Re: [R] 3D plot of a bivariate normal distribution

[Spencer Graves]

Have you considered "contour", "persp", and "image", in package(base) 
and "contourplot", "levelplot" in package(lattice)?  See the 
documentation and Venables and Ripley (2002) Modern Applied Statistics 
with S, 4th ed. Springer).

hope this helps.  spencer graves

Rafael Bertola wrote:
Hi,

I've used the Mathematica to produce 3D graphics, contour plots of a
bivariate normal distribution
Now I want make these graphics in R, but i do not know how.
I would like to:
- Plot a 3D graph for some different variance matrix
- Plot the contour plots
- Find and try to plot (in the 3d graph ou contour plot) the (1-a)%
confidence region based in a chi-square(a) with the degrees of freedom
equal a 2 or bigger.
Below is the Mathematica Notebook that i've used until now

<< "Graphics`PlotField`"

NB[x_,y_]:=(1/((2 Pi)*Sqrt[a*b*(1-c^2)]))*Exp[(-1/(2*(1-c^2)))*( 
          ((x-u)/Sqrt[a])^2 + ((y-v)/Sqrt[b])^2 
			- 2*c(((x-u)/Sqrt[a])((y-v)/Sqrt[b]))
								)]

{{a,c}, {c,b}} = {{1,0}, {0,1}};  The covariance Matrix
{u,v} = {0,0};                    Mean vector
Plot3D[NB[x,y],{x,-1.5,1.5},{y,-1.5,1.5},
		AxesLabel->{x,y,z},
		BoxRatios->{1,1,1}];
ContourPlot[NB[x,y],{x,-1,1},{y,-1,1},
		Axes->True, 
		AxesLabel->{x,y}];

3d graph rotation
Do[
        Plot3D[NB[x,y],{x,-1.5,1.5},{y,-1.5,1.5},
                PlotPoints->20,
                Mesh ->False,
                SphericalRegion ->True,
                Axes ->None,
                Boxed ->False,
                ViewPoint->{2 Cos[t], 2 Sin[t], 1.3},
                BoxRatios->{1,1,1}
        ],{t, 0, 2Pi-2Pi/36, 2Pi/36}]
Thanks, 
Rafael
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