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$X,Y$ are two Random Variables.
$Var(X)=1\; Var(Y)=2$.
$\rho(X,Y)=\frac16$ ($\rho=\frac{cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$).

How I'm calculating $Var(X-2Y+8)$?

I'm using the formula:
$$Var\left( \sum_{i=1}^n X_i\right)=\sum_{i=1}^n Var(X_i)+2\cdot \sum_{1\le i<j\le n} cov(X_i,X_j)$$ So, I'll get: $$Var(X-2Y+8)=Var(X)+Var(-2Y)+Var(8)+2(cov(X,-2Y)+cov(X,8)...)$$ How I'm continue from here?
$Var(8)=0$, but what with $cov(X,8)$?

Thank you!!

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    $\begingroup$ what do you think cov(X,8) is? Using the that fact $cov(X,Y) = \rho \sqrt{var(x) var(y)}$... $\endgroup$ – Lost1 May 12 '14 at 9:04
  • $\begingroup$ @Lost1, it's depends on the random variable... no? $\endgroup$ – CS1 May 12 '14 at 9:05
  • $\begingroup$ What is $Var(8)$? It's $0$? $\endgroup$ – CS1 May 12 '14 at 9:06
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Make the computation easier by eliminating the constant in the variance. We have $$\text{Var}(X-2Y+8)=\text{Var}(X-2Y)=\text{Var}(X) + 4\text{Var}(Y)+2\text{Cov}(X,-2Y)\\=\text{Var}(X) + 4\text{Var}(Y)-4\text{Cov}(X,Y)=\text{Var}(X)+4\text{Var}(Y)-4\rho\sqrt{\text{Var}(X)\text{Var}(Y)}\,\,,$$ where the third equality comes from the fact that $\text{Cov}(\cdot,\cdot)$ is linear in the first argument and symmetric (note that the vector space of random variables and the covariance function form an inner product space). To build upon your calculation, instead of the ellipse, you should have $+\,\text{Cov}(-2Y,8)$, however, the covariances $\text{Cov}(X,8)$ and $\text{Cov}(-2Y,8)$ are both zero because of the application of the formula you provided for the correlation $\rho$ (refer to the comment of @Lost1). In fact, this formula $$\rho(X_1,X_2)=\frac{\text{Cov}(X_1,X_2)}{\sqrt{\text{Var}(X_1)\text{Var}(X_2)}}$$ applies to any random variables $X_1,X_2$.

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    $\begingroup$ I think you need $\mathbf{Cov} (X, -2Y)$ $\endgroup$ – Alex May 12 '14 at 9:10

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