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This article provides a comprehensive overview of random variables, including their definition, properties, and various examples. It covers concepts such as distribution functions, probability density functions, continuous-type and discrete-type random variables, as well as specific distributions like the normal distribution, exponential distribution, and more.
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3. Random Variables (Fig.3.1)
Random Variables If belongs to the associated field F, then the probability of A is well defined . in that case we can say : Random Variable (r.v):A finite single valued function that maps the set of all experimental outcomes into the set of real numbers R is said to be a r.v, if the set is an event for every x in R. (3-1)
Random Variables • X :r.v, B represents semi-infinite intervals of the form • The Borel collection B of such subsets of R is the smallest -field of subsets of R that includes all semi-infinite intervals of the above form. • if X is a r.v, then is an event for every x. (3-2)
Distribution Function The role of the subscript X in (3-3) is only to identify the actual r.v. is said to the Probability Distribution Function associated with the r.v X. (3-3)
Properties of a PDF if g(x) is a distribution function, then (i) (ii) if then (iii) for all x. (3-4)
Properties of a PDF Supposedefined in (3-3) (i) and (ii) I f then the subset Consequently the event since implies As a result =>the probability distribution function is nonneg ative and monotone nondecreasing. (3-5) (3-6) (3-7)
Properties of a PDF (iii) Let and consider the event since using mutually exclusive property of events we get But and hence (3-8) (3-9) (3-10) (3-11)
Properties of a PDF Thus But the right limit of x, and hence i.e., is right-continuous, justifying all properties of a distribution function. (3-12)
Additional Properties of a PDF (iv) If for some then This follows, since implies is the null set, and for any will be a subset of the null set. (v) We have (vi) The events and are mutually exclusive and their union represents the event (3-13) (3-14) (3-15)
Additional Properties of a PDF (vii) Let and From (3-15) or Thus the only discontinuities of a distribution function occur at points where is satisfied. (3-16) (3-17) (3-18) (3-19)
continuous-type & discrete-type r.v • X is said to be a continuous-type r.v if And from => • If is constant except for a finite number of jump discontinuities (piece-wise constant; step-type), then X is said to be a discrete-type r.v. If is such a discontinuity point, then (3-20)
Fig. 3.2 Example • Example 3.1: X is a r.v such that Find Solution: For so that and for so that (Fig.3.2) at a point of discontinuity we get
Fig.3.3 Example Example 3.2: Toss a coin. Suppose the r.v X is such that Find Solution: For so that at a point of discontinuity we get
Example Example:3.3 A fair coin is tossed twice, and let the r.v X represent the number of heads. Find Solution: In this case
Fig. 3.4 Example From Fig.3.4,
Probability density function (p.d.f) (3-21) Since from the monotone-nondecreasing nature of => for all x
Fig. 3.5 Probability density function (p.d.f) • will be a continuous function, if X is a continuous type r.v • if X is a discrete type r.v as in (3-20), then its p.d.f has the general form (Fig. 3.5) • represent the jump-discontinuity points in As Fig. 3.5 shows represents a collection of positive discrete masses, and it is known as the probability mass function (p.m.f ) in the discrete case.
(a) (b) Probability density function (p.d.f) From (3-23), Since => the area under in the interval represents the probability in (3-22). (3-22)
Fig. 3.7 Continuous-type random variables (3-23) • Normal (Gaussian): Where the notation will be used to represent (3-23).
Fig. 3.8 Continuous-type random variables • Uniform: (Fig. 3.8) (3.24)
Fig. 3.9 Continuous-type random variables • Exponential: (Fig. 3.9) (3.25)
Exponential distribution • Assume the occurences of nonoverlapping intervals are independent, and assume: • q(t): the probability that in a time interval t no event has occurred. • x: the waiting time to the first arrival • Then we have: P(x>t)=q(t) • t1 and t2 : two consecutive nonoverlapping intervals,
Exponential distribution • Then we have: q(t1) q(t2) = q(t1+t2) • The only bounded solution is: So the pdf is exponential. • If the occurrences of events over nonoverlapping intervals are independent, the corresponding pdf has to be exponential.
Memoryless property of exponential distribution • Let . Consider the events and . Then
Fig. 3.10 Continuous-type random variables 4. Gamma: if (Fig. 3.10) If an integer (3-26)
Continuous-type random variables • The exponential random variable is a special case of gamma distribution with • The (chi-square) random variable with n degrees of freedom is a special case of gamma distribution with
Fig. 3.11 Continuous-type random variables 5. Beta: if (Fig. 3.11) where the Beta function is defined as Beta distribution with a=b=1 is the uniform distribution on (0,1). (3-27) (3-28)
6. Chi-Square: if (Fig. 3.12) Note that is the same as Gamma 7. Rayleigh: if (Fig. 3.13) 8. Nakagami – m distribution: (3-29) Fig. 3.12 (3-30) Fig. 3.13 (3-31)
9. Cauchy: if (Fig. 3.14) 10. Laplace: (Fig. 3.15) 11. Student’s t-distribution with n degrees of freedom (Fig 3.16) (3-32) (3-33) (3-34) Fig. 3.15 Fig. 3.14 Fig. 3.16
12. Fisher’s F-distribution (3-35)
Fig. 3.17 Discrete-type random variables 1. Bernoulli: X takes the values (0,1), and 2. Binomial: if (Fig. 3.17) (3-36) (3-37) The probability of k successes in n experiments with replacement (in ball drawing)
Fig. 3.18 Discrete-type random variables 3. Poisson: if (Fig. 3.18) (3-38)
Discrete-type random variables • Poisson distribution represents the number of occurrences of a rare event in a large number of trials. • Pk Increasing with k from 0 to λ and decreasing after that.
4. Hypergeometric: The probability of k successes in n experiments without replacement (ball drawing) 5. Geometric: if (3-39) (3-40)
6. Negative Binomial: ~ if 7. Discrete-Uniform: (3-41) (3-42)
Polya’s distribution: A box contains a white balls and b black balls. A ball is drawn at random, and it is replaced along with c balls of the same color. If X represents the number of white balls drawn in n such draws, find the probability mass function of X.
Solution: : probability of drawing k successive white balls :probability of drawing k white balls ,followed by n – k black balls (3-43) (3-44)
Polya distribution: probability of getting k white balls in n draws. if draws are done with replacement, then c = 0 and (3-45) simplifies to the binomial distribution (3-45)
if the draws are conducted without replacement, Then c = – 1 in (3-45), and it gives which represents the hypergeometric distribution. (3-46)
Finally c = +1 gives (replacements are doubled) (3-47) is Polya’s +1 distribution. (3-47)