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I have a random vector $X = ( X_1 , X_2 )$ that has a bidimensional normal repartition with mean $0$ and covariation matrix :

$$ \Sigma = \left( \begin{array}{ccc} 1 & q \\ q & 1 \end{array} \right) $$

I need to calculate $q$, which is

$$ q = \mathrm{Corr}(X_1,X_2)$$

How can I calculate $\mathrm{Corr}(X_1,X_2)$ ?

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1 Answer 1

By definition

$ \mathrm{corr}(X_1,X_2) = {\mathrm{cov}(X_1,X_2) \over {\sigma_{X_1} \sigma_{X_2}}}$

You can get $\mathrm{cov}(X_1,X_2)$, $\sigma_{X_1}$ and $\sigma_{X_2}$ from $\Sigma$ (see covariance matrix )

$\Sigma_{11}$ = $\mathrm{cov}(X_1,X_1)$ = $\sigma^2_{X_1}$ = $1$

$\Sigma_{22}$ = $\mathrm{cov}(X_2,X_2)$ = $\sigma^2_{X_2}$ = $1$

But

$\Sigma_{12}$ = $\mathrm{cov}(X_1,X_2)$

which is already $q$

So,

$ \begin{eqnarray} \mathrm{corr}(X_1,X_2) &=& {\mathrm{cov}(X_1,X_2) \over {\sigma_{X_1} \sigma_{X_2}}} \\ q &=& {q \over {1 * 1}} \\ 1 &=& 1 \end{eqnarray}$

It's trivially true, since $\mathrm{cov}(X_1,X_2) = q = \mathrm{Corr}(X_1,X_1) \Leftrightarrow \sigma_{X_1} = \sigma_{X_2} = 1$

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