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I have the following problem:

Random variable $X = \begin{pmatrix}x_1\\x_2\end{pmatrix}$ has bivariate normal distribution with means $\mu = \begin{pmatrix}3\\5\end{pmatrix}$ and covariance matrix $\sum = \begin{pmatrix}4 & -2\\-2 & 4\end{pmatrix}$

$y = 2x_1 - 3x_2$

Find probability that $y > 4$

I have a solution but not sure if it's correct:

So I calculate $\mu_y = 2*3 - 3*5 = -9$

$var(y) = var(2x_1) + var(-3x_2)+2cov(2x_1,-3x_2) = 16+36+24=76$

Then $\sigma_y = \sqrt{var(y)} = 8.7178$

With $\mu_y$ and $\sigma_y$ I calculate $P(y>4)$ using Z-table, which gives the answer:

$1 - \phi(1.72) = 0.0427$

Does that makes sense?

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  • $\begingroup$ Method looks good to me. I have no $Z$-table at my disposal. $\endgroup$
    – drhab
    Commented May 9, 2016 at 17:34

1 Answer 1

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Your approach is correct, though I think your math is wrong. If $x\sim \mathcal{N}(\mu,\Sigma)$, and $y=A x$, then $y\sim\mathcal{N}(A\mu,A\Sigma A^T)$. In your problem, $A=\begin{pmatrix}2&-3\end{pmatrix}$, so $y\sim\mathcal{N}(-9, 76)$ as you note. But $\frac{4-\mu_y}{\sigma_y}=\frac{13}{\sqrt{76}}=1.4912$, not $1.72$. So $$p(y>4)=1-\Phi(\frac{13}{\sqrt{76}})=0.06795\dots$$

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