A question of how to transform data to a desired mean and standard deviation has been answered here. However I would like to understand the properties that make this possible. How does one prove that the following is appropriate?



The answers you quoted goes as follows :

Suppose you start $\{x_i\}$ with mean $m_1$ and non-zero standard deviation $s_1$ and you want to arrive at a similar set with mean $m_2$ and standard deviation $s_2$.

Then multiplying all your values by $\frac{s_2}{s_1}$ will give a set with mean $m_1 \times \frac{s_2}{s_1}$ and standard deviation $s_2$.

Now adding $m_2 - m_1 \times \frac{s_2}{s_1}$ will give a set with mean $m_2$ and standard deviation $s_2$.

So a new set $\{y_i\}$ with $$y_i= m_2+ (x_i- m_1) \times \frac{s_2}{s_1} $$ has mean $m_2$ and standard deviation $s_2$.

You woulEd get the same result with the three steps: translate the mean to $0$, scale to the desired standard deviation; translate to the desired mean.

To proove that the transformation : $$y_i= m_2+ (x_i- m_1) \times \frac{s_2}{s_1} $$

will give you desired standard deviation and mean, just compute the standard deviation and mean of this expression of the $y_i$'s :

$$\mathbb{E}(y_i) = \mathbb{E}(m_2+ (x_i- m_1) \times \frac{s_2}{s_1}) = m_2+ (\mathbb{E}(x_i)- m_1) \times \frac{s_2}{s_1} = m_2$$ and :

$$\mathbb{V}(y_i) = \mathbb{V}(m_2+ (x_i- m_1) \times \frac{s_2}{s_1}) = (\mathbb{V}(x_i) \times \frac{s_2^2}{s_1^2} = s_1^2 \frac{s_2^2}{s_1^2} = s_2^2$$

Remebre that the expectation is linear and that the variance is quadratic, meaning that for a random variable $X$ and scalars $a,b$,

$$\mathbb{E}(aX+b) = a\mathbb{E}(X)+b \text{ and } \mathbb{V}(aX+b) = a^2 \mathbb{V}(X)$$.

For the sake of completeness, remember also that the standard deviation is the square root of the variance..

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