IIT JEE , AIEEE Forum - Mathematics , Algebra - Let x and y be uniformly distributed, independent random variables on [0,1

sneha.bagri

Let x and y be uniformly distributed, independent random variables on [0,1]. The probability that the distance between x and y is less than 1/2 is

1) 2/3

2)1/4

3)3/4

4)1/3

Its urgent.

budokai_tenkaichi_returns

i know it .

its geometrical probability ..

just find the ratio of area 1 by area 2

area 1 : graph plotted by |x-y|<1/2

area 2 : area of square ,,..centre origin .and sides of length 2 units ...

budokai_tenkaichi_returns

hope u r gettin it

budokai_tenkaichi_returns

i m gettin 1/3 ..maybe

sneha.bagri

i m not gettin u

budokai_tenkaichi_returns

ok do u know sumthing abt geometrical probability ..

(maybe my ans is wrong ..but my approach is correct)

sneha.bagri

no i dont

budokai_tenkaichi_returns

ok ..

then lemme ...

wait for a while ..

budokai_tenkaichi_returns

accha ..is my ans rite

budokai_tenkaichi_returns

dont read this

Geometric
Probability mass function
Cumulative distribution function
Parameters	$0< p \leq 1$ success probability (real)
Support	$k \in \{1,2,3,\dots\}\!$
Probability mass function (pmf)	$(1 - p)^{k-1}\,p\!$
Cumulative distribution function (cdf)	$1-(1 - p)^k\!$
Mean	$\frac{1}{p}\!$
Median	$\left\lceil \frac{-\log(2)}{\log(1-p)} \right\rceil\!$ (not unique if $- log(2) / log(1 - p)$ is an integer)
Mode	1
Variance	$\frac{1-p}{p^2}\!$
Skewness	$\frac{2-p}{\sqrt{1-p}}\!$
Excess kurtosis	$6+\frac{p^2}{1-p}\!$
Entropy	$-\frac{1-p}{p}\ln(1-p)-\ln p\!$
Moment-generating function (mgf)	$\frac{pe^{t}}{1-(1-p) e^t}\!$
Characteristic function	$\frac{pe^{it}}{1-q\,e^{it}}\!$ (where $q = 1 - p$ )

In probability theory and statistics, the geometric distribution is either of two discrete probability distributions:

the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ...}, or

the probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... }.

Which of these one calls "the" geometric distribution is a matter of convention and convenience.

These two different geometric distributions should not be confused with each other. Often, the name shifted geometric distribution is adopted for the latter one (distribution of the number Y); however, to avoid ambiguity, it is considered wise to indicate which is intended, by mentioning the range explicitly.

If the probability of success on each trial is p, then the probability that k trials are needed to get one success is

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

for k = 1, 2, 3, ....

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

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

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

In either case, the sequence of probabilities is a geometric sequence.

For example, suppose an ordinary die is thrown repeatedly until the first time a "1" appears. The probability distribution of the number of times it is thrown is supported on the infinite set { 1, 2, 3, ... } and is a geometric distribution with p = 1/6.

Moments and cumulants

The expected value of a geometrically distributed random variable X is 1/p and the variance is (1 − p)/p²:

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

Similarly, the expected value of the geometrically distributed random variable Y is $(1 - p) / p$ , and its variance is $(1 - p) / p 2$ :

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

Let $μ = (1 - p) / p$ be the expected value of Y. Then the cumulants $κ n$ of the probability distribution of Y satisfy the recursion

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

Parameter estimation

For both variants of the geometric distribution, the parameter p can be estimated by equating the expected value with the sample mean. This is the method of moments, which in this case happens to yield maximum likelihood estimates of p.

Specifically, for the first variant let $k 1,..., k n$ be a sample where $k_i \geq 1$ for $i = 1,..., n$ . Then p can be estimated as

$\widehat{p} = \left(\frac1n \sum_{i=1}^n k_i\right)^{-1}. \!$

In Bayesian inference, the Beta distribution is the conjugate prior distribution for the parameter p. If this parameter is given a Beta(α, β) prior, then the posterior distribution is

$p \sim \mathrm{Beta}\left(\alpha+n,\ \beta+\sum_{i=1}^n (k_i-1)\right). \!$

The posterior mean $E[p]$ approaches the maximum likelihood estimate $\widehat{p}$ as α and β approach zero.

In the alternative case, let $k 1,..., k n$ be a sample where $k_i \geq 0$ for $i = 1,..., n$ . Then p can be estimated as

$\widehat{p} = \left(1 + \frac1n \sum_{i=1}^n k_i\right)^{-1}. \!$

The posterior distribution of p given a Beta(α, β) prior is

$p \sim \mathrm{Beta}\left(\alpha+n,\ \beta+\sum_{i=1}^n k_i\right). \!$

Again the posterior mean $E[p]$ approaches the maximum likelihood estimate $\widehat{p}$ as α and β approach zero.

Other properties

The probability-generating functions of X and Y are, respectively,

$G_X(s) = \frac{s\,p}{1-s\,(1-p)}, \!$

$G_Y(s) = \frac{p}{1-s\,(1-p)}, \quad |s| < (1-p)^{-1}. \!$

Like its continuous analogue (the exponential distribution), the geometric distribution is memoryless. That means that if you intend to repeat an experiment until the first success, then, given that the first success has not yet occurred, the conditional probability distribution of the number of additional trials does not depend on how many failures have been observed. The die one throws or the coin one tosses does not have a "memory" of these failures. The geometric distribution is in fact the only memoryless discrete distribution.

Among all discrete probability distributions supported on {1, 2, 3, ... } with given expected value μ, the geometric distribution X with parameter p = 1/μ is the one with the largest entropy.

The geometric distribution of the number Y of failures before the first success is infinitely divisible, i.e., for any positive integer n, there exist independent identically distributed random variables Y₁, ..., Y_n whose sum has the same distribution that Y has. These will not be geometrically distributed unless n = 1; they follow a negative binomial distribution.

The decimal digits of the geometrically distributed random variable Y are a sequence of independent (and not identically distributed) random variables. For example, the hundreds digit D has this probability distribution:

$\Pr(D=d) = {q^{100d} \over {1 + q^{100} + q^{200} + \cdots + q^{900}}},$

where q = 1 − p, and similarly for the other digits, and, more generally, similarly for numeral systems with other bases than 10. When the base is 2, this shows that a geometrically distributed random variable can be written as a sum of independent random variables whose probability distributions are indecomposable.

Related distributions

The geometric distribution Y is a special case of the negative binomial distribution, with r = 1. More generally, if Y₁,...,Y_r are independent geometrically distributed variables with parameter p, then

$Z = \sum_{m=1}^r Y_m$

follows a negative binomial distribution with parameters r and p.

If Y₁,...,Y_r are independent geometrically distributed variables (with possibly different success parameters $p (m)$ ), then their minimum

$W = \min_{m} Y_m\,$

is also geometrically distributed, with parameter p given by

1 -	∏	(1 - p^(m)).
	m

Suppose 0 < r < 1, and for k = 1, 2, 3, ... the random variable X_k has a Poisson distribution with expected value r^k/k. Then

$\sum_{k=1}^\infty k\,X_k$

has a geometric distribution taking values in the set {0, 1, 2, ...}, with expected value r/(1 − r).

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Moments and cumulants

Parameter estimation

Other properties

Related distributions