As far as I know, the Inverse Mill's ratio, $\lambda(x)=\phi(x)/\Phi(x)$, is decreasing in $x$. Thus, I am curious now whether $\lambda(x)$ is in fact strictly decreasing in $x$.

To see this, I derived the derivative of the inverse Mill's ratio: \begin{align*}\lambda(x)&=\phi(x)/\Phi(x) \\[2pt] \Rightarrow \lambda'(x)&=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2} \\[2pt] &=-x\lambda(x)-\lambda(x)^2\quad\because \phi(x)'=-x\phi(x)\end{align*} Here, I have no idea how to show whether $\lambda(x)<0$ or $\lambda(x)\leq0$.

Thus, is the inverse Mill's ratio in fact "strictly" decreasing in $x$?

As far as I know, the Inverse Mills ratio, $\lambda(x)=\phi(x)/\Phi(x)$, is decreasing in $x$. Thus, I am curious now whether $\lambda(x)$ is in fact strictly decreasing in $x$.

To see this, I derived the derivative of the inverse Mills ratio: \begin{align*}\lambda(x)&=\phi(x)/\Phi(x) \\[2pt] \Rightarrow \lambda'(x)&=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2} \\[2pt] &=-x\lambda(x)-\lambda(x)^2\quad\because \phi(x)'=-x\phi(x)\end{align*} Here, I have no idea how to show whether $\lambda(x)<0$ or $\lambda(x)\leq0$.

Thus, is the inverse Mills ratio in fact "strictly" decreasing in $x$?

As far as I know, the Inverse Mill's ratio, $\lambda(x)=\phi(x)/\Phi(x)$, is decreasing in $x$. Thus, I am curious now whether $\lambda(x)$ is in fact strictly decreasing in $x$.

To see this, I derived the derivative of the inverse Mill's ratio: \begin{align*}\lambda(x)&=\phi(x)/\Phi(x) \\[2pt] \Rightarrow \lambda'(x)&=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2} \\[2pt] &=-x\lambda(x)-\lambda(x)^2\quad\because \phi(x)'=-x\phi(x)\end{align*} Here, I have no idea how to show $\lambda(x)\leq0$ (note the weak inequality).

Thus, is the inverse Mill's ratio really a decreasing function? if it is, is the ratio in fact "strictly" decreasing in $x$?

As far as I know, the Inverse Mill's ratio, $\lambda(x)=\phi(x)/\Phi(x)$, is decreasing in $x$. Thus, I am curious now whether $\lambda(x)$ is in fact strictly decreasing in $x$.

To see this, I derived the derivative of the inverse Mill's ratio: \begin{align*}\lambda(x)&=\phi(x)/\Phi(x) \\[2pt] \Rightarrow \lambda'(x)&=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2} \\[2pt] &=-x\lambda(x)-\lambda(x)^2\quad\because \phi(x)'=-x\phi(x)\end{align*} Here, I have no idea how to show whether $\lambda(x)<0$ or $\lambda(x)\leq0$.

Thus, is the inverse Mill's ratio in fact "strictly" decreasing in $x$?