LYM Inequality question Suppose that $F ⊂ P(n)$ is a set system containing no chain with $k + 1$ sets.
Prove that $\sum\limits_{r=1}^n \frac{|F_{r}|}{n \choose r} ≤ k$,
where $F_{i} = F \cap [n]^{(i)}$
for each i.
($[n]^{(i)}$ is the set of all subsets of $[n]$ which have size $i$)
It's clearly very similar to the LYM equality but I can't seem to work out whether I'm supposed to use the LYM equality or adapt the proof for it.
 A: Pick a permutation $ \sigma $ of $ 1,2,3, \cdots , n$ uniformly at random.
Let's define a new set
$$ B = \left\{ { \{\phi ,\sigma (1), \sigma (2), \sigma (3), ..., \sigma (i) \}| i \le n}\right\}$$.
Note that $$ |B \cap F| \le k$$ 
Because $ F $ cannot contain a chain of length $(k+1)$ and every subset of $B$ forms a chain.
Now,  define a random variable $ X_f$ such that 
$$X_{f} =
\left\{
 \begin{array}{ll}
  1  & \mbox{if $ f \in B$ } \\
  0 & \mbox{if } \text{otherwise}
 \end{array}
\right.$$
Let's define another random variable $ X$
Such that $$ X =|B \cap F|= \sum_{f \in F}^{} X_f$$
Note that for a given $ f$, the probability that $f \in B$ is simply 
$$\frac{1}{\binom{n}{|f|}}$$
This is because it is equally likely to be any of the $\binom{n}{|f|}$ subsets of $1,2,...,n$
So we have,
$$ E[X] = \sum_{f \in F}^{} \frac{1}{\binom{n}{|f|}}$$
But we already know that
$$|B \cap F| \le k$$
Hence, 
$$E[X] = \sum_{f \in F}^{} \frac{1}{\binom{n}{|f|}} \le k$$
This is essentially a slight modification of the proof of the LYM inequality which can be found here
