Show that this inequality involving expectations holds 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 $D &lt; E$ if and only if 
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} &lt; \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)} $$

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

$$\alpha = \frac{p\mathbb{E}(\min(K, X)|G) + (1-p) \mathbb{E}(\min(K, X)|B)}{p\mathbb{E}(X|G) + (1-p)\mathbb{E}(X|B)} $$ and then I am stuck.
It is only true if we assume something more. Let's $X\sim U(-4,2), K=1$, $G$- $X&gt;=0$, $B$- $X&lt;0$. In this case $D=2.5&lt;4.5=E$, but 
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)}=-\frac{1}{4} &gt;-\frac{1}{2}= \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$

Let's start with $D&lt;E$
\begin{equation}
\mathbb{E}(\min(K,X)|G) - \mathbb{E}(\min(K,X)|B)&lt;  \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 $\mathbb{E}(X) &gt; 0$ so we can multiply both sides by $\mathbb{E}(X)$. 

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

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

\begin{equation}
\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)
 &lt;
\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:
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} &lt; \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$

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 $E(X)&lt;0$ you'll have to change the inequality signs respectively after the multiplication, and then after the division).
It is only true if we assume something more. Let's $X\sim U(-4,2), K=1$, $G$- $X&gt;=0$, $B$- $X&lt;0$. In this case $D=2.5&lt;4.5=E$, but 
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)}=-\frac{1}{4} &gt;-\frac{1}{2}= \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$

Let's start with $D&lt;E$
\begin{equation}
\mathbb{E}(\min(K,X)|G) - \mathbb{E}(\min(K,X)|B)&lt;  \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 $\mathbb{E}(X) &gt; 0$ so we can multiply both sides by $\mathbb{E}(X)$. 

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

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

\begin{equation}
\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)
 &lt;
\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:
$$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} &lt; \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$

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 $E(X)&lt;0$ you'll have to change the inequality signs respectively after the multiplication, and then after the division).