Hi,
Please, what is the derivative of
d/dx{[1-Q(sqrt(x))]^N}
derivative involving Q-function.pdf
Thanks
Hi,
Please, what is the derivative of
d/dx{[1-Q(sqrt(x))]^N}
derivative involving Q-function.pdf
Thanks
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.