0
$\begingroup$

So the problem is, given $X\sim N(1,2^2)$, $Y=e^X$, if the cdf of standard normal distribution is $\Phi$, to show that

$$\Phi\left(\frac{\ln Y-1}{2}\right)\sim U[0,1],$$

where $U[0,1]$ is the uniform distribution defined on $[0,1]$.

What I can tell now is that

$$\frac{\ln Y-1}{2}=\frac{X-1}{2}=Z\sim N(0,1^2),$$

is a standard normal distribution. So basically the problem is reduced to showing that $$\Phi(Z)\sim U[0,1].$$

But as you can see in here, the cdf of snd (the red curve) is not "uniform" on $[0,1]$. Rather $N(-2,0.5)$ (the green curve) is. So I'm certainly misunderstanding here something. Please help me to understand.

$\endgroup$
2
$\begingroup$

$\Phi (Z)$ is uniformly distributed on $(0,1)$: $P(\Phi (Z) \leq u)=P(Z \leq \Phi^{-1}(u))=\Phi (\Phi^{-1}(u))=u$ for all $u \in (0,1)$. I have used the fact that $\Phi$ is a continuous strictly increasing function.

$\endgroup$
  • $\begingroup$ I feel like I'm blind. Thank you for nice answer. +1 $\endgroup$ – user642721 Mar 12 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.