I have two normally distributed random variables, X and Y.
X has mean 66 and standard deviation 6.
Y has mean 77 and standard deviation 7.
The correlation between the random variables is given as 0.8.
The task is to find the probability $$P(X > 0.7Y)$$
This is my attempt:
Create the random variable $$D = 0.7Y – X$$
and calculate $$P(D < 0)$$
The variable D should also be normally distributed with these properties:
$$\mu_D = 0.7 \times 77 – 66 = -12.1$$
$$\sigma^2_D = 0.7^2 \times 49 + 36 + 2 \times 0.7 \times -1 \times 0.8 \times 7 \times 6 = 12.97$$
Then, $$P(D < 0) = F_Z\left(\frac{0 + 12.11}{\sqrt{12.97}}\right) \approx F_Z(3.36)$$
But, this is not the answer I'm supposed to get (which is instead $$F_Z(1.17)$$ according to the textbook). Where am I going wrong?
Best Answer
You did not go wrong anywhere, your book did. Whomsoever worked out the answer in the book missed the negative sign on the $2ab\rho\sigma_X\sigma_Y$ and calculated $$\text{var}(0.7Y - X) = 60.01 + 47.04 = 107.05$$ instead of $$\text{var}(0.7Y - X) = 60.01 - 47.04 = 12.97.$$ Thus, the book answer is $\displaystyle F_Z\left(\frac{12.11}{\sqrt{107.05}}\right) = F_Z(1.17)$ instead of $\displaystyle F_Z\left(\frac{12.11}{\sqrt{12.97}}\right) = F_Z(3.36)$.