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if $X$ and $Y$ are stochastic variables with $\operatorname{Var}(X)=1.34$ and $\operatorname{Cov}(X,Y) = 0.64$, find $\operatorname{Cov}(2X, 3X+2Y)$. No ideas on this one, as I don't see any way of combining the formulas I know to figure this out. I would greatly appreciate some hints

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    $\begingroup$ The answer that you "accepted" is really lousy. I posted another one. $\endgroup$ – Michael Hardy Sep 14 '18 at 16:49
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$$\int_{\Omega} (2X - 2\mu_x) (3X - 3\mu_x + 2Y - 2\mu_y) d P = 6 \int_{\Omega} (X - \mu_x) (X - \mu_x) d P + 4 \int_{\Omega} (X - \mu_x) (Y - \mu_y) d P = 6 Var(X) + 4 Cov(X,Y),$$ where $\mu_x := \mathbb E(X)$, $\mu_y := \mathbb E(Y)$, and using the linearity of $\mathbb E$.

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\begin{align} \operatorname{Cov}(2X, 3X+2Y) & = 2\operatorname{Cov}(X,3X+2Y) \\[10pt] & = 2\big(\operatorname{Cov}(X,3X)+\operatorname{Cov}(X,2Y) \big) \\[10pt] & = 2\big(3\operatorname{Cov}(X,X) + 2\operatorname{Cov}(X,Y)\big) \\[10pt] & = 6\operatorname{Var}(X) + 4\operatorname{Cov}(X,Y). \end{align}

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  • $\begingroup$ Thank you, that was indeed a more simple solution. I'm new to this subject however, and I'm a little bit confused about how you factor out 2 from $Cov(2X, 3X+2Y)$ but you only factor it out from the "x-component". -> $2Cov(X, 3X+2Y)$. In the next step you do the same, with the factors 3 and 2 which are only present in one of the "components". Why is this legal? $\endgroup$ – novo Sep 15 '18 at 9:56
  • $\begingroup$ @novo : That is a part of what is meant by "bilinearity of covariance", i.e. it is linear in each component separately. Suppose $\operatorname E X=\mu$ and $\operatorname E Y = \nu$ and $c$ is a constant, i.e. $c$ is not random. Then $\operatorname E(cX) = c\mu$ and $$\begin{align} & \operatorname{cov}(cX,Y) = \operatorname E\big( (cX - c\mu)(Y-\nu)\big) = \operatorname E\big( c(X-\mu)(Y-\nu)\big) \\ \\ = {} & c\operatorname E\big((X-\mu)(Y-\nu)\big) = c\operatorname{cov}(X,Y). \end{align}$$ $\endgroup$ – Michael Hardy Sep 15 '18 at 18:21
  • $\begingroup$ The other aspect of bilinearity is that $\operatorname{cov}(U+V, Y) = \operatorname{cov}(U,Y) + \operatorname{cov}(V,Y). \qquad$ $\endgroup$ – Michael Hardy Sep 15 '18 at 18:21
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Hint: Use bilinearity of covariance!

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  • $\begingroup$ $+1,$ but I wonder whether someone who understands this answer would phrase the question the way this one is phrased. $\endgroup$ – Michael Hardy Sep 14 '18 at 16:54
  • $\begingroup$ @Michael Hardy : +1 too for you! I see your point and clearly your example is a good step by step guide, which may be exactly what the OP wanted.... I just thought it may be good to just leave a clue.... $\endgroup$ – ckrk Sep 14 '18 at 16:56
  • $\begingroup$ Often leaving a terse clue is the best option, but sometimes the question should make you suspect that the poster won't understand the clue. This seems like such a case. A clue works only if the poster understands it. $\endgroup$ – Michael Hardy Sep 15 '18 at 18:13
  • $\begingroup$ . . . . and now you will note that the comments under my answer clearly show that the poster had not yet learned what bilinearity of covariance means. $\endgroup$ – Michael Hardy Sep 15 '18 at 18:23

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