Can a Normal Distribution be specified by its mean and cubic deviation

normal distributionstandard deviationstatistics

I recently read that:

a normal distribution is completely specified by its mean and standard deviation.

That makes a lot of sense. But I was wondering isn't it also true that it could be completely specified by its mean and the cubic deviation? Or quadratic one? Or even the mean deviation?

If we consider the standard deviation formula:
$$\sigma = (\frac{1}{N} \sum_i \lvert x_i – avg \rvert^\color{red}{p})^{1/\color{red}{p}}$$
Then:

p = 1: mean deviation.
p = 2: standard deviation.
p = 3: cubic deviation. I just made this name up.
p = 4: quartic deviation. I just made this name up.
p = 2.3456789: any positive non-integer value of p.

Can any of those deviations completely specify a normal distribution, in addition to the mean value of course?

Best Answer

From the Wikipedia entry, we know that the $p$-th central absolute moment of a $N(\mu,\sigma^2)$ random variable $X$ is $$E[|X-\mu|^p]=\sigma^p\frac{2^{p/2}\Gamma(\frac{p+1}2)}{\sqrt\pi}.$$ If we know this number, and know $p$, we can determine $\tau=\sigma^p$ and then determine $\sigma=\tau^{1/p}$. It might seem paradoxical, but even when $p=1$ this is sufficient to yield $\sigma$.

Related Solutions

[Math] Standard deviation of mean of a set of numbers, which are imprecise

You have defined two distinct standard deviations, describing different things.

The first one describes the standard deviation of the average itself - that is, a measure of how accurately we know the average, ignoring the spread of the set of measurements.

The second one describes the standard deviation of the set of measurements, ignoring the confidence of each measurement.

I assume that what you seek is the standard deviation of the set of all possible sets of measurements, where each measurement has a distribution described by the individual mean and standard deviation. That's a slightly more complicated problem.

Recall that the variance is defined as $$ \text{Var}(X) = E(X^2)-E(X)^2 $$ Now, if $$ X=\frac1N\sum_{i=1}^N X_i $$ where each $X_i\sim \mathcal{N}(x_i,\sigma_i)$, then $E(X)$ is just the average of the $x_i$ values. However, $E(X^2)$ is the average of the expected values of $X_i^2$. And so, we have $$ E(X_i^2) = \text{Var}(X_i)+E(X_i)^2 = \sigma_i^2+x_i^2 $$ and the average value is the sum of the average variance and the average of the $x_i^2$ values.

From here, it is easy to see that the final variance is quite simply the sum of the average of the measurement variances and the variance in the measurement values. That is, taking the square root to get the final standard deviation, $$ \sigma = \sqrt{\frac1N\left(\sum_i \left[\sigma_i^2+(x_i-\mu)^2\right]\right)} $$ where $\mu=\frac1N \sum_i x_i$.

[Math] How to the standard deviation be interpreted when the range is partially impossible

You have assumed a normal distribution for a data which cannot be negative. It doesn't make sense at all. You can use lognormal distribution instead. It is used in Black-Scholes model for pricing options. (Stock prices cannot be negative)

Obviously, I can't tell you if your sample fits such a distribution if I don't have access to the full dataset.

r-script:

require(MASS)
hist(x, freq=F)
fit<-fitdistr(x,"log-normal")$estimate
lines(dlnorm(0:max(x),fit[1],fit[2]), lwd=3)

(x is a sample vector)

Obviously, your sample is way too small here.

Best Answer

Related Solutions

[Math] Standard deviation of mean of a set of numbers, which are imprecise

[Math] How to the standard deviation be interpreted when the range is partially impossible

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