So, I just was reading Hogg and Craig(page 20 of the book) and came across the following inequality for arbitrary events $C_1, C_2, \dots , C_k$:
$p1 ≥ p2$.
Where, $$p1 = P(C1) + P(C2) + P(C3) +P(C4)+\dots +P(C_k)$$ and $$ p2 = P(C1 ∩ C2) + P(C1 ∩ C3) + P(C1 ∩ C4)+ P(C2 ∩ C3)+ P(C2 ∩ C4)+ P(C3 ∩ C4)+ \cdots \cdots + P(C_i\cap C_j )+ \cdots\cdots$$
But this is obviously false(which can be seen by letting $k=4$ and $C_1 = C_2 = C_3 = C_4$)
There are other results in the book of a similar nature( like $p2\ge p3$ etc..) that can be disproved by a similar argument.
What am i missing?
(Also, any hints on proving the general result $p1\ge p2\ge p3 \dots\ge p_k$ would be greatly appreciated! )
the page that i'm talking about is shown below:
We may take $C_1,\ldots,C_n$, $n\geq 1$ some events. Then the inclusion-exclusion principle states $$\mathbb P\Big(\bigcup_{i=1}^nC_i\Big)=\sum_{k=1}^n(-1)^{k-1}\sum_{J\subset\{1,2,\ldots,n\}\\ |J|=k}\mathbb P\Big(\bigcap_{j\in J}C_j\Big).$$
If all $C_i=C$, then the formula reduces (note, there are $\binom{n}{k}$ subsets $J$ of size $k$): $$\mathbb P(C)=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\mathbb P(C) =\mathbb P(C)\Big(1-\sum_{k=0}^n\binom{n}{k}(-1)^k1^{n-k}\Big)=\mathbb P(C)(1+0^n)=\mathbb P(C).$$
For your specific "counter example", we have 4 singletons, 6 intersections of two sets, 4 of three and 1 of one. That gives as expected $4-6+4-1=1$.
The author then defines $p_i:=\sum_{J\subset\{1,2,\ldots,n\}\\|J|=i}\mathbb P(\bigcap_{j\in J}C_j)$.
But $p_i\geq p_{i+1}$ does not hold in general, as you showed for $n=4$ and $C=C_1=C_2=C_3=C_4$, but rather only for $i\geq n/2$. It has to do with the monotonicity of the binomial coefficients. Only in that case you can use $\mathbb P(\bigcap _{j\in J}C_j)\leq \mathbb P(C_i)$ for any $i\in J$. Then $p_{i+1}$ has fewer summands than $p_i$.
So the author hasn't given the completly correct answer.