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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?

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  • $\begingroup$ I see... but I don't see why. The formula I have to find the variance of D does not use squares. It's just $$a^2\sigma^2 + b^2\sigma^2 + 2abCorr(X, Y)\sigma\sigma$$ (I'm being sloppy with the subscripts here, but yeah). $\endgroup$
    – vicvicvic
    Jan 15 '12 at 16:26
  • $\begingroup$ Actually, you are right using that formula (my answer was incorrect, I forgot to compute the new covariance....). $\endgroup$ Jan 15 '12 at 16:40
  • $\begingroup$ 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? $\endgroup$
    – vicvicvic
    Jan 15 '12 at 16:43
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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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  • $\begingroup$ Hm. The formula I have is $$D = aX + bY$$ and I used a = -1, b = 0.7. Then (from my book), $$\sigma^2_D = a^2\sigma^2_X + b^2\sigma^2_Y + 2ab\rho\sigma_X\sigma_Y$$ $\endgroup$
    – vicvicvic
    Jan 15 '12 at 16:35
  • $\begingroup$ 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? $\endgroup$
    – vicvicvic
    Jan 15 '12 at 18:42

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