# Prove whether a particular function is concave

Given the following equation: $$V(w) = - \frac{\alpha}{2} \left[ y_1(w) + y_2(w) + \int _{-\infty}^{+\infty} \vert y_1(w) - y_2(w) - x\vert f_{T1}(x)dx\right] \\- \beta \int _{w - y_1(w)} ^{+\infty} \left( y_1(w) + x - w\right) f_{T2}(x)dx$$

and the information that

• $\frac{\partial y_2(w)}{\partial w} \geq 0$ and $\frac{\partial y_1(w)}{\partial w}=0$
• $F_{k}(x)=\int\limits_{-\infty}^{x} f_{k}(t)dt$ is $k$'s cumulative distribution function which is continuous on a compact support where $k=\{T_1,T_2\}$ and $0 \leq f_k \leq 1$ is $k$'s probability density function; and
• $\alpha, \beta > 0$

I would like to prove whether $V(w)$ is (strictly) concave, or if it is not concave, whether it is quasi-concave.

Attempts: I tried to see if the function has a second derivative that is negative.

First, I derived: $$\frac{\partial V(w)}{\partial w} = -\alpha y_2^{\prime}(w)\left[1-F_{T_1}\left(y_1(w)-y_2(w)\right) \right] + \beta\left[1-F_{T_2}(w-y_1(w))\right]$$

and

\begin{align} \frac{\partial }{\partial w} \left(\frac{\partial V(w)}{\partial w}\right) &= -\alpha y_2^{\prime}(w)\left[1-F_{T_1}\left(y_1(w)-y_2(w)\right) \right] + \beta\left[1-F_{T_2}(w-y_1(w))\right] \\ &= -\alpha \left\{ y_2^{\prime \prime}(w)\left[1-F_{T_1}\left(y_1(w)-y_2(w)\right) \right] + (y_2^{\prime}(w))^2 f_{T_1}\left(y_1(w)-y_2(w)\right) \right\} \\ \quad \quad \quad &- \beta f_{T_2} \left( w-y_1(w)\right) \end{align}

However, I could not sign $y_2^{\prime \prime}(w)$ without additional information.

• since y1 is constant as a function of w, your first integral is constant, and irrelevant. Your characterization of f_k doesn't make sense to me, F_k cannot be a function of x. What have you tried?
– John
Oct 22, 2014 at 18:06
• @John I edited the question in light of your comments which are spot on.
– Duna
Oct 22, 2014 at 18:34
• @John I now included attempts I made
– Duna
Oct 22, 2014 at 19:02

Edit: (based upon edits to question)

Ah, well, you clearly have more knowledge than you put into your original question, so perhaps you are holding out on more information? That said, there's still hope, you still need properties of $y_2(w)$, but basically, you need to guarantee the following...

$$V''(w) \le 0$$

Notice that $y_1(w)$ is just a constant. Let's call that constant $y_1$. Then you're left with:

$$0 \le \alpha \{y''_2(w) [1 - F_{T_1}(y_1 - y_2(w))] + (y'_2(w))^2 f_{T_1}(y_1-y_2(w))\} - \beta f_{T_2}(w-y_1)$$

You're done if $y''_2 \ge 0$.

If you can't guarantee that, then you need to show... $$\frac{\beta}{\alpha} f_{T_2}(w-y_1) \le y''_2(w) [1 - F_{T_1}(y_1 - y_2(w))] + (y'_2(w))^2 f_{T_1}(y_1-y_2(w))$$

Ugh. Good luck with that... :) The above assumes you computed the derivatives correctly. I don't think you did. The absolute values are going to wreak havoc, and I see no evidence of that in your derivative.

• I guess it should be a - before $\beta$ in your equation. By definition, each $f$ is non-negative and $[1-F(.)] \geq 0$. Thus, if $y_2^{\prime \prime}(w) \geq 0$ then the function is can be said concave.
– Duna
Oct 23, 2014 at 14:48