Easily Solve Math: Geometric Distributions

Wednesday, February 27, 2013

Geometric Distributions

Introduction:

Let we will discuss about the geometric distributions. The geometric distributions should be either of two discrete probability distributions in statistics and probability theory. The probability allocation of number X of Bernoulli trials desired to obtain one success, beard on the set { 1, 2, 3, ...}. The probability allocation of number Y = X − 1 of failures previous to the first success, maintained on the set { 0, 1, 2, 3, ... }

More about geometric distributions:

The two distinct geometric distributions does not mystified with each other.
Most commonly, name shifted geometric allocation is accepted for the former one.
But, to keep away from ambiguity, it is measured wise to point out which is planned, by mentioning the range explicitly.
If the probability of success on every check should be p, then the probability that kth check is the first success is,

$\Pr(X = k) = (1 - p)^{k-1}\,p\,$

Where, k = 1, 2, 3, ....

Consistently, if the probability of success on each check is p, then the probability that there are k failures before the first success is

$\Pr(Y=k) = (1 - p)^k\,p\,$

Where, k = 0, 1, 2, 3, ....

In both case, the series of probabilities is a geometric series.

Example:

Assume a normal die is thrown frequently until the first time a "1" appears. The probability allocation of the number of times it is thrown is holded on the endless set { 1, 2, 3, ... }. They should have a geometric allocation with p = 1/6.

Moments and cumulants:

The predictable value of a geometrically allocated random variable X is 1/p , variance is (1 − p)/p2

$\mathrm{E}(X) = \frac{1}{p}, \qquad\mathrm{var}(X) = \frac{1-p}{p^2}.$

Likewise, the expected value of the geometrically dispersed random variable Y is (1 − p)/p, and its variance is (1 − p)/p2

$\mathrm{E}(Y) = \frac{1-p}{p}, \qquad\mathrm{var}(Y) = \frac{1-p}{p^2}.$

Let μ = (1 − p)/p be the expected value of Y and then the cumulants κn of the probability distributions of Y satisfy the recursion

$\kappa_{n+1} = \mu(\mu+1) \frac{d\kappa_n}{d\mu}.$

Outline of proof:

That the expected value is (1 − p)/p can be shown in the following way. Let Y be as above. Then

$\begin{align} \mathrm{E}(Y) & {} =\sum_{k=0}^\infty (1-p)^k p\cdot k \\ & {} =p\sum_{k=0}^\infty(1-p)^k k \\ & {} = p\left[\frac{d}{dp}\left(-\sum_{k=0}^\infty (1-p)^k\right)\right](1-p) \\ & {} =-p(1-p)\frac{d}{dp}\frac{1}{p}=\frac{1-p}{p}. \end{align}$

Easily Solve Math

Wednesday, February 27, 2013

Geometric Distributions

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