Lecture#17.pdf - 13 The Norma or Gaussian Distribution The...

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Unformatted text preview: 13 The Norma] or Gaussian Distribution The normal or Gaussian density function, which has the familiar bell—shaped curve, is omnipresent throughout the world of statistics and probability. If X is a Gaussian random 1variable then its probability density function fflx}, sometimes denoted by Nip... 02}, is l fxtx} = Miami} = —e-‘I-M’W for — oo «5. x a: no. {1.1} chE 142 Continuous Distribution Functions It may be shown that expectation is ELY] = p. and die variance Var[X] = o2. Thus the mean and variance are the actual parameters of the distribution. The normal density function for selected values of its parameters is shown in Figure 14. 113 ll? illi 115 {L4 113 ll! ill u . .I -- I'_ . —I5 —4 —2 ll 2 4 ti Figure 14. The normal distribution for different values of a. As is apparent from this figure. the normal distribution is symmetric about its mean value u. It is short and flat when tr is large and tall and skinny when or is small. As we shall see momentarily, approximately 63% of the area under fflx] lies within one standard deviation of its mean while 95% lies within two standard deviations. The cumulative distribution function for a random variable that is normally distributed has no closed form, i.e., we cannot write Fxtx} in the form of an equation as we have been able to do with other distributions. This means that we must have recourse to precomputed tables. Usually these tables provide data relating to the standard normal distribution NIIEI. I], having mean i} and variance 1. The probability density function of a standard normal random variable is obtained by substituting the values p. = i} and o2 = 1 into Equation {7.1} . Because of its importance. and to avoid confusion, the probability density function of a standard normal random variable X is often denoted 43510:} and its cumulative distribution function by Ibflx]. Mathematical tables generally provide die values of 'DIIEJC] in increments of (101 from x = {l [where ore] = {1.5m to J: = 3 {where 111(3) = EHQET}. Some tables include values of I up to x = 5 (@{5} = 0.999999'3]. Since a standard normal random variable X has variance and standard deviation equal to i, die quantity 111(1) — dirt—l} is the probability that X lies within one standard deviation of its mean. ${2}—¢={—2] is the probability that it lies within two standard deviations of its mean. and so on. The symmetry of the normal distribution allows us to use the standard tables to find ¢{x} for negative arguments. For example, the symmetry of the standard normal distribution around zero implies that Prob{X 5 —l} = P‘robL'ltr 31}andhence Pmb{X 5 —l] = l — Prob{X 51]. vaiously. this holds more generally: for any nonnegative value or. we have Prob{X 5 —-:r] = 1 —Prob{X 5 or]. The values provided by tables of die standard normal distribution can be used to find the values of all normal distributions. Nflr, o 2}. Indeed, if a random variable X has the distribution Ntu. o2}. then the random variable Z = [X — ulfo' has a standard normal distribution. To see this, observe 13 The Normal or Gaussian Distribution 143 that Problz E z] = Probltx — Julia 5 z} = 11mlI-{X E crz+rtl ”1+“ 1 '1 '1 _ E—tI—ul He J): — _m «321:3 I l = f e‘Ffldt _m 271w for t = {x — “we. Thus, if X has a normal distribution with expectation p. and standard deviation or, then. for any constant or, Pmbll' 5 a] = PmbtiX — elf-tr 5 to — NIH} = Pmbtz 5 to: — sift-r]- In words, the probability that X, a normally distributed random variable, is less dran or is equal to the probability that the standard normal random variable is less dran {or — “no. l[libsen-re also that if a random variable X has distribution Nut. o3}, then the derived random variable Y = 0X + b has a normal distribution with mean pry = arr + b and standard deviation of = Ialor. i.e.. Y is More + b. [Halit- Example 16 Assume that test scores in CSCST‘Q are normally distributed with mean pr = 33 and standard deviation or = 8. Since we wish to use the normal distribution, we shall treat die test scores as real numbers rather than integer values. We wish to compute the probability of scores between T5 and '95. w¢955X5sa=4hmr5om—PmmX5in = Prob{Z 5 (95 — says} — Prob{Z 5 [TS — 331:3} =nmw5Lfl—amm5—n =oun_opn = ossaz — {1158? = onus. Observe that if we were to replace '95 with 91 in the above analysis, we would effectively compute the probability of scores drat lies within one standard deviation of die mean (i.e., scores in the range 83 :I: E}. This yields PIDb{T5 5 Jr' 5 91} = Prob{Z 51}_ Prob{Z 5 _1:. =o.s413 _o.rssr = (1.5326. This result is not restricted to die example given above. Approximately two—drirds of the outcomes associated with normally distributed random variable lie within one standard deviation of the mean. Furthermore, since @{2} — tilt—2) = (1.9"??3 — {l — GHTTS} = [1.9546, approximately 95% of the outcomes of a normally distributed random variable lie within two standard deviations of the mean. We shall now compute the moment generating function for the normal distribution. We first consider the standard normal case. Let the random variable Z be normally distributed with mean {1 and variance 1. Then Mzo): Ere”) = f Eezfztzla'z = f —t'_'4:| 5“ 1 —tz:—Efiz+61—Hzlfl f” l —r -9): m1 2 = e dz = e z “i “i dz [on 1.? 2H —oc| 1.? 2H m l ear—zz‘fldz 2H F 1 =86 ,r: J—Erfi—Smdz = gate f lezldz = 231”. 144 Continuous Distribution Functions Now let X be a normally distributed random variable with mean u and variance 02 and let Z be standard normal.Then X = u+oZand Mum = Eran = Emma] = Wendi] = ewMzrioS] = eweflrziafllz = Eaewlalgr Let us now consider a random variable X constructed as a sum of normally distributed random variables. Recall from Equation [5.3). that if XI. X2. . . .. X” are n independent random variables andX=X|+Xg+---+thhen Mflfll = Elgexlgaxz _ _ £81.] 2 Mx1ifllM11{3}"‘MI.{3l- If these n random variables are annually distributed, with expectations n,- and variances or,-2 respectively. then Mflfl] = Eme+arfazrs Eme+a=§azr2 Ehawfisln = E(itL+nz+---+.u.‘16'+licri+cr§+---+cr,E‘r|-rlr2_ Settinga = fl1+fl2+'”+#fl ando‘2 = o|2+o§+ ---+o";;’iI itbecomesappar‘entthatX is normally distributed with mean u and variance o2. This result concerning sums of normally distributed random variables leads us to an important application of the normal distribution. It turns out first the normal distributionI in certain cases may be used as an approximation to the discrete binomial distribution. [fthe random variable X has a binomial distribution for which both It and die mean value np are large. then the density Function of X is close to that of the normal distribution. In other words, the binomial distribution becomes more and more normal as n becomes large. We provide the following theorem. stated without proof. Theorem '13.] [De Moivre-Laplace] Let X1. X2. .... X" be n independent andI identically distributed Bernoulli random variables eaelr fairing the value I with probability p and the value [1 with probabilityq = l — p. Let ” X —E[X] X = E Xi and X“ = —. a. «sum Then 1 b 1 lim Probe-r: re ct: =_ e' flax. .l'I—HJICI { _ _ ] JE [a Observe that X is a binomial random variable and denotes the number of successes in n independent Bernoulli trials with expectation up and standard deviation June. The random variable X‘ is the normalized sum drat takes the values k—np i5: . k=fl,l....,n, 1.3311“! with probabilities Prob{X* = .5} = cgpqu-t. s = o,1,...,n. Example I? As an exarrrple of the use of the normal distribution in approximating a binomial distribution. let us compare the results obtained for binomial distribution with n = lll and p = [1.5 and with die normal distribution with mean p. = np = 5 and o = .tnpq = 62.5. We shall just TA The Gamma Distribution 145 check a single valueI Prob{X 5 3]. For the binomial distribution we have 3 Z coo, kro.5*o.51“-* = ossss. H For the normal distribution. we obtain Prob{X 5 S} = Prob{Z 51:3 — SLIHJZj} = Prob{Z 5 l.3'9?4] = (1.97”. The approximation is not particularly accurate. but the reason is drat H is not sufficiently large. It is possible to add a (real-rtirl'rJamit)I correction which attempts to adjust for the fact that we are approximating a discrete distribution widr a continuous one. However. for small values of n it is relatively easy to compute binomial coefficients. This task becomes much harder for large :1. but it is precisely these cases that die approximation by fire normal distribution beconres more accurate. One final comment on the normal distribution: the central limit theorem. which is discussed in Section 3.5, states that the sum of n independent random variables tends to the normal distribution in the limit as n —> 00. This is an important result in many branches of statistics and probability for it asserts drat no matter which distribution the in random variables have. so long as they are independent. the distribution of their sum can be approxinrated by fire nomral distribution. Thus our previous comments concerning approximating a binomial distribution with a normal distribution should not be too surprising since a binomial random variable is a sum of n independent Bernoulli random variables. ...
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