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With $\tau=3,\ell=4$, and all the information available:

For $X_1, X_2$ two random variables, with $X_1\sim N(\tau, \ell+1)$ and $X_2\sim N(-5\ell, 2\tau+1)$, and $Cov(X_1,X_2)=0$.
What is the distribution of $X_3=3X_1-2X_2$?

If I knew $X_1, X_2$ were independent, it would just be adding up means and variances, but as $Cov(X_1,X_2)=0$ does in general not imply independence – how do I calculate this?

(Side question: Does using $\tau,\ell$ in both normal distributions imply correlation?)

Jaleks
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    You're correct to say that $Cov(X_1, X_2) = 0$ does not, in general, imply independence. However, in the case that $X_1$ and $X_2$ are Normally distributed, uncorrelated _does_ imply independence. – jcken Feb 14 '22 at 08:55
  • Thanks – and why? – Jaleks Feb 14 '22 at 08:58
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    If you write down the joint pdf of $(X_1, X_2)$, $f_X(x)$, with $Cov(X_1, X_2) = \text{diag}(\sigma_1^2, \sigma_2^2)$ this can be factorised as $f_X(x) = f_{X_1}(x_1)f_{X_2}(x_2)$ which implies independence. Try the proof yourself with the mean vector being zero. If you can do this, try adding in a non-zero mean vector. If you get stuck, open a new question (or browse pre-existing questions) for some help! – jcken Feb 14 '22 at 09:15
  • So I would end up with $X_3=N(3(\tau)-2(-5\ell), (\ell+1)-(2\tau+1)) = N(9+40, 5-7)$ – or would I have to multiply the variances also by 2 resp.3? (Sorry if having no clue…) – Jaleks Feb 14 '22 at 09:55
  • The variance in your final answer is negative which is a huge alarm bell. Remember that $Var(aX + bY) = a^2 Var(X) + b^2 Var(Y)$ when $X$ and $Y$ are uncorrelated. – jcken Feb 14 '22 at 10:24
  • (OMG) Thanks for your great, leading, help. So $X_3=N\big(3(\tau)-2(-5\ell), \quad 3^2(\ell+1)+(-2)^2(2\tau+1)\big) = N\big(9+40, 9\cdot 5 + 4\cdot 7\big)$ would be more promising… – Jaleks Feb 14 '22 at 10:38
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    @jcken [When $X_1$ and $X_2$ are *jointly* normal](https://stats.stackexchange.com/a/30205/247274) – Dave Feb 14 '22 at 10:45
  • @Dave: I think jcken cited the general addition rule for independent random variables? – Jaleks Feb 14 '22 at 11:09
  • Uncorrelated implies independent for jointly normal $X_1$ and $X_2$. If the distribution is not jointly Gaussian, then uncorrelated need not imply independent. [Wikipedia](https://en.m.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent) has some nice examples showing this. – Dave Feb 14 '22 at 11:34
  • I see. So the covarainace would have to be taken into account, in summation, but the final result stays the same, because it is zero. Or do I still miss something? – Jaleks Feb 14 '22 at 12:02

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