| To: | Agin.Patrick@hydro.qc.ca |
|---|---|
| Subject: | Re: expected value of a normal variable with integrate() |
| From: | Sundar Dorai-Raj <sundar.dorai-raj@PDF.COM> |
| Date: | Mon, 22 Dec 2003 15:23:18 -0600 |
| Cc: | s-news@lists.biostat.wustl.edu |
| In-reply-to: | <3B7812F13593D411979200508BE3A75C10A23380@msxcentral2.hydro.qc.ca> |
| Organization: | PDF Solutions, Inc. |
| References: | <3B7812F13593D411979200508BE3A75C10A23380@msxcentral2.hydro.qc.ca> |
| Reply-to: | sundar.dorai-raj@PDF.COM |
| User-agent: | Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.4) Gecko/20030624 Netscape/7.1 (ax) |
You need to rescale the density to integrate to one. What you have below only integrates to 0.5. Try integrate(function(x) 2 * x * dnorm(x), -Inf, 0) Regards, Sundar Agin.Patrick@hydro.qc.ca wrote: Hi,Could anyone tell me why the two following expressions don't give the same result:integrate(function(x) x*dnorm(x),-Inf,0) equals -0.3989 andtmp <- rnorm(1000000); mean(tmp[tmp<0]) approximately equals -0.79 (two times -0.3989)I noticed that -0.3989 is the result of mean(pmin(0,tmp)), the mean of all the normal deviates with a cap of zero.If I'm interested in the expected value of all the negatives under the standard normal, the good result seems to be -0.79. Am I wrong? Why the integrate() expression above doesn't give this number?Thank you, Patrick P.S. I'm using S-Plus 4.5 on Windows XP. |
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