Why do we take the square root of variance to arrive at standard deviation?

Why do we take the square root of variance to arrive at standard deviation?

The variance is the average of the squared differences from the mean. Because of this squaring, the variance is no longer in the same unit of measurement as the original data. Taking the root of the variance means the standard deviation is restored to the original unit of measure and therefore much easier to interpret.

What is the relation between variance and expectation?

Given a random variable, we often compute the expectation and variance, two important summary statistics. The expectation describes the average value and the variance describes the spread (amount of variability) around the expectation.

Why square for variance and then square root back again for standard deviation?

Because the differences are squared, the units of variance are not the same as the units of the data. Therefore, the standard deviation is reported as the square root of the variance and the units then correspond to those of the data set.

When calculating the standard deviation Why do we square the differences between each observation and the mean only to later take the square root out of it?

Squaring the difference from the mean has a couple of reasons. Variance is defined as the 2nd moment of the deviation (the R.V here is (x−μ)) and thus the square as moments are simply the expectations of higher powers of the random variable.

How do you calculate population mean?

The population mean is the mean or average of all values in the given population and is calculated by the sum of all values in population denoted by the summation of X divided by the number of values in population which is denoted by N.

How to estimate the expected value and variance of a random variable?

I have asked this in a general way here: Approximating the expected value and variance of the function of a (continuous univariate) random variable . I have also read answers and coments to this question: Variance of powers of a random variable , but I think it refers to integer powers, which is not my case.

What is the expectation of a Cauchy random variable?

A Cauchy random variable takes a value in (−∞,∞) with the fol- lowing symmetric and bell-shaped density function. f(x) = 1 π[1+(x−µ)2] The expectation of Bernoulli random variable implies that since an indicator function of a random variable is a Bernoulli random variable, its expectation equals the probability.

How to calculate the expected value of a distribution?

Otherwise the standard tool to approximate the expectation is to use the Taylor series of y = X − 1, 1 + y = 1 − y / 2 + y 2 / 8 − y 3 / 16 + 5 y 4 / 128 … and since you already have the the MGF of X, you should be able to calculate that quickly…

How to find the expected value of var ( X )?

As whuber points out in comments, all we need to find is E ( X), since Var ( X) = E ( X) − E ( X) 2. if you are only interested in the upper bound of the expectation, you can use Jensen’s Inequality to immediately upper bound E [ X] ≤ E [ X], if E [ X] is sufficiently close to 1, the approximation would be quite good..