Lectures prepared by:Elchanan MosselYelena Shvets

Introduction to probability

Stat 134 FAll 2005

Berkeley

Follows Jim Pitman’s book:

ProbabilitySection 5.3

The normalization of the normal

Recall: N(0,1) has density f(x) = Ce-1/2x2

•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:

Question: what is the value of C?Answer:

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:

X

YR

Rotational invariance

r r+dr

•Note that R 2 (r,r + dr) if (X,Y) is in the annulus A(r,r+dr) of circumference 2 r, and area 2 r dr.

• In A(r,r+dr) we have: f(x,y) » C2 e-1/2 r2 . Hence:

•Therefore the density of R is:

•So: x

y

The Variance of N(0,1)

•By the change of variables formula S = R2 ~ Exp(1/2):

•The probability distribution of R is called the Rayleigh distribution. It has the density

.0

.1

.2

.3

.4

.5

.6

.7

.8

.9

1.0

1.1

0 1 2 3 4

• 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,).

• 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?

Questions:

• What’s the radius of the bull’s eye?

Radial Distancer

• The hitting distance R has Rayleigh distribution. Therefore:

• What’s the radius r of the bull’s eye?

P(A) ¼ 0.5

A

• 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

VariablesSuppose 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 such that

Z = cos X + sin Y.

Linear Combinations of Independent Normal

Variables•Z = cos X + sin Y.

•By rotational symmetry:

P(x<Z<x+x) = P(x<X<x+x)

•So: Z ~ N(0,1).

X

Y

sin

Y

cos X

Z

x

x x x

x

Linear Combinations of Independent Normal

VariablesIf 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(, ) and Y » N(, ) then

So X + Y » N( + , 2 + 2 2).

N independent Normal VariablesClaim:

If X1 ,…, XN are independent N(i,i2) variables then

Z = X1+X2+…+XN » N(1+…+N, (12+…+N

2) ).

Proof: By induction. Base case is trivial: Z1 = X1

»N(1,12)

Assuming the claim for N-1 variables we get

ZN-1 » N(1 +…+ N-1, (12+…+N-1

2) ) .

•Now: ZN = ZN-1 + XN , where XN and ZN-1 are independent Normal variables. So by the previous result:

ZN»N(1+..+N, (12+…+N

2) ) .

square Distribution

•Claim: The joint density of n independent N(0,1) variables is:

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.

Note: The density is spherically symmetric it depends on the radial distance:

•Claim:

square Distribution

This distribution is also called the -square distribution with n degrees of freedom.

•Claim: The distribution of R2

satisfies:

Applications of 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 ith type (N1+…+Nm = n). •Then for large n:

10 draws with replacement

Pb=6/20; Pi=4/20; Pc=10/20.

Nb=3; Ni=1; Nc=6.

has approximately a -square distribution with m-1 degrees of freedom.

R22 = (3–3)2/3 + (1-2)2/2 + (6-

5)2/5 =1/2 + 1/5 = 0.7

Note: The claim allows to “test” to what extent an outcome is consistent with an a priory guess about the actual probabilities.

10 draws with replacement

Pb=6/20; Pi=4/20; Pc=10/20.

Nb=3; Ni=1; Nc=6.

= (3–3)2/3 + (1-2)2/2 + (6-5)2/5 =1/2 + 1/5 = 0.7

square Distribution

df\area 0.995 0.99 0.975 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.025 0.01 0.0051 0.00004 0.00016 0.00098 0.00393 0.01579 0.10153 0.45494 1.3233 2.70554 3.84146 5.02389 6.6329 7.879442 0.01003 0.0201 0.05064 0.10259 0.21072 0.57536 1.38629 2.77259 4.60517 5.99146 7.37776 9.21034 10.59663

2 = 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.

square Example

Sandals Sneakers

Leather shoes

Boots Other Totals

Male observed

6 17 13 9 5 50

Male expected

9.5 11 10 12.5 7

Female observed

13 5 7 16 9 50

Female expected

9.5 11 10 12.5 7

Total 19 22 20 25 14 100

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.

–square Example

(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

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

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