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Suppose $x_1\sim\mathcal N(2,0.5),x_2\sim \mathcal N(2,3),$ and $x_3\sim \mathcal N(2.5,7)$ with correlations $\rho_{(1,2)}=0.3,\rho_{(1,3)}=0.1,$and $\rho_{(2,3)}=0.4.$ What is the distribution of

  1. $x_1+x_2+x_3?$

  2. $x_1+ (3\times x_2)+x_3?$

  3. $x_1+x_2+(0.5\times x_3)?$

    Solutions:

  4. The distribution of $x_1+x_2+x_3 =\mathcal{N}(6.5,15.275)$.

  5. The distribution of $x_1+(3\times x_2)+x_3=\mathcal{N}(10.5,30.077)$

  6. The distribution of $x_1+x_2+(0.5\times x_3)=\mathcal{N}(5.25,8.005)$

    But I want to know what is the distribution of $(2\times x_1)-(3\times x_2)-x_3$ and $x_1+x_2-(2\times x_3)?$

    The distribution of $(2\times x_1)-(3\times x_2)-x_3$ is $\mathcal{N}(-4.5,41.8407).$

The distribution of $x_1 +x_2-(2\times x_3) $ is $\mathcal{N}(-1,24.1544)$

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  • $\begingroup$ Are they jointly normal? $\endgroup$
    – gunes
    Commented Apr 24, 2019 at 12:29
  • $\begingroup$ @gunes, They are jointly normal in the sense that if $x_1=1$, then $x_2=-0.4494897447$ and $x_3=-1.24165.$ $\endgroup$ Commented Apr 25, 2019 at 8:46

1 Answer 1

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If they're assumed to be jointly normal, a situation which normally should be given to you, we already know that any linear combination of jointly normal RVs is a univariate normal. In that case, we only need to figure out the mean and variance. Your answer for (1) is therefore correct. For the other options you'll apply the following formulas:

$$E[aX+bY+cZ]=a\mu_x+b\mu_y+c\mu_z$$ $$\begin{align}\operatorname{var}(aX+bY+cZ) &=a^2\sigma_x^2+b^2\sigma_y^2+c^2\sigma_z^2+2(ab\rho_{xy}\sigma_x\sigma_y+ac\rho_{xz}\sigma_x\sigma_z+bc\rho_{yz}\sigma_y\sigma_z)\end{align}$$ In (1), you implemented and obtained this for $a=b=c=1$.

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  • $\begingroup$ I think the expression for the variance is not correct. See en.wikipedia.org/wiki/Multivariate_normal_distribution and online.stat.psu.edu/stat505/lesson/4/4.2 or wolfram for examples. The term with the correlation coefficients has a negative sign, not a positive sign. $\endgroup$
    – jpcgandre
    Commented Aug 17, 2023 at 10:43
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    $\begingroup$ If we decrease it to two variables, we should have $$\operatorname{var}(X+Y)=\operatorname{cov}(X+Y, X+Y)=\operatorname{var}(X)+2\operatorname{cov}(X,Y)+\operatorname{var}(Y)$$ Therefore, it should be correct. $\endgroup$
    – gunes
    Commented Aug 17, 2023 at 19:26
  • $\begingroup$ Yes it is correct. I was confusing joint distribution with sums or differences! Thank you $\endgroup$
    – jpcgandre
    Commented Aug 18, 2023 at 15:30

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