Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 http://prntscr.com/33si9s) 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)

share|improve this question

1 Answer 1

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: $$ X|(Y=y)\sim N\left(\mu_x+\frac{\sigma_x}{\sigma_y}\rho(y-\mu_y),(1-\rho^2)\sigma_x^2\right). $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.