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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?

  • 0
    Well, I'm starting to feel stupid here. We've been taught to consider $D = aX + bY$ which in this case should be $D = -X + 0.7Y$, but then I still get $2ab\rho\sigma\sigma = 2 \cdot 0.7 \cdot -1 \cdot 0.8 \cdot 6 \cdot 7$ which is negative?2012-01-15

2 Answers 2

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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 $\displaystyle F_Z\left(\frac{12.11}{\sqrt{12.97}}\right) = F_Z(3.36)$.

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I can't see where you went wrong. It seems right to me: $\eqalign{ \text{Var}(.7Y-X)&=\text{Var}(.7Y)+\text{Var}(-X)+2\text{Cov}( -X, .7Y)\cr &=(.7)^2\text{Var}( Y)+\text{Var}( X)+2(-1)(0.7)\text{Cov}( X, Y)\cr &=(.7)^2\text{Var}( Y)+\text{Var}( X)+2(-1)(0.7) [ \text{Corr}( X, Y)\sigma_X\sigma_Y]\cr &=(.7)^2 \cdot 7^2+6^2+2(-1)(0.7)(0.8)\cdot 6\cdot7\cr &=12.97. } $

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    I'm not discounting the possibility that my book is wrong; it has an awful lot of errors... If you stand by your latest answer?2012-01-15