Home page for accesible maths 2.6 Expectation and Related Summaries 2.6.3 Properties of Summary Measures 2.6.5 Higher Moments

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

2.6.4 Standardisation

If $X$ has expectation $\mu_{X}$ and standard deviation $\sigma_{X}<\infty$ then the random variable $Y$ defined as

\displaystyle Y=\frac{X-\mu_{X}}{\sigma_{X}}

has $\operatorname{\mathsf{E}}\left[{Y}\right]=0$ and ${\operatorname{\mathsf{Var}}}\left[{Y}\right]=1$ for any $\mu_{X}$ and $\sigma_{X}<\infty$ .

Proof.

Using properties already discussed,

1.

$\operatorname{\mathsf{E}}\left[{Y}\right]=\operatorname{\mathsf{E}}\left[{% \frac{X-\mu_{X}}{\sigma_{X}}}\right]=\frac{\operatorname{\mathsf{E}}\left[{X}% \right]-\mu_{X}}{\sigma_{X}}=0$ ,
2.

${\operatorname{\mathsf{Var}}}\left[{Y}\right]={\operatorname{\mathsf{Var}}}% \left[{\frac{X-\mu_{X}}{\sigma_{X}}}\right]=\frac{{\operatorname{\mathsf{Var}}% }\left[{X}\right]}{\sigma_{X}^{2}}=\frac{\sigma_{X}^{2}}{\sigma_{X}^{2}}=1.$

∎

The process of subtracting the mean and then dividing by the standard deviation is known as standardisation.

Conversely, if $Y$ has has $\operatorname{\mathsf{E}}\left[{Y}\right]=0$ and ${\operatorname{\mathsf{Var}}}\left[{Y}\right]=1$ then the random variable $X=\mu+\sigma Y$ has $\operatorname{\mathsf{E}}\left[{X}\right]=\mu$ and ${\operatorname{\mathsf{Var}}}\left[{X}\right]=\sigma^{2}$ .