The Mystery of the ARROW: Can Someone Explain?

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SUMMARY

The discussion focuses on the derivation of the ARROW formula, specifically the mathematical transformation involving the exponential function. The key steps include completing the square for the expression involving variables x and y, leading to the separation of exponentials. The final form presented is e^{-\frac{(x- \rho y)^2}{2\sigma(1-\rho)^2}}e^{-\frac{y^2(1- \rho^2)}{2\sigma(1-\rho)^2}}. This derivation is crucial for understanding the statistical properties of correlated variables in multivariate distributions.

PREREQUISITES
  • Understanding of multivariate normal distributions
  • Familiarity with exponential functions in statistics
  • Knowledge of completing the square in algebra
  • Basic concepts of correlation coefficients, specifically the symbol ρ (rho)
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  • Study the properties of multivariate normal distributions
  • Learn about the implications of correlation coefficients in statistical modeling
  • Explore the method of completing the square in various mathematical contexts
  • Investigate the application of exponential functions in statistical inference
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Mathematicians, statisticians, and data scientists interested in the derivation and application of the ARROW formula in statistical analysis and modeling of correlated variables.

tuanle007
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can someone explain to me how did it derived to the ARROW...
I am not quite sure how that happened.thanks
 

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[tex]e^{-\frac{x^2+ y^2- 2\rho xy}{2\sigma(1-\rho)^2}}[/itex]<br /> They complete the square:<br /> [tex]x^2- 2\rho xy+ \rho^2y^2- \rho^2y^2+ y^2= (x- \rho y)^2+ (y^2- \rho^2 y^2)[/tex]<br /> so the exponential becomes<br /> [tex]e^{-\frac{(x-\rho y)^2- (y^2- \rho^2 y^2)}{2\sigma(1-\rho)^2}}[/itex]<br /> Then separate the exponentials:<br /> [tex]e^{-\frac{(x- \rho y)^2}{2\sigma(1-\rho)^2}}e^{-\frac{(y^2- \rho^2 y^2}{2\sigma(1-\rho)^2}}[/tex]<br /> and finally factor out y<sup>2</sup> in the last exponential<br /> [tex]e^{-\frac{(x- \rho y)^2}{2\sigma(1-\rho)^2}}e^{-\frac{y^2(1- \rho^2}{2\sigma(1-\rho)^2}}[/tex][/tex][/tex]
 
Thank You So Much!
 

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