Lectures prepared by: Elchanan Mossel Yelena Shvets

Lectures prepared by:Elchanan MosselYelena Shvets

The normalization of the normal Recall:N(0,1) has density f(x) = Ce-1/2x2 Question: what is the value of C? Answer: • We will calculate the value of C using X ,Y » N(0,1) that are independent. • (X,Y) have joint density • f(x,y) = C2 e-1/2 ( x2 + y2); • And:

Rotational Invariance • Note: The joint density f(x,y) = C2 e-1/2 ( x2 + y2) is rotationally invariant – the height depends only on the radial distance from (0,0) and not on the angle. Let: Y R X

Rotational invariance • Note that R 2 (r,r + dr) if (X,Y) is in the annulus A(r,r+dr) of circumference 2p r, and area 2p r dr. • In A(r,r+dr) we have: f(x,y) » C2 e-1/2 r2 . Hence: • Therefore the density of R is: y r r+dr • So: x

The Variance of N(0,1) • The probability distribution of R is called the Rayleigh distribution. It has the density • By the change of variables formula S = R2 ~ Exp(1/2): • Therefore the Variance of N(0,1) is given by:

Radial Distance • A dart is thrown at a circular target by an expert. • The point of contact is distributed over the target so that approximately 50% of the shots are in the bull’s eye. • Assume that the x and y-coordinates of the hits measured from the center, are distributed as (X,Y), where X,Y are independent N(0,1). Questions: • What’s the radius of the bull’s eye? • What’s the % of the shots that land within the radius twice that of the bull’s eye? • What’s the average distance of the shot from the center?

Radial Distance • What’s the radius r of the bull’s eye? r A • The hitting distance R has Rayleigh distribution. Therefore: P(A) ¼ 0.5 • What’s the % of the shots that land within the radius twice that of the bull’s eye?

Radial Distance • What’s the approximate average distance of the shot from the center? • The average is given by: (by symmetry,)

Linear Combinations of Independent Normal Variables • Suppose that X, Y » N(0,1) and independent. • Question: • What is the distribution of Z = aX + bY ? • Solution: • Assume first that a2+b2 = 1. • Then there is an angle q such that • Z = cosq X + sin q Y.

Linear Combinations of Independent Normal Variables • Z = cosq X + sin q Y. • By rotational symmetry: • P(x<Z<x+Dx) = P(x<X<x+Dx) • So: Z ~ N(0,1). Y q Z sin q Y cos q X q X D x x x D x x

Linear Combinations of Independent Normal Variables • If Z = aX + bY, where a and b are arbitrary, we can define a new variable: • So Z’» N(0,1) and Z » N(0,  a2 + b2 2). If X» N(m, s2) and Y » N(l, t2) then So X + Y » N(l + m,  s2 + t2 2).

N independent Normal Variables • Claim: • If X1 ,…, XN are independent N(mi,si2) variables then • Z = X1+X2+…+XN»N(m1+…+mN, (s12+…+sN2)). • Proof: By induction. Base case is trivial: Z1 = X1»N(m1,s12) • Assuming the claim for N-1 variables we get • ZN-1» N(m1 +…+ mN-1, (s12+…+sN-12)) . • Now: ZN = ZN-1 + XN , where XNand ZN-1areindependent Normal variables. So by the previous result: • ZN»N(m1+..+mN, (s12+…+sN2)) .

c-square Distribution • Claim: The joint density of n independent N(0,1) variables is: Note: The density is spherically symmetric it depends on the radial distance: • Claim: This follows from the fact that a shell of radius r and thickness dr in n dimensions has volume cn rn-1dr, where cn denotes the surface area of a unit sphere.

c-square Distribution • Claim: The distribution of R2 • satisfies: This distribution is also called the c-square distribution with n degrees of freedom.

Applications of c-square Distribution • Claim: • Consider an experiment that is • repeated independently n times • where the ith outcomes has the probability pi for 1 · i · m. • Let Ni =# of outcomes of the ithtype (N1+…+Nm = n). • Then for large n: Pb=6/20; Pi=4/20; Pc=10/20. 10 draws with replacement Nb=3; Ni=1; Nc=6. R22 = (3–3)2/3 + (1-2)2/2 + (6-5)2/5 =1/2 + 1/5 = 0.7 has approximately a c-square distribution with m-1 degrees of freedom.

c-square Distribution • Note: The claim allows to “test” to what extent an outcome is consistent with an a priory guess about the actual probabilities. Pb=6/20; Pi=4/20; Pc=10/20. 10 draws with replacement Nb=3; Ni=1; Nc=6. c2 = (3–3)2/3 + (1-2)2/2 + (6-5)2/5 =1/2 + 1/5 = 0.7 c2 = 0.7 and the probability of observing a statistic of this size or larger is about 60%, so the sample is consistent with the box.

c-square Example We have a sample of male and female college students and we record what type of shoes they are wearing. We would like to test the hypothesis that men and women are not different in their shoe habits, so we set the expected number in each category to be the average of the two observed values.

c –square Example M/Sandals: ((6 - 9.5)2/9.5) =1.289 M/Sneakers: ((17 - 11)2/11) =3.273 M/L. Shoes: ((13 - 10)2/10) =0.900 M/Boots: ((9 - 12.5)2/12.5) =0.980 M/Other: ((5 - 7)2/7) =0.571 F/Sandals: ((13 - 9.5)2/9.5) =1.289 F/Sneakers: ((5 - 11)2/11) =3.273 F/L. Shoes: ((7 - 10)2/10) =0.900 F/Boots: ((16 - 12.5)2/12.5) =0.980 F/Other: ((9 - 7)2/7) =0.571 (Again, because of our balanced male/female sample, our row totals were the same, so the male and female observed-expected frequency differences were identical. This is usually not the case.) The total chi square value for Table 1 is 14.026 the number of degrees of freedom is 4. This gives This allows to reject the null hypothesis

Lectures prepared by: Elchanan Mossel Yelena Shvets