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Please consider the following example:

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In the second column, I standardize VAR1, subtracting the mean and dividing by the standard deviation. In the third column, I normalize VAR1 with the formula:

$$\frac{x - \min(x)}{\max(x)-\min(x)}$$

In the fourth column I normalize the standardized values of VAR1 (using the above mentioned formula, on column 2).

Can you tell me why columns 3 and 4 are the same?

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The easy answer is that standardization and normalization are linear transformations of the data and any line is determined by two distinct points on it. Since columns 3 and 4 are constructed to include the points $(\min(\text{data}), 0)$ and $(\max(\text{data}), 1)$ (namely, $(2,0)$ and $(95, 1)$), they must result from identical transformations.

The algebraically rigorous answer manipulates the formulas. Recall that standardization of data $\mathbf{x} = x_1, x_2, \ldots, x_n$ replaces each $x_i$ with

$$y_i = \frac{x_i - \bar x}{s_x}$$

where $\bar x$ usually is the (arithmetic) mean of the data and $s_x$ often is a standard deviation of the data. (Generally, $\bar x$ may be any estimate of a central value and $s_x$ may be any estimate of the spread, but such generality is usually not intended.)

On the other hand, normalization replaces each $x_i$ with

$$z_i = \frac{x_i - \min(\mathbf{x})}{\max(\mathbf x) - \min(\mathbf x)}.$$

But since $s_x \gt 0$, division by $s_x$ is an increasing function. Subtracting any constant like $\bar x$ similarly is an increasing function. Therefore the extremes of $\mathbf x$ correspond to the extremes of $\mathbf y$. Consequently, normalization of $\mathbf y = y_1, y_2, \ldots,y_n$ produces

$$\eqalign{ z_i^\prime &= \frac{y_i - \min(\mathbf{y})}{\max(\mathbf y) - \min(\mathbf y)} \\ &= \frac{\frac{x_i - \bar x}{s_x} - \min_j(\frac{x_j - \bar x}{s_x})}{\max_j(\frac{x_j - \bar x}{s_x}) - \min_j(\frac{x_j - \bar x}{s_x})} \\ &= \frac{{x_i - \bar x} - \min_j({x_j - \bar x})}{\max_j({x_j - \bar x}) - \min_j({x_j - \bar x})} \\ &= \frac{x_i - \min(\mathbf{x})}{\max(\mathbf x) - \min(\mathbf x)} \\ &= z_i. }$$

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