[R] densities with overlapping area of 0.35

[Moshe Olshansky]

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:
> >
> http://www.nabble.com/densities-with-overlapping-area-of-0.35-tp19384741p19384741.html
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> Nabble.com.
> > 
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