[Math] Standardizing A Random Variable That is Normally Distributed

normal distributionprobabilityrandom variablesstatistics

To standardize a random variable that is normally distributed, it makes absolute sense to subtract the expected value $\mu$ , from each value that the random variable can assume–it shifts all of the values such that the expected value is centered at the origin. But how does dividing by the standard deviation play a role in the standardization of a random variable? That part is not as intuitive to me as is subtracting $\mu$.

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Let $X \sim N(\mu,\sigma^2)$.

Let $Y = \large\frac{X-\mu}{\sigma}$.

$E[Y] = \large\frac{E[X] - \mu}{\sigma} = \large\frac{\mu-\mu}{\sigma} = 0$

$\text{Var}(Y) = \large\frac{1}{\sigma^2}\text{Var}(X) = \large\frac{1}{\sigma^2}\sigma^2 = 1$.

So that $Y \sim N(0,1)$.

This is precisely why we subtract the mean and divide by the standard deviation.

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[Math] Expected value of a normally distributed random variable

Write $V = \mu + \sigma Z$, where $Z$ is the standard normal random variable. Then $$ \mathbb{E}(W) = r \mathbb{E}\left( \left(\mu + \sigma Z\right)^3 \right) = r \left( \mu^3 +3 \mu^2 \sigma \mathbb{E}(Z) + 3 \mu \sigma^2 \mathbb{E}(Z^2) + \sigma^3 \mathbb{E}(Z^3) \right) $$ Since $Z$ is symmetric, i.e. $Z \stackrel{d}{=} -Z$, it follows that $\mathbb{E}(Z) = 0$ and $\mathbb{E}(Z^3)=0$. It only remains to evaluate $$ \mathbb{E}(Z^2) = \int_{-\infty}^\infty z^2 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-z^2/2} \mathrm{d} z = \sqrt{\frac{2}{\pi}} \int_0^\infty z^2 \mathrm{e}^{-z^2/2} \mathrm{d} z \stackrel{z^2 = x}{=} \frac{1}{\sqrt{2\pi}} \underbrace{\int_0^\infty \sqrt{x} \mathrm{e}^{-x/2} \mathrm{d} x}_{2^{3/2} \Gamma(3/2) = \sqrt{2\pi}} = 1 $$ Now combine these results to readily obtain $\mathbb{E}(W)$.

[Math] Does standardizing a random variable that is not normally distributed change the underlying distribution

Your histograms look different because your stats package is binning the data. Normalization of a data set by subtracting its mean and dividing by its standard deviation doesn't really change the shape of the data set; it's only scaling the data and shifting the center to achieve a mean of zero and a sd of 1. If you want to confirm this, do a normal QQ plot of the pre data and a normal QQ plot of the post data: you'll see the plots look pretty much the same. The R command

qqnorm(x)

will produce a normal quantile-quantile plot of your data set x, which compares the shape of your data to a normal distribution.

If you really want to change the shape of your data, you need to apply a nonlinear transformation, such as the log transformation (which will pull in a long right tail), or, less drastically, a square root function. (There are many other possibilities.)

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[Math] Expected value of a normally distributed random variable

[Math] Does standardizing a random variable that is not normally distributed change the underlying distribution

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