How to show that the function is concave when second derivative is inconclusive ?

How can I show that the following function f(λ)f(\lambda) is concave for λ≥hz1+σ2\lambda \geq hz_1 + \sigma^2 ?

f(λ)=1−exp(hz1+σ2−λz1b)+exp(hz0+σ2−λz0b)f(\lambda) = 1- \exp(\frac{hz_1 + \sigma^2 – \lambda}{z_1b}) + \exp(\frac{hz_0+\sigma^2-\lambda}{z_0b})

where z1>z0z_1 > z_0, b>0b>0 and σ2>0\sigma^2 >0.

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I have considered the second derivative test, but can’t figure how to prove the second derivative ≤0\leq 0.
– King008
2 days ago

  

 

could you show your work on the second derivative?
– LinAlg
2 days ago

  

 

The second derivative comes w.r.t λ\lambda comes to be
– King008
2 days ago

  

 

The second derivative w.r.t λ\lambda comes to be f″(λ)=−1z21b2exp(hz1+σ2−λz1b)+1z20b2exp(hz0+σ2−λz0b)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\lambda and it is strictly concave for \lambda \geq hz_1+\sigma^2\lambda \geq hz_1+\sigma^2. I have also plotted the second derivative for this range of \lambda\lambda and it is all negative. I just need help with proving it. Thanks.
– King008
2 days ago

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1 Answer
1

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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}

\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)>0f”(\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})

\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}

\frac{\sigma^2 – \lambda}{z_1 b} > \frac{\sigma^2 – \lambda}{z_0 b}

Since \lambda > \sigma^2\lambda > \sigma^2, and z_1 b > z_0 bz_1 b > z_0 b, this is indeed valid. So, the function is convex for \lambda > \sigma^2\lambda > \sigma^2.

  

 

Thanks a lot. This helps.
– King008
yesterday

  

 

Please accept the answer if it is to your full satisfaction.
– LinAlg
3 hours ago