Ytukyg

Let there be two events (which are disjoint and a partition of the sample space) $G$ and $B$ where $p = Pr(G)$ and $1-p = Pr(B)$. Let $X$ be a random variable and $K$ be a positive constant. Let $D = mathbb{E}[min(K, X)|G] - mathbb{E}[min(K, X)|B]$ and $E = alpha mathbb{E}(X|G) - alpha mathbb{E}(X|B)$ where $alpha = frac{mathbb{E}[min(K,X)]}{mathbb{E}(X)}$.

Show that $D < E$ if and only if
$$frac{mathbb{E}(min(K, X)|G)}{mathbb{E}(min(K, X)|B)} < frac{mathbb{E}(X|G)}{mathbb{E}(X|B)} $$

What I've tried so far is to write $alpha$ as

$$alpha = frac{pmathbb{E}(min(K, X)|G) + (1-p) mathbb{E}(min(K, X)|B)}{pmathbb{E}(X|G) + (1-p)mathbb{E}(X|B)} $$ and then I am stuck.

It is only true if we assume something more. Let's $Xsim U(-4,2), K=1$, $G$- $X>=0$, $B$- $X<0$. In this case $D=2.5<4.5=E$, but
$$frac{mathbb{E}(min(K, X)|G)}{mathbb{E}(min(K, X)|B)}=-frac{1}{4} >-frac{1}{2}= frac{mathbb{E}(X|G)}{mathbb{E}(X|B)}. $$

Let's start with $D<E$
begin{equation}
mathbb{E}(min(K,X)|G) - mathbb{E}(min(K,X)|B)< frac{mathbb{E}(min(K,X))}{mathbb{E}(X)} big( mathbb{E}(X|G) - mathbb{E}(X|B)big)
end{equation}
Let's assume that $mathbb{E}(X) > 0$ so we can multiply both sides by $mathbb{E}(X)$.

Also, as you noticed, $mathbb{E}(min(K, X)) = pmathbb{E}(min(K, X)|G) + (1-p) mathbb{E}(min(K, X)|B)$ and $mathbb{E}(X)= pmathbb{E}(X|G) + (1-p)mathbb{E}(X|B)$. Now we have:
begin{equation}
begin{split}
p mathbb{E}(X|G)mathbb{E}(min(K,X)|G)&- &pmathbb{E}(X|G)mathbb{E}(min(K,X)|B) + \
+(1-p) mathbb{E}(X|B)mathbb{E}(min(K,X)|G)&- (1-p&) mathbb{E}(X|B)mathbb{E}(min(K,X)|B)
\ <\
pmathbb{E}(X|G)mathbb{E}(min(K, X)|G)&- &pmathbb{E}(X|B)mathbb{E}(min(K, X)|G) + \ +(1-p)mathbb{E}(X|G)mathbb{E}(min(K, X)|B)&-(1-p&)mathbb{E}(X|B)mathbb{E}(min(K, X)|B)
end{split}
end{equation}

As you can see, we can subtract some components and we are left with:
begin{equation}
begin{split}
(1-p) mathbb{E}(X|B)mathbb{E}(min(K,X)&|G)-pmathbb{E}(X|G)mathbb{E}(min(K,X)|B)
\ <\
(1-p)mathbb{E}(X|G)mathbb{E}(min(K, X)&|B)-pmathbb{E}(X|B)mathbb{E}(min(K,X)|G)
end{split}
end{equation}

Now we add to both sides $pbigg(mathbb{E}(X|B)mathbb{E}(min(K,X)|G)+mathbb{E}(X|G)mathbb{E}(min(K,X)|B)bigg)$ to got:

begin{equation}
mathbb{E}(X|B)mathbb{E}(min(K,X)|G)
<
mathbb{E}(X|G)mathbb{E}(min(K, X)|B)
end{equation}

If we divide by $mathbb{E}(min(K, X)|B)mathbb{E}(X|B)$ we got:
$$frac{mathbb{E}(min(K, X)|G)}{mathbb{E}(min(K, X)|B)} < frac{mathbb{E}(X|G)}{mathbb{E}(X|B)}. $$

We only have to assure, that both $mathbb{E}(min(K, X)|B), mathbb{E}(X|B)$ have same sign.

If one or three out of $mathbb{E}(X),mathbb{E}(min(K, X)|B), mathbb{E}(X|B)$ have negative sign, we won't have the dependence. Otherwise it's ok (if $E(X)<0$ you'll have to change the inequality signs respectively after the multiplication, and then after the division).