# Difference between revisions of "Eleisha's Segment 7: Central Tendency and Moments"

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<math> -2 \int_{-\infty}^a -2(x-a)p(x) dx + \int_a^\infty -2(x-a)p(x)dx = 0 </math> | <math> -2 \int_{-\infty}^a -2(x-a)p(x) dx + \int_a^\infty -2(x-a)p(x)dx = 0 </math> | ||

− | <math> -2 \int_{-\infty}^{\infty} xp(x)dx + 2a\int_{-\infty}^{\infty} p(x) dx | + | <math> -2 \int_{-\infty}^{\infty} xp(x)dx + 2a\int_{-\infty}^{\infty} p(x) dx = 0 </math> |

− | |||

− | <math> -2 | + | <math> -2\langle x \rangle + 2a(1) = 0 </math> |

+ | |||

+ | <math> a = \langle x \rangle </math> | ||

## Revision as of 11:20, 22 February 2014

** To Calculate: **

1. Prove the result of slide 3 the "mechanical way" by setting the derivative of something equal to zero, and solving.

You want to minimize the function: **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta^2 = \langle (x-a)^2 \rangle }**

Take the derivative and set it equal to zero.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\Delta^2}{da} \left[\int_{-\infty}^{\infty} (x-a)^2p(x)dx\right] = 0 }**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\Delta^2}{da} \left[ \int_{-\infty}^{a} (x-a)^2p(x)dx + \int_{a}^{\infty} (x-a)^2p(x)dx \right] = 0 }**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2 \int_{-\infty}^a -2(x-a)p(x) dx + \int_a^\infty -2(x-a)p(x)dx = 0 }**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2 \int_{-\infty}^{\infty} xp(x)dx + 2a\int_{-\infty}^{\infty} p(x) dx = 0 }**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2\langle x \rangle + 2a(1) = 0 }**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = \langle x \rangle }**

2. Give an example of a function p(x), with a maximum at x=0, whose third moment **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M_3}**
exists, but whose fourth moment **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M_4 }**
doesn't exist.

3. List some good and bad things about using the median instead of the mean for summarizing a distribution's central value.

Using the median gives you the central value of the distribution where half of the probability distribution is to the left of this value and half is to the right. This could be useful in particular for skewed distributions. When summarizing the distribution of some data the median is not as affected by outliers as compared to the mean. The mean gives you the expected value of a particular distribution. This may be different from the median.

** To Think About: **

1. This segment assumed that p(x) is a known probability distribution. But what if you know p(x) only experimentally. That is, you can draw random values of x from the distribution. How would you estimate its moments?

2. High moments (e.g., 4 or higher) are algebraically pretty, but they are rarely useful because they are very hard to measure accurately in experimental data. Why is this true?

3. Even knowing that it is useless, how would you find the formula for I_8, the eighth semi-invariant?

**Back To ** Eleisha Jackson