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After months of study I still do not get it. I apologize (see: Estimating values knowing their Pearson's r and their means and standard deviations)

Imagine, for example, I have two bivariate normal populations:

  • A: mean 100, standard deviation 10
  • B: mean 100, standard deviation 10
  • Correlation: 0.9

How can I calculate the probability of obtaining a value of "95" or greater sampling from both A and B (that is, to obtain two "95"s or greater at the same time)?

From a simulation I performed in R, the probability might be around 62.9%

Edit: Thank you, user2974951. The probability might be around 69.2 %

Edit: Thank you Xi'an. Now it is "Correlation: 0.9"

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    $\begingroup$ Note that Pearson's r should be replaced by correlation since this is a distribution and not the outcome of a sample. $\endgroup$ – Xi'an Dec 19 '18 at 13:59
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This quantity is exactly provided by the integral $$\int_{95}^\infty\int_{95}^\infty \frac{1}{2\pi\times 10\times\sqrt{1-0.9^2}} \exp\left\{-\frac{\frac{(x-100)^2}{100}-\frac{2\times 0.9\times(x-100)(y-100)}{100}+\frac{(y-100)^2}{100}}{2(1-0.9^2)}\right\}\text{d}x\text{d}y$$ or $$\int_{95}^\infty\int_{95}^\infty \frac{1}{200\pi\sqrt{1-0.9^2}} \exp\left\{-\frac{(x-100)^2-1.8(x-100)(y-100)+(y-100)^2}{200(1-0.9^2)}\right\}\text{d}x\text{d}y$$ or yet $$\int_{-.5}^\infty\int_{-.5}^\infty \frac{1}{2\pi\sqrt{1-0.9^2}} \exp\left\{-\frac{x^2-1.8xy+y^2}{2(1-0.9^2)}\right\}\text{d}x\text{d}y$$ The inner (or conditional) integral is provided by Mathematica:

enter image description here

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  • $\begingroup$ Thank you so much, Xi'an. I did not find a similar approach in any of the text books I examined. Is it an advanced/obscure approach? $\endgroup$ – statisticianwannabe Dec 19 '18 at 13:45
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    $\begingroup$ This is the very definition of the bivariate Normal distribution, since the integrand in the first equation is the bivariate Normal density. No advanced or obscure reference needed! $\endgroup$ – Xi'an Dec 19 '18 at 13:58
  • $\begingroup$ What a blunder. I must have overlooked the conceptual bases of the bivariate normal distribution, my bad, I apologize. Anyway: I solved the first and last integral and I obtained 6.282483369042868 and 0.06282508180143374. Am I missing something? $\endgroup$ – statisticianwannabe Dec 19 '18 at 15:32
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According to my simulation (using multivariate normal density) with sample size 1 million the result is a little higher

M=c(100,100)
R=matrix(c(1,0.9,0.9,1),nrow=2)
V=c(10,10)%*%t(c(10,10))*R

library(mvtnorm)
mean(rmvnorm(1e6,M,V)>=95)
[1] 0.6917805
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  • $\begingroup$ Thank you, I edited my message with your simulation results $\endgroup$ – statisticianwannabe Dec 19 '18 at 13:42

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