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

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: $\displaystyle \Delta^2 = \langle (x-a)^2 \rangle$

Take the derivative and set it equal to zero.

$\displaystyle \frac{d\Delta^2}{da} \left[\int_{-\infty}^{\infty} (x-a)^2p(x)dx\right] = 0$

$\displaystyle \frac{d\Delta^2}{da} \left[ \int_{-\infty}^{a} (x-a)^2p(x)dx + \int_{a}^{\infty} (x-a)^2p(x)dx \right] = 0$

$\displaystyle -2 \int_{-\infty}^a -2(x-a)p(x) dx + \int_a^\infty -2(x-a)p(x)dx = 0$

$\displaystyle -2 \int_{-\infty}^{\infty} xp(x)dx + 2a\int_{-\infty}^{\infty} p(x) dx = 0$

$\displaystyle -2\langle x \rangle + 2a(1) = 0$

$\displaystyle a = \langle x \rangle$

2. Give an example of a function p(x), with a maximum at x=0, whose third moment $\displaystyle M_3$ exists, but whose fourth moment $\displaystyle M_4$ 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.

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