Normal Distribution ch5

Normal Distribution The random variable X has a normal distribution, N(μ, σ2), if its p.d.f. is defined by where the mean μ∈(-∞, ∞), and variance σ2∈(0, ∞). If Z is N(0, 1), Z is called as a standard normal distribution. M(t)= M’(t)= E(X)=

Examples Ex.5.2-3: If Z is N(0,1), then from Table Va on p.686, Φ(1.24)=P(Z≦1.24)=0.8925, P(1.24≦Z≦2.37)=Φ (2.37)-Φ (1.24)=0.9911-0.8925=0.0986 P(-2.37≦Z≦-1.24)= P(1.24 ≦Z≦ 2.37)=0.0986 From Table Vb (right-tail) on p.687, (the negative z is derived from –Z) P(Z>1.24)=0.1075 P(Z≦-2.14)=P(Z≧2.14)=0.0162 Ex.5.2-4: If Z is N(0,1), then find a & b from Table Va & Vb. P(Z≦a)=0.9147 => a=1.37 P(Z≧b)=0.0526 => b=1.62 Percentiles: The 100(1-α) percentile is Zα, where P(Z≧Zα)=α=P(Z≦-Zα) Z1-α=-Zα (the upper 100αpercent point) The 100p percentile is πp, where P(X≦πp)=p p=1-α =>πp=Z1-p=-Zp

Conversion: N(μ,σ2) ) ⇒ N(0,1) Thm5.2-1: If X is N(μ,σ2), then Z=(X-μ)/σis N(0,1). Usage: Ex.5.2-6: X is N(3,16). Ex.5.2-7: X is N(25,36). Find c such that P(|X-25|≤c) =0.9544. Within 2*σabout the mean, X and Z share the same the probability. P(4X8)

Conversion N(μ,σ2) ⇒ χ2(1) Ex.5.2-8: If Z is N(0,1), k=? So Thm5.2-2: If X is N(μ,σ2), then V=[(X-μ)/σ]2=Z2is χ2(1). Pf: V=Z2, Z is N(0,1). Then G(v) of V is G(V) Ex.5.2-8: If Z is N(0,1), k=? So From the chi-square table with r=1, k2=3.841, so k=1.96.

Q-Q Plot Given a set of observations, how to determine its distribution? If its population is large, relative frequency histogram can be used. For small samples, a q-q plot can be used to check. Q-Q plot: say against a “theoretical” normal distribution N(0,1). The mean and variance are obvious, N(0,1): the quantile z1-p. The sample is suspected as N(μ,σ2): the quantile qp. The ideal curve in the plot would be qp=μ+σz1-p. If the actual curve is approximated to a straight line, the distribution of the data is verified as a normal distribution. The reciprocal of the slope is σ. The interception on the x-axis is μ. (Ref. Ex.4.4-9 & Fig.4.4-3 on p.201)

Statistics on Normal Distributions Thm5.3-1: X1,…,Xn are the outcomes on a random sample of size n from the normal distribution N(μ,σ2). The distribution of the sample mean is N(μ,σ2/n). Pf: Thm5.3-2: Z1,…,Zn are independent and all have N(0,1); Then, W=Z12+…+Zn2is χ2(n). Thm5.2-2: If X is N(μ,σ2), then V=[(X-μ)/σ]2=Z2 is χ2(1). Thm4.6-3: Y=X1+…+Xn is χ2(r1+…+rn)= χ2(n) in this case. MX(t)

Example Fig.5.3-1: p.d.f.s of means of samples from N(50,16). is N(50, 16/n).

Theoretical Mean and Sample Mean Cly5.3-1: Z1,…,Zn are independent and have N(μi,σi2),i=1..n; Then, W=[(Z1-μ1)/σi2]2+…+[(Zn-μn)/σn2]2 is χ2(n). Thm5.3-3: X1,…,Xn are the outcomes on a random sample of size n from the normal distribution N(μ,σ2). Then, The sample mean & variance are indep. is χ2(n-1) Pf: (a) omitted; (b): As theoretical mean is replaced by the sample mean, one degree of freedom is lost! MV(t)

Linear Combinations of N(μ,σ2) Ex5.3-2: X1,X2,X3,X4 are a random sample of size 4 from the normal distribution N(76.4,383). Then P(0.711W 7.779)=0.9-0.05=0.85, P(0.352 V 6.251)=0.9-0.05=0.85 Thm5.3-4: If X1,…,Xn are n mutually indep. normal variables with means μ1,…,μn & variances σ12,…,σn2, then the linear function has the normal distribution Pf: By moment-generating function, … Ex5.3-3: X1:N(693.2,22820) and X2:N(631.7,19205) are indep. Find P(X1>X2) Y=X1-X2 is N(61.5,42025). P(X1>X2)=P(Y>0)=

Box-Muller Transformation Ex5.3-4: X1 and X2 have indep. Uniform distributions U(0,1). Consider Two indep. U(0,1) ⇒ two indep. N(0,1)!!

Distribution Function Technique Ex.5.3-5: Z is N(0,1), U is χ2(r), Z and U are independent. The joint p.d.f. of Z and U is χ2(r+1)

Student’s T Distribution Gossett, William Sealy published “t-test” in Biometrika1908 to measure the confidence interval, the deviation of “small samples” from the “real”. Suppose the underlying distribution is normal with unknown σ2. Fig.5.3-2: The T p.d.f. becomes closer to the N(0, 1) p.d.f. as the number of degrees of freedom increases. tα(r) is the 100(1-α) percentile, or the upper 100αpercent point. [Table VI, p.658] f(t)= Only depends on r!

Examples Ex: Suppose T has a t distribution with r=7. From Table VI on p.688, Ex: Suppose T has a t distribution with r=14. Find a constant c, such that P(|T|<c)=0.9

Central Limit Theorem Ex4.6-2: X1,…,Xn are a random sample of size n from a distribution with mean μand variance σ2; then The sample mean: Thm5.4-1: (Central Limit Theorem) If is the mean of a random sample X1,…,Xn of size n from some distribution with a finite mean μand a finite positive variance σ2, then the distribution of is N(0, 1) in the limit as n →∞. Even if Xi is not N(μ,σ2). W= if n is large W

More Examples Ex: Let denote the mean of a random sample of size n=15 from the distribution whose p.d.f. is f(x)=3x2/2, -1<x<1. μ=0, σ2=3/5. Ex5.4-2: Let X1,…,X20 be a random sample of size 20 from the uniform distribution U(0,1). μ=½, σ2=1/12; Y=X1+…+X20. Ex5.4-3: Let denote the mean of a random sample of size n=25 from the distribution whose p.d.f. is f(x)=x3/4, 0<x<2 μ=1.6, σ2=8/75.

How large of size n is sufficient? If n=25, 30 or larger, the approximation is generally good. If the original distribution is symmetric, unimodal and of continuous type, n can be as small as 4 or 5. If the original is like normal, n can be lowered to 2 or 3. If it is exactly normal, n=1 or more is just good. However, if the original is highly skew, n must be quite large. Ex5.4-4: Let X1,…,X4 be a random sample of size 4 from the uniform distribution U(0,1) with p.d.f. f(x)=1, 0<x<1. μ=½, σ2=1/12; Y=X1+X2. Y=X1+…+X4.

Graphic Illustration Fig.5.4-1: Sum of n U(0, 1) R.V.s  N( n(1/2), n(1/12) ) p.d.f.

Skew Distributions Suppose f(x) and F(x) are the p.d.f. and distribution function of a random variable X with mean μ and variance σ2. Ex5.4-5: Let X1,…,Xn be a random sample of size n from a chi-square distribution χ2(1). Y=X1+…+Xn is χ2(n), E(Y)=n, Var(Y)=2n. n=20 or 100 →N( , ).

Graphic Illustration Fig.5.4-2: The p.d.f.s of sums of χ2(1), transformed so that their mean is equal to zero and variance is equal to one, becomes closer to the N(0, 1) as the number of degrees of freedom increases.

Simulation of R.V. X with f(x) & F(x) Random number generator will produce values y’s for U(0,1). Since F(x)=U(0,1)=Y, x=F-1(y) is an observed or simulated value of X. Ex.5.4-6: Let X1,…,Xn be a random sample of size n from the distribution with f(x), F(x), mean μand variance σ2. 1000 random samples are simulated to compute the values of W. A histogram of these values are grouped into 21 classes of equal width. f(x)=(x+1)/2, F(x)=(x+1)2/4, f(x)=3x2/2, F(x)=(x3+1)/2, -1<x<1; μ=1/3, σ2=2/9. -1<x<1; μ=0, σ2=3/5. N(0,1)

Approximation of Discrete Distributions Let X1,…,Xn be a random sample from a Bernoulli distribution with μ=p and σ2=npq, 0<p<1. Thus, Y=X1+…+Xn is binomial b(n,p). →N(np,npq) as n →∞. Rule: n is “sufficiently large” if np5 and nq5. If p deviates from 0.5 (skew!!), n need to be larger. Ex.5.5-1: Y, b(10,1/2), can be approximated by N(5,2.5).  Ex.5.5-2: Y, b(18,1/6), can be hardly approx. by N(3,2.5), ∵3<5.

Another Example Ex5.5-4: Y is b(36,1/2). Correct Probability: Approximation of Binomial Distribution b(n,p): Good approx.!

Approximation of Poisson Distribution Approximation of a Poisson Distribution Y with mean λ: Ex5.5-5: X that has a Poisson distribution with mean 20 can be seen as the sum Y of the observations of a random sample of size 20 from a Poisson distribution with mean 1. Correct Probability: Fig.5.5-3: Normal approx. of the Poisson Probability Histogram

Bivariate Normal Distribution The joint p.d.f of X : N(μX,σX2)and Y : N(μY,σY2) is Therefore, A linear function of x. A constant w.r.t. x.

Examples Ex.5.6-1: Ex.5.6-2

Bivariate Normal:ρ=0 ⇒ Independence Thm5.6-1: For X and Y with a bivariate normal distribution with ρ, X and Y are independent iffρ=0. So are trivariate and multivariate normal distributions. When ρ=0,