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Let X be the heigh of the father and Y the height of the son. The two random variables distributed with bivariate normal distribution, as demonstrated by Pearson in 1900. If E [X] = 68 inches and E [Y] = 69 inches, σx=σy = 2 and p = 0.5 : Find if the son of a father who has a height of 80 inches, is taller than his father.

My solution:

Calculate: X|(Y=80)~Ν ( , ) (by using this formula Compare the mean that i will find above with 80 that is the height of the son.

I am not sure that this is the right solution.Any ideas? (Hints/Answers)

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up vote 1 down vote accepted

The question seems to be asking you for the probability that the son with a father of height 80 inches is taller than that father. The answer is $$ \int_{80}^\infty f(x|Y=80)dx $$ where $f(x|Y=y)$ is the PDF of the normal distribution on the RHS below: $$ X|(Y=y)\sim N\left(\mu_x+\frac{\sigma_x}{\sigma_y}\rho(y-\mu_y),(1-\rho^2)\sigma_x^2\right). $$

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