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I am new user at this site need help in my research. I want to find the derivative of CDF of inverse gaussian distribution w.r.t. to parameters $\lambda$ and $\mu$. The PDF and CDF of inverse gaussian distribution is given as

$f(x;\lambda, \mu )= \sqrt(\frac{\lambda}{2\pi x^3})e^{- \frac{\lambda(x-\mu^2)}{2\mu^2 x}} , x>0, \lambda>0, \mu>0 $ $ F(x|\lambda, \mu)= \Phi\Bigg[\sqrt{\frac{\lambda}{x}}(\frac{x}{\mu}-1)\Bigg]+ e^\frac{2\lambda}{\mu}\Phi\Bigg[-\sqrt{\frac{\lambda}{x}}(1+\frac{x}{\mu})\Bigg] $

I am working on this problem. I would appreciate it if you would like to give me any help.

Thanks

Dev

Dev
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  • Hint: Use the chain rule and $\Phi'(x) = \frac{1}{\sqrt{2\pi}}\exp( -\tfrac{1}{2}x^2)$ – gammatester Jul 06 '17 at 12:50
  • $F(x; \lambda, \mu) = \Phi(a)+\Psi \Phi(b)$ – Dev Jul 06 '17 at 13:05
  • May I write my CDF in this form

    $F(x; \lambda, \mu) = \Phi(a)+\Psi \Phi(b)$

    where $a = \sqrt{\frac{\lambda}{x}}\left( \frac{x}{\mu}-1 \right) $,$ b = -\sqrt{\frac{\lambda}{x}}\left( \frac{x}{\mu}+1\right)$

    $\frac{\partial F}{\partial\mu } = \Phi^{'}(a)\frac{\partial a}{\partial\mu}-\Psi\Phi^{'}(b)\frac{\partial b}{\partial\mu}-\Phi(b)\frac{\partial \Psi}{\partial\mu}$. Is it correct or not ? Please give your comments.

     – Dev Jul 06 '17 at 13:19
  • Your $\Psi$ depends on $\lambda, \mu$. You must use the product rule for $\Psi \Phi.$ – gammatester Jul 06 '17 at 14:10
  • Thank u so much for your help gammatester. – Dev Jul 06 '17 at 15:01

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