If H(b) $\leq$ b then $\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b) - [1-b]db $?

Question

$H$ is a CDF over $[0,1]$

I want to prove that if $H(b) \leq b$ then :

$$\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b) - [1-b]db $$ for all $k \in [0,1]$

I think this is true, but a rigorous proof does seem difficult to me. Can anyone help me to prove this? or does any theorem helps?

Answer 1

I assume $h(b)=H'(b)$ is a density, $v$ is a constant and $H(b) \leq b \quad\forall b$ \begin{align} &H(b) \leq b \implies H(b)-1 \leq b-1 \implies 1-H(b)\geq1-b \implies -(1-H(b))\leq-(1-b) \end{align}

Substituting in the equation you get the following inequality.

$\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b)h(b) - [1-b]db$

Now we need to show that substituting $h(b)$ for $1$ gives us the desired inequality. We know that $\int_0^k(v-b)db = vk -\frac{k^2}{2}$ then:

\begin{align} \int_0^k (v-b)h(b)db &= v\int_0^kh(b)db-\int_0^kbh(b)db\\ &= v[H(k)-H(0)]-\int_0^kbh(b)db\\ &= v(H(k)-H(0))-bH(b)|_0^k +\int_0^kH(b)db\\ &\leq(v-k)H(k)-H(0)v +\int_0^kbdb\\ &H(b)\leq b\; \forall b\; \wedge \text{H(.) is a cdf}\implies H(0)=0 \\ &\leq (v-k)k+\frac{k^2}{2}\\ &=vk-\frac{k^2}{2} \end{align} then the following holds whenever $v \geq k$ $$\int_0^k (v-b)h(b)db \leq vk-\frac{k^2}{2} = \int_0^k(v-b)db$$

Answer 2

It is obvious that

$$\int_0^k ((v-b)h(b) - (1-H(b)))db \leq \int_0^k ((v-b) - (1-b))db$$ if and only if

$$\int_0^k ((v-b)h(b) + H(b))db \leq \int_0^k ((v-b) +b)db$$ or equivalently

$$\int_0^k ((v-b)h(b) + H(b))db \leq \int_0^k vdb.$$

We will work with the LHS. We have

$$\int_0^k ((v-b)h(b) + H(b))db =vH(k)+\int_0^k(H(b)-bh(b))db.$$

Now, using integration by parts, it is

$$\int_0^kbh(b)db=[bH(b)]_0^k-\int_0^kH(b)db=kH(k)-\int_0^kH(b)db.$$ Thus

$$\int_0^k ((v-b)h(b) + H(b))db =vH(k)-kH(k).$$ On the other hand, the RHS is

$$\int_0^k vdb=vk.$$ So, the given inequality is equivalent to

$$vH(k)-kH(k)\le kv.$$ In other words

$$-kH(k)\le v(k-H(k)).$$ Now we see that

$$-kH(k)\le 0 \le v(k-k)\le v(k-H(k)),$$ where we have used $H(k)\le k \implies-k\le -H(k).$