1. ## Derivative involving Q-function

Hi,

Please, what is the derivative of

d/dx{[1-Q(sqrt(x))]^N}

derivative involving Q-function.pdf

Thanks

2. ## Re: Derivative involving Q-function

Use the Chain Rule and the Fundamental Theorem of Calculus:

$\dfrac{d}{dx}\left[1-Q(\sqrt{x})\right]^N = -N\left[1-Q(\sqrt{x})\right]^{N-1}\dfrac{dQ(\sqrt{x})}{dx}$

Edit: For the derivative of the $Q$ function, you have:

$\displaystyle Q(\sqrt{x}) = \dfrac{1}{2\pi}\int_{\sqrt{x}}^\infty e^{-u^2/2}du$

So, when you take the derivative, you are just using the fundamental theorem of calculus, and again, the chain rule.

3. ## Re: Derivative involving Q-function

Interesting function.

4. ## Re: Derivative involving Q-function

Originally Posted by SlipEternal
Use the Chain Rule and the Fundamental Theorem of Calculus:

$\dfrac{d}{dx}\left[1-Q(\sqrt{x})\right]^N = -N\left[1-Q(\sqrt{x})\right]^{N-1}\dfrac{dQ(\sqrt{x})}{dx}$

Edit: For the derivative of the $Q$ function, you have:

$\displaystyle Q(\sqrt{x}) = \dfrac{1}{2\pi}\int_{\sqrt{x}}^\infty e^{-u^2/2}du$

So, when you take the derivative, you are just using the fundamental theorem of calculus, and again, the chain rule.
@ SlipEternal, thanks for your help.

Using the relation between Q-function $Q(x)$ and the cdf $\Phi(x)$ of the normal standard Gaussian distribution, will it be appropriate to do $$\dfrac{dQ(\sqrt{x})}{dx}=\dfrac{d\left[1-\Phi(\sqrt{x})\right]}{dx}$$ $$\dfrac{dQ(\sqrt{x})}{dx}=-\dfrac{d\Phi(\sqrt{x})}{dx}$$. Also, using the relation between the cdf $\Phi(x)$ and pdf $\phi(x)$ of the normal standard Gaussian distribution, then $$-\dfrac{d\Phi(\sqrt{x})}{dx}=-\phi(\sqrt{x})=-\dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{1}{2}{\sqrt{x}^2}}$$

Thanks.