2
$\begingroup$

I am reading Robert Hogg's (Introduction to Mathematical Statistics) EM algorithm.

In example 6.6.1 (page 370 in the 7th version), please help to explain how the following integral

$$\int_a^\infty(z-\theta_0)\frac{1}{\sqrt{2\pi}}\frac{\exp \left\{{-(z-\theta_0)^2/2}\right \}}{1-\Phi(a-\theta_0)}dz$$

is equal to

$$\frac{1}{1-\Phi(a-\theta_0)}\phi(a-\theta_0)$$

where $\phi(x)=(2\pi)^{-1/2}\exp\left\{-x^2/2\right\}$

or the book made some mistakes here?

I also think this post might give a little help.

Thanks

$\endgroup$

1 Answer 1

4
$\begingroup$

We know that $\int U'\exp \{U\} = \exp \{U\}$ then $$\int_a^\infty (z-\theta_0)\exp\{-(z-\theta_0)^2/2\} dz = \left[-\exp\{-(z-\theta_0)^2/2\}\right]_a^\infty = \exp\{-(a-\theta_0)^2/2\}$$

$$\int_a^\infty (z-\theta_0)\frac{1}{\sqrt{2\pi}}\frac{\exp\{-(z-\theta_0)^2/2\}}{1-\Phi(a-\theta_0)} dz =\\\frac{1}{\sqrt{2\pi}}\frac{1}{1-\Phi(a-\theta_0)}\exp\{-(a-\theta_0)^2/2\}\\ = \frac{1}{1-\Phi(a-\theta_0)}\frac{1}{\sqrt{2\pi}}\exp\{-(a-\theta_0)^2/2\}\\=\frac{1}{1-\Phi(a-\theta_0)}\phi(a-\theta_0)$$

$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$
    – Deep North
    Commented Jun 23, 2019 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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