Proving that a maximum exists when finding the optimal criterion in SDT

Question

When deriving that there is an optimal criterion in the Signal Detection Theory literature that maximizes Proportion Correct (PC), we usually arrive at the following expression by setting the first derivative of PC with respect to $\lambda$ to zero. In other words:

From :

$ PC = s Z(d'-\lambda) + (1-s) Z(\lambda)$

To :

$ \frac{\partial PC}{\partial \lambda} = -s z(d'-\lambda)+(1-s) z (\lambda) = 0 $

by setting $\frac{\partial PC}{\partial \lambda} = 0$ where $z$ is the pdf of a univariate zero mean gaussian with $(\mu,\sigma^2)=(0,1)$, and $Z$ is its respective cdf. $s$ is the proportion of target present trials, so $s$ and $1-s$ are both bounded $s\in[0,1]$.

The problem now is that I can't seem to analytically prove that a maximum exists by finding the second derivative of the above expression and showing that within a certain range it is negative (thus showing there is a maximum via concavity). I know that the optimal value is (but can't seem to prove that it is a maximum (vs a minimum):

$\lambda^* = \frac{d'}{2} - \frac{1}{d'}\log(\frac{s}{1-s})$

I computed the second derivative and am stuck at the following equation, without knowing how to show that it will be negative within a certain domain of $\lambda$:

$\frac{\partial^2 PC}{\partial \lambda^2} = -\frac{(d'-\lambda)^3}{2}\text{exp}(-(1/2)(d'-\lambda)^2)+\frac{\lambda^3}{2}\text{exp}(-(1/2)\lambda^2)$

Question: For what values of $\lambda$ is $\frac{\partial^2 PC}{\partial \lambda^2} < 0$ ?

Answer 1

After correcting for a change of sign addressed in the comments, when we compute the second derivative we get:

$\frac{\partial^2 PC}{\partial \lambda^2} = (-s)z(d'-\lambda)(d'-\lambda)-(1-s)z(\lambda)\lambda$

From here it follows that $\frac{\partial^2 PC}{\partial \lambda^2}<0$ only if $d'>\lambda$ as all the terms in the expression above are positive, which is equivalent to the case in which the Hit Rate is greater than 0.5 or 50%. Recall that $d'=Z^{-1}(HR) + \lambda$.