}(1-p)^{n-(i +1)}$, But this is exactly what we’ve already shown at the beginning.
(inductive step): Now suppose that for some $i=k\in\{0,...,n-2\}$ we have $p^k(1-p)^{n-k }>p^{k+1}(1-p)^{n-(k +1)}$, And we’ll prove that for $i=k+1\in\{1,...,n-1\}$ we have $p^{k+1}(1-p)^{n-(k+1) }>p^{k+2}(1-p)^{n-(k +2)}$:
(Note: Note that we ensure in the inductive step that both $k$ and $k+1$ are within $\{0,...,n-1\}$ and do not get out of the bounds)
Since $0\lt p$ and since we’ve already shown that $0\lt 1-p$, We have $0\lt \frac{p}{1-p}$, Now since by the induction hypothesis we have $p^k(1-p)^{n-k }\gt p^{k+1}(1-p)^{n-(k +1)}$, We get by multiplying this inequality by $\frac{p}{1-p}\gt 0$ that:
$p^{k+1}(1-p)^{n-(k+1)}=\frac{p}{1-p}p^k(1-p)^{n-k }\gt \frac{p}{1-p}p^{k+1}(1-p)^{n-(k +1)}=p^{k+2}(1-p)^{n-(k +2)}$ as was to be shown.
Thus we see from cases (case 1) and (case 2) that it is always the case that $\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }>p^{i+1}(1-p)^{n-(i +1)}$ as was to be shown.
Q.E.D.
(Prop.2):
Let $p\in \{\frac{1}{2}\}$, $n\in\mathbb{N}$, and $ i\in\{0,...,n-1\}$, We’ll prove that $p^i(1-p)^{n-i }=p^{i+1}(1-p)^{n-(i +1)}$:
Since $p\in\{\frac{1}{2}\}$, We get that $p=\frac{1}{2}$, And thus $1-p=1-\frac{1}{2}=\frac{1}{2}=p$, But this implies that $p^{i+1}(1-p)^{n-(i +1)}=p^i\cdot p\cdot(1-p)^{n-i-1}=p^i\cdot (1-p)\cdot (1-p)^{n-i-1}=p^i(1-p)^{n-i}$
as was to be shown.
Q.E.D.
(Prop.3):
Let $p\in (\frac{1}{2},1)$ and $n\in\mathbb{N}$, We’ll prove that $\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }<p^{i+1}(1-p)^{n-(i +1)}$:
Since $\frac{1}{2}<p<1$, We get that $-1<-p<-\frac{1}{2}$, And this implies that $0=1-1<1-p<1-\frac{1}{2}=\frac{1}{2}$, And thus $1-p\in (0,\frac{1}{2})$, Now let’s define $q:=1-p$, And we get that $1-q=1-(1-p)=1-1+p=p$ and that $q\in (0,\frac{1}{2})$, And thus by (Prop.1) we conclude that (Fact.1) $\forall i\in\{0,...,n-1\}, q^i(1-q)^{n-i }\gt q^{i+1}(1-q)^{n-(i +1)}$,
Now we’ll prove that (Fact.2) $\forall i\in\{0,...,n-1\}, p^{n-i}(1-p)^{i }\gt p^{n-(i+1)}(1-p)^{i+1}$:
Let $i\in\{0,...,n-1\}$, Then by (Fact.1) we get that $q^i(1-q)^{n-i }\gt q^{i+1}(1-q)^{n-(i +1)}$, But since $q^i(1-q)^{n-i }=(1-p)^i p^{n-i}=p^{n-i}(1-p)^i$ and since $q^{i+1}(1-q)^{n-(i +1)}=(1-p)^{i+1}p^{n-(i +1)}=p^{n-(i+1)}(1-p)^{i+1}$, We conclude that $p^{n-i}(1-p)^{i }\gt p^{n-(i+1)}(1-p)^{i+1}$ as was to be shown.
Now we’ll finnish the proof by showing that $\forall i\in\{0,...,n-1\}, p^i(1-p)^{n-i }<p^{i+1}(1-p)^{n-(i +1)}$:
Let $i\in\{0,...,n-1\}$, Then $0\leq i\leq n-1$, And thus $-n+1\leq -i\leq 0$, Which implies that $1=n-n+1\leq n-i\leq n$, And thus $0=1-1\leq n-i-1\leq n-1$, But this implies that $n-i-1\in\{0,...,n-1\}$, And thus we can conclude by (Fact.2), That $p^{n-(n-i-1)}(1-p)^{n-i-1 }\gt p^{n-(n-i -1+1)}(1-p)^{n-i -1+1}$, But since $p^{n-(n-i-1)}(1-p)^{n-i-1 }=p^{i+1}(1-p)^{n-(i +1)}$ and since $p^{n-(n-i -1+1)}(1-p)^{n-i -1+1}=p^i(1-p)^{n-i}$, We conclude that $p^i(1-p)^{n-i}\lt p^{i+1}(1-p)^{n-(i +1)}$ as was to be shown.
Q.E.D.