1
$\begingroup$

How can I solve the following integral?

$$\int_{-\infty}^\infty \prod_{i=1}^n \bigg( 1 - \Phi\left(\frac{c - \mu_i}{\sigma_i}\right) \bigg) \frac{1}{\sigma_Y}\phi \bigg(\frac{c-\mu_Y}{\sigma_Y} \bigg) \,\mathrm dc$$

where Φ(⋅) is the CDF of Normal distribution and ϕ(⋅) is the pdf of Normal distribution.

even when $n=1$ I cannot solve it. I would appreciate if you could help me either for the case that $n=1$ or the above general case.

$\endgroup$
6
  • 2
    $\begingroup$ I know this is from statistics, but you may want to define your terms for the nonspecialists who still may be able to help you. $\endgroup$ Commented Feb 5, 2013 at 1:12
  • $\begingroup$ IF $\Phi(\cdot)$ is the CDF of Normal RV and $\phi(\cdot)$ is the pdf of Normal RV, did you try $\frac{1}{\sigma} \phi \bigg( \frac{c-\mu}{\sigma} \bigg)dc=d \Phi \bigg(\frac{c-\mu}{\sigma} \bigg)$ $\endgroup$ Commented Feb 5, 2013 at 1:34
  • $\begingroup$ thank you I edited the question $\endgroup$ Commented Feb 5, 2013 at 1:38
  • $\begingroup$ No I did not try this. How can this substitution solve this integral? $\endgroup$ Commented Feb 5, 2013 at 1:44
  • $\begingroup$ Got something from my answer below? $\endgroup$ Commented Feb 9, 2013 at 11:31

1 Answer 1

2
$\begingroup$

Considering independent random variables $X_i$ and $Y$, gaussian with mean and variance $(\mu_i,\sigma_i^2)$ and $(\mu_Y,\sigma_Y^2)$ respectively, this integral is $$ \mathbb P(\min\limits_{1\leqslant i\leqslant n}X_i\geqslant Y). $$ If $n=1$, its value is $$ \Phi\left(\frac{\mu_1-\mu_Y}{\sqrt{\sigma_1^2+\sigma_Y^2}}\right). $$ For $n\geqslant2$, I see no reason to expect some simple explicit formulas in the general case.

$\endgroup$
1

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.