Integration of a normal variable

[RandomIntegral]

Hey guys,

I have been thinking about the last equation for some time, but I do not understand its equality. Partial integration wasn't helping me either. Could somebody help me out?

Thanks in advance!

Thomas

 

[daon]

What is phi, and Phi? I'm suspect either a formula has been derived previously and/or you are expected to use integration by parts. You haven;t defined anything, so unless someone has used this same notation before it will be hard to help.

 

[RandomIntegral]

Oh sorry. Big Phi is the cdf of the Normal distribution and small phi is the density function. Integration by parts did not work for me. The formula has not derived previously. I looked for an explanation in a textbook but it only said "as one can easily see". What would you suggest?

 

[daon]

I'm curious, which book is this from? I can't seem to find a straightforward way to prove this without knowing certain things. Integration by parts yields a relatively neat solution, but only if you know something about the integral of the CDF.

It is in fact not "easy to see", without some background beforehand, and the only proof I have found involves the use of logarithms among other things. This is called the Inverse Mills Ratio. This is not even in the index of my undergrad advanced statistics class.

 

[RandomIntegral]

Econometric Analysis of Cross Section and Panel Data, Jeffrey Wooldridge, p.522

Solution is:
dphi(z)/dz = - z * phi(z) (trivial if you look at the definition of normal density function)
yielding (by fundamental theorem of calculus)
int[z*phi(z)]= - phi(z)
insert limits c to inf = - phi(inf) - (-phi(c)) = phi(c), Voilà!

Can you elaborate on your integration by parts solution? I am curious about that!

 

[daon]

Since \(\displaystyle \int \phi = \Phi\)

\(\displaystyle \int z \phi(z) dz = z \Phi(z) - \int \Phi(z) dz\). The latter of which cannot be expressed algebraically, and seems to produce a nonsensical result. I'm out of my field though, so I'll digress.

By the "easy to see" reference, I expected it to be a general result of PDFs, not specific to the normal distribution.

 

[RandomIntegral]

Yeah, that's similar to what I got stuck with as well.
I cannot be generel result as it states that the expected value of cut distrubtion equals the density of the cut point. This is easily contradicted by a coinflip/a uniform density variable.
Yet, there might be a class of density functions for which that result (or a similar) holds.