MeanAndVariance: Difference between revisions
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Var(X) = \int (x^2 - 2 x \mu + \mu^2) f(x) dx = \int x^2 f(x) d - 2 \mu \int x f(x) dx + \mu^2 \int f(x) dx = \int x^2 f(x) d - 2 \mu^2 + \mu^2 = \int x^2 f(x) dx - \mu^2 | Var(X) = \int (x^2 - 2 x \mu + \mu^2) f(x) dx = \int x^2 f(x) d - 2 \mu \int x f(x) dx + \mu^2 \int f(x) dx = \int x^2 f(x) d - 2 \mu^2 + \mu^2 = \int x^2 f(x) dx - \mu^2 | ||
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or just | |||
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\sigma^2 = Var(X) = \int x^2 f(x) dx - \mu^2 | |||
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Revision as of 21:23, 24 May 2017
Summary
This page covers mean and variance definitions in detail.
Mean
Continuous Random Variables
If we have a continuous random variable $ X $ with a probability density function $ f(x) $, the mean and variance are given by:
$ \mu = E[X] = \int x f(x) dx $
(where the integral is over the range of x values)
Discrete Random Variable
The mean of a discrete random variable $ X $ with discrete values $ x_i, 1 \leq i \leq n $ and a probability mass function $ p_i $ is given by the expression:
$ \mu = E[X] = \sum_{i=1}^{n} p_i x_i $
Note that by definition, the probability mass function must sum to 1:
$ \sum_{i=1}^{n} p_i = 1 $
If we assume a uniform probability for each value, then the probability mass function of component i is just:
$ p_i = \frac{1}{n} $
Variance
Continuous Random Variable
The variance of a continuous random variable is given by:
$ \sigma^2 = Var(X)^2 = Var(X) = \int (x - \mu)^2 f(x) dx $
Note that this can be expanded and simplified,
$ Var(X) = \int (x^2 - 2 x \mu + \mu^2) f(x) dx = \int x^2 f(x) d - 2 \mu \int x f(x) dx + \mu^2 \int f(x) dx = \int x^2 f(x) d - 2 \mu^2 + \mu^2 = \int x^2 f(x) dx - \mu^2 $
or just
$ \sigma^2 = Var(X) = \int x^2 f(x) dx - \mu^2 $
which is equivalent to saying:
$ Var(X) = E[X^2] - E[X]^2 $