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I have a problem and looked at this forum to find this thread: Probability with covariance matrix

However, I am looking for the solution involving covariance matrix with three variables.

Let $ X_1, X_2, X_3 $ be normal with the following covariance matrix:

$ \Sigma = \begin{pmatrix} 6 & -3 & -5\\ -3 & 7 & -2 \\ -5 & -2 & 3 \end{pmatrix} $

Find the probability that: $ P(X_1 - X_2 ≥1 \cap X_1-X_3 ≥ 1) $

I get the probability for either one of the legs the following way: $ P(X_1 - X_2 ≥1) $: $ Var(Y)=Var(X_1 )+Var(X_2 )−2Cov(X_1 ,X_2 )= 6+7-2*(-3)=19$

$ P(Y≥1)=1−P(Y<1)=1−Φ(\frac{1}{\sqrt{19}}) $

Similarly for $ P(X_1 - X3 ≥1)$. But these two events are clearly correlated and hence I cannot multiply them to get the final probability I am looking for. I thought about it and did not figure out how to incorporate the correlation between $X_1 ,X_2$ into the calculation. Does anyone know this?

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If we have the joint distribution, $f_{\underline X}(\underline x)$ of the random veczor variable, $X_,X_2,X_3$ and if we have to calculate the probability that $\underline X=(X_1,X_2,X_3)\in A\subset R^3$, where $A$ is a nice (Borel) set then the solution is the following integral:

$$\iiint_Af_{\underline X}(\underline x)\ d\underline x.$$

In our case

$$f_{\underline X}(\underline x)=\frac1{\sqrt{(2\pi)^3\det \Sigma}}e^{-\frac12(\underline x-\underline m)^T\Sigma^{-1}(\underline x-\underline m)}$$ where $\underline m$ is the expectation vector, $\Sigma$ is the covariance matrix and it is only impliciteky given that the joint distribution is normal.

Furthermore

$$A=\{\underline x=(x_1,x_2,x_3):x_1-x_2\geq 1, x_1-x_3\geq 1\}.$$

Unfortunately $\underline m$ is not given. Even if the same was given the integral can only be evaluated by some numeric integration method, that is, an easy closed formula cannot be given.

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