I want to prove that the function $x \mapsto \Phi(\Phi^{-1}(x) + \lambda)$ defined for $x \in [0,1]$ is concave for any $\lambda \geq 0$. $\Phi$ is the cumulative distribution function of a standard normal random variable

$$\Phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{t^2}{2}}\,dt$$

Since this function is smooth, I differentiated it twice with respect to $x$ and if I didn't make an error I need to show that $$x - (\Phi^{-1}(x) + \lambda)\sqrt{2\pi}e^{\frac{x^2}{2}} < 0$$ whenever $x \in [0,1]$. I don't know how to show this. I am also not sure if this is the best way to go.

  • 1
    $\begingroup$ A slightly different strategy: show that the first derivative is decreasing. Writing $\phi$ for the standard normal density, the first derivative of your function is $\phi(\Phi^{-1}(x)+\lambda)/\phi(\Phi^{-1}(x))$. The logarithm of this, multiplied by 2, is $[\Phi^{-1}(x)]^2-[\Phi^{-1}(x)+\lambda]^2$. $\endgroup$ Commented May 29, 2016 at 17:47
  • $\begingroup$ @JohnDawkins Shouldn't the denominator be $\phi(x)$ instead of $\phi(\Phi^{-1}(x))$? $\endgroup$
    – Calculon
    Commented May 29, 2016 at 17:50
  • $\begingroup$ The derivative of $\Phi^{-1}(x)+\lambda$ is $1/[\phi(\Phi^{-1}(x))]$. $\endgroup$ Commented May 29, 2016 at 18:02
  • $\begingroup$ @JohnDawkins Oh right, sorry. That makes the second derivative in my question wrong as well. Thanks a lot. $\endgroup$
    – Calculon
    Commented May 29, 2016 at 18:05

1 Answer 1


We may write \begin{align} \Phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{t^2}{2}}\,dt= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{-\frac{t^2}{2}}\,dt+\frac{1}{\sqrt{2\pi}}\int_0^x e^{-\frac{t^2}{2}}\,dt \end{align} consequently, $\Phi'(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$.

Let $y=\Phi^{-1}(x)$, so $\Phi(y)=x$, differentiate both side w.r. to $x$, we get $\Phi'(y)\frac{dy}{dx}=1$, thus $$\frac{dy}{dx}=\frac{d}{dx}\Phi^{-1}(x)=\frac{1}{\Phi'(y)}=\frac{1}{\Phi'(\Phi^{-1}(x))}$$

Let $g(x)=\Phi(\Phi^{-1}(x) + \lambda)$, so that \begin{align} g'(x)=\Phi'( \Phi^{-1}(x) + \lambda )\cdot \frac{d}{dx}\Phi^{-1}(x) &=\Phi'( \Phi^{-1}(x) + \lambda )\cdot \frac{1}{\Phi'(\Phi^{-1}(x))} \\ &=\frac{1}{\sqrt{2\pi}}e^{-\frac{(\Phi^{-1}(x) + \lambda )^2}{2}} \cdot \frac{1}{\frac{1}{\sqrt{2\pi}}e^{-\frac{(\Phi^{-1}(x) )^2}{2}}} \\ &=e^{-\frac{(2\lambda\Phi^{-1}(x) + \lambda^2 )}{2}}. \end{align} Alos, \begin{align} g''(x)&=-e^{-\frac{(2\lambda\Phi^{-1}(x) + \lambda^2 )}{2}} \cdot \lambda\frac{d}{dx}\Phi^{-1}(x) \\ &= -e^{-\frac{(2\lambda\Phi^{-1}(x) + \lambda^2 )}{2}} \cdot \lambda \frac{1}{\frac{1}{\sqrt{2\pi}}e^{-\frac{(\Phi^{-1}(x) )^2}{2}}} <0, \,\, \forall x\in \mathbb{R} \,\, {\text{and thus}}\,\, \forall x\in[0,1] \end{align} which means that $g(x)$ is concave on $[0,1]$.


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