R help - Jun 2009 - Interpreting R -results for Bivariate Normal

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.

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

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

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]


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




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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
> 
> 
> 
> 

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