By Wolfgang Karl Härdle, Zdeněk Hlávka
This publication provides the instruments and ideas of multivariate information research. It demonstrates the appliance of straightforward calculus and easy multivariate tools in actual lifestyles events, and lines greater than 2 hundred solved workouts.
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Extra info for Multivariate Statistics, Hardle
What sign do you expect the covariance to have? The empirical covariance is −3732. It is negative as expected since heavier cars tend to consume more gasoline and this leads to lower mileage. The negative covariance corresponds to a negative slope that could be observed in a scatterplot. It is very diﬃcult to judge the strength of the dependency between weight and mileage on the basis of the covariance. A more appropriate measure is the correlation which is a scale independent version of the covariance.
Why does this not apply to the following situation: X ∼ N (0, 1), Cov(X, X 2 ) = EX 3 − EXEX 2 = 0 − 0 = 0 but obviously X 2 is totally dependent on X? It is easy to show that independence of two random variables implies zero covariance: Cov(X, Y ) = E(XY ) − EXEY indep. = EXEY − EXEY = 0. The opposite is true only if X and Y are jointly normally distributed which can be checked by calculating the joint density and the product of the marginals. From above we see that, for standard normally distributed random variable X, we have Cov(X, X 2 ) = 0.
It is easy to show that independence of two random variables implies zero covariance: Cov(X, Y ) = E(XY ) − EXEY indep. = EXEY − EXEY = 0. The opposite is true only if X and Y are jointly normally distributed which can be checked by calculating the joint density and the product of the marginals. From above we see that, for standard normally distributed random variable X, we have Cov(X, X 2 ) = 0. In this example, zero covariance does not imply independence since the random variable X 2 is not normally distributed.