[R] Interpreting R -results for Bivariate Normal
Peter Dalgaard
p.dalgaard at biostat.ku.dk
Mon Jun 8 08:39:29 CEST 2009
beetle2 wrote:
> HI Guys,
> I know that this forum is not for homework but I am trying to interpret R
> output code.
> I was just wondering if someone might be able to help.
Well, if homework, we can only give hints.
>
> I have been given the following.
>
> For (X1,X2) distributed bivariate normal with parameters
>
> mu1 = 5.8
> mu2 = 5.3
> sd1 = sd2 = 0.2
>
> and p = 0.6
That's Greek letter rho, not p, I expect.
>
> The r-code and inpit/output are as follows
>
> input
>
> m <- 5.3 + 0.6*(6.3 - 5.8)
> s <- sqrt(0.2^2*(1-0.6^2))
> q <- seq(5.12,6.08,0.16)
>
> print(rbind(q,pnorm(mean=m,sd=sd=s,q=q)))
>
> output
>
> q 5.1200 5.280 5.44 5.6 5.76 5.92 6.1
> 0.0013 0.023 0.16 0.5 0.84 0.98 1
>
> I have been asked to interpret
>
> E[X2|X1 = 6.3] and varE[X2|X1 = 6.3]
>
>
> I take it that s<- = 0.16 is the standard variation
Standard _deviation_
> So I am assuming that varE[X2|X1 = 6.3] = 0.16^2 = .0256
>
> m <- 5.3 + 0.6*(6.3 - 5.8) = 5.6 this the Expected value of E[X+Y]
What makes you think that? E[X+Y] is mu1+mu2.
The m calculation could also have been written
x1 <- 6.3
m <- 5.3 + 0.6*(x1 - 5.8)
Now go back to your text book and read up on the formulas that connect
correlation and regression coefficients.
>
>
> I see from the output that this would be correct because the probability of
> 5.6 = 0.5
>
>
> to interpret E[X2|X1 = 6.3] I can't see it in the output. And I'm not sure
> how to find the conditional probabilty from the output.
>
> Any help would be greatly appreciated
>
>
>
>
--
O__ ---- Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907
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