Show that the following expressions are true:i. ∑ni=1(xi − ¯x) = 0ii. ∑ni=1(xi − ¯x)2 = ∑ni=1 x2i − n ¯x2iii. ∑ni=1(xi − ¯x)(yi − ¯y) = ∑ni=1 xiyi − n ¯x ¯

Question

Show that the following expressions are true:

i. $\sum_{i=1}^{n}(x_i - \bar{x}) = 0$ ii. $\sum_{i=1}^{n}(x_i - \bar{x})^2 = \sum_{i=1}^{n} x_i^2 - n \bar{x}^2$ iii. $\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y}) = \sum_{i=1}^{n} x_iy_i - n \bar{x} \bar{y}$

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Solution

Sure, let's go through these one by one:

i. ∑ni=1(xi − ¯x) = 0

This is true because ¯x is the mean (average) of all xi. So, when you subtract the mean from each xi and then sum them all up, the result is zero. This is because the sum of the deviations of the data points from their mean is always zero.

ii. ∑ni=1(xi − ¯x)2 = ∑ni=1 x2i − n ¯x2

This is a mathematical property of variances. The left side of the equation is the definition of variance, which is the sum of the squared deviations from the mean. The right side of the equation is another way to express the same thing, by expanding the square in the left side of the equation.

iii. ∑ni=1(xi − ¯x)(yi − ¯y) = ∑ni=1 xiyi − n ¯x ¯y

This is the definition of covariance, which measures how much two variables vary together. The left side of the equation is the sum of the product of the deviations of x and y from their means. The right side of the equation is another way to express the same thing, by expanding the product in the left side of the equation.

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Show that the following expressions are true:i. ∑ni=1(xi − ¯x) = 0ii. ∑ni=1(xi − ¯x)2 = ∑ni=1 x2i − n ¯x2iii. ∑ni=1(xi − ¯x)(yi − ¯y) = ∑ni=1 xiyi − n ¯x ¯

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