# Tag Info

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Since $\rho_{AB} = 1$ and $\rho_{BC}=-1$, we have $A = xB+y$ and $B = -zC+w$, for some constant real $w,x>0,y,z>0$. Hence, $$A = xB+y = x(-zC+w)+y = -zxC + (xw+y),$$ which implies $\rho_{AC} = -1$.

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(1) In statistics and probability, the word 'correlation' has a very specific technical meaning, so you should avoid using it for just any kind of 'association'. (2) You are asking whether your data show evidence that in the population from which you sampled, the events $A=\{\text{studied}\}$ and $B = \{\text{took exam}\}$ are associated (dependent). (3) ...

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Pearson's (product moment) correlation coefficient may be defined for nonconstant samples $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$ by taking an inner product of the vectors with their mean values subtracted from each entry and "normalized" to have unit length. To see this, start with the following "standard" definition (which can be rearranged in many ...

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As Michael said, the fact that the correlation coefficient is between $-1$ and $1$ can be shown as a special case of the Cauchy-Schwarz inequality (see https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Rn ). There is a geometrical interpretation where you see random variables as vectors. Then the covariance is a scalar product, the standard ...

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Hints: Actually $\sigma^2_x=8$ so $\sigma_x \approx 2.83$. It is $\sigma_{f(x)}$ which is about $5.66$. You should then be able to calculate $\sigma^2_y$ (an integer) since it is $f(x)+\epsilon$, assuming the $x$ and $\epsilon$ are independent. That gives you $\sigma_y$. The covariance $\sigma_{f(x) y}$ is equal to the variance of $f(x)$, again ...

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