Probability:Normal distribution

Question

My question is a prove question as follows.

I have solved it in full length which i am uploading as images below.

Now my doubt is that in page 3, is the derivation of psi(-x) wrong? or is there some other mistakes too? Please help.

Answer 1

The problem looks wrong. $\Phi(x)$ is an even function of $x$. Therefore $\int_{-x}^x\Phi(u)du=2\int_0^x\Phi(u)du=2\psi(x)$.

The correct statement would have $\psi(x)=\int_{-\infty}^x\Phi(u)du$ Using the fact that $\Phi(x)$ is even, $2\psi(x)=\int_{-\infty}^x \Phi(u)du+\int_{-x}^\infty \Phi(u)du=\int_{-x}^0\Phi(u)du +\int_0^x\Phi(u)du +\int_{-\infty}^0\Phi(u)du +\int_0^\infty\Phi(u)du=\int_{-x}^x\Phi(u)du+1$.

Answer 2

I think you just forgot to write an integral and an integration limit seems to be wrong.

In the context of the normal distribution $\Phi$ stands for the cumulative distribution function:

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

Then, with $\psi (x) = \frac{1}{\sqrt{2\pi}}\int_{\color{blue}{-x}}^x e^{-\frac{t^2}{2}}\, dt$ you get $$2\Phi(x) - 1 = \psi(x)$$

Note that because of symmetry you get: $$\Phi(x) = \frac{1}{2}\left(1- \psi(x)\right) + \psi(x) \Rightarrow 2\Phi(x) = 1+\psi(x)$$