densities with overlapping area of 0.35
Just a correction: if we take X+2a then everything is OK (the curves intersect at a), so a = 0.9345893 is correct but one must take X ~ N(0,1) and Y ~N(2*a,1).
--- On Tue, 9/9/08, Moshe Olshansky <m_olshansky at yahoo.com> wrote:
From: Moshe Olshansky <m_olshansky at yahoo.com> Subject: Re: [R] densities with overlapping area of 0.35 To: r-help at r-project.org, "Lavan" <rsumithran at yahoo.com> Received: Tuesday, 9 September, 2008, 12:37 PM Let X be normally distributed with mean 0 and let f be it's density. Now the density of X+a will be f shifted right by a. Since the density is symmetric around mean it follows that the area of overlap of the two densities is exactly P(X>a) + P(X<-a). So if X~N(0,1), we want P(X>a) + P(X<-a) = 2P(X<-a) = 0.35, so P(X<-a) = 0.175 which yields -a = qnorm(0.175) = -0.9345893, so a = 0.9345893. --- On Tue, 9/9/08, Lavan <rsumithran at yahoo.com> wrote:
From: Lavan <rsumithran at yahoo.com> Subject: [R] densities with overlapping area of 0.35 To: r-help at r-project.org Received: Tuesday, 9 September, 2008, 12:11 PM Hi, I like to generate two normal densities such that the overlapping area between them is 0.35. Is there any code/package
available
in R to do that?? Regards, Lavan -- View this message in context:
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