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How can I show that the following function $f(\lambda)$ is concave for $\lambda \geq hz_1 + \sigma^2$ ?

$f(\lambda) = 1- \exp(\frac{hz_1 + \sigma^2 - \lambda}{z_1b}) + \exp(\frac{hz_0+\sigma^2-\lambda}{z_0b})$

where $z_1 > z_0$, $b>0$ and $\sigma^2 >0$.

King008
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  • I have considered the second derivative test, but can't figure how to prove the second derivative $\leq 0$. – King008 Oct 21 '16 at 07:46
  • could you show your work on the second derivative? – LinAlg Oct 21 '16 at 12:06
  • The second derivative comes w.r.t $\lambda$ comes to be – King008 Oct 21 '16 at 12:41
  • The second derivative w.r.t $\lambda$ comes to be $f''(\lambda) = -\frac{1}{z_1^2 b^2}\exp (\frac{hz_1 + \sigma^2 - \lambda}{z_1 b}) + \frac{1}{z_0^2 b^2}\exp (\frac{hz_0 + \sigma^2 - \lambda}{z_0 b})$. I have plotted the original function against $\lambda$ and it is strictly concave for $\lambda \geq hz_1+\sigma^2$. I have also plotted the second derivative for this range of $\lambda$ and it is all negative. I just need help with proving it. Thanks. – King008 Oct 21 '16 at 12:47

1 Answers1

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$ \begin{align} f''(\lambda) &= -\frac{1}{z_1^2 b^2}\exp (\frac{hz_1 + \sigma^2 - \lambda}{z_1 b}) + \frac{1}{z_0^2 b^2}\exp (\frac{hz_0 + \sigma^2 - \lambda}{z_0 b}) \\ &> -\frac{1}{z_0^2 b^2}\exp (\frac{h}{b} + \frac{\sigma^2 - \lambda}{z_1 b}) + \frac{1}{z_0^2 b^2}\exp (\frac{h}{b} + \frac{\sigma^2 - \lambda}{z_0 b}) \\ &= \frac{1}{z_0^2 b^2}\left( \exp (\frac{h}{b} + \frac{\sigma^2 - \lambda}{z_0 b})- \exp (\frac{h}{b} + \frac{\sigma^2 - \lambda}{z_1 b}) \right) \end{align} $

The first factor is positive, so the second factor needs to be positive as well for $f''(\lambda)>0$, which translates to:

$$ \exp (\frac{h}{b} + \frac{\sigma^2 - \lambda}{z_1 b}) > \exp (\frac{h}{b} + \frac{\sigma^2 - \lambda}{z_0 b}) $$ $$ \frac{\sigma^2 - \lambda}{z_1 b} > \frac{\sigma^2 - \lambda}{z_0 b} $$ Since $\lambda > \sigma^2$, and $z_1 b > z_0 b$, this is indeed valid. So, the function is convex for $\lambda > \sigma^2$.

LinAlg
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