MATH 251, Probability and Statistics I, Fall 2005, Wed Oct. 19, Day 23After class

Read rest of Sec. 4.4 . Skip 4.5 for now; 5.1 next.
Hand in:  Sec. 4.4
A) Read "Prestidigitator of Digits", on reserve + outside my door (Due Monday): IPS presents the "relative frequency" foundation for probability.  The same probability rules apply to "personal probabilities," (IPS p. 299) where an individual assigns fractions between 0 and 1 to the outcomes, giving some sort of "likelihood" that the outcome will happen. (they still have to sum to 1, P(not A) = 1-P(A), and P(A or B) = P(A) + P(B) if A and B have no outcomes in common.) 
Diaconis is again at Stanford, after a couple of years at Cornell.
--What is the capitalized name given in the article to this kind of personal probability approach? 
--How does Diaconis feel when he sits in front of his paper? 

Sec. 4.4, p. 306 ff
Handout on Algebra of means and variances: Do A, 
  Postpone B, C, D. If you are willing to work ahead, the big labor in the handout is a bunch of computations (finding distributions, calculating means and variances from scratch) that can be done now, and don't depend on knowing the "Algebra" laws.  You can do all of that now;   then finishing the handout (checking the algebra rules) will be short.

DO 4.66 Benford's law (This is properly about probability as long-term average; put in the Law of Large Numbers section for some reason...)
DO 4.62 mean of triangular
Postpone the rest:
4.654 typing errors again.  Use your results from 4.60
4.69, 4.70 object length, distribution
B) X and Y are independent, µx = 1  µY = 4, sigmax = 2,  sigmaY =4
 W = 3X - 2Y + 1.  Find mean and s.d.
4.77 special case of corr. = 1.  (Lecture did an example of this. Now use algebra on the general rule 3)
4.73, 75  position + attach 
4.76  average two measurements (Pay attention to this; we'll use this pattern again) 

Read, discuss 
Play with  Applet, called Expected value (Mean of a Random Variable) not Law of Large Numbers as book p.296  says.
4.71, 72 independent or not?

Postpone 4.67
4.67 four coinflips Note, I did this analysis in class.

 

Optional 
 
 
Homework questions?
   Dice?  Compare to triangular.
Sec. 4.4, continued.  Mean and variance of probability distributions of R.V. 's:
mean µx  p.292 , variance sigmax2 & std. dev. sigmax p. 300 (X discrete) parallel to those for data.
(Sometimes the mean of X is called the "expected value of X")
Weight each value by its probability (Probability = long-run proportion) .
Discrete  µx = sum (xi pi)    sigmax2 = sum[ (xi x )2 pi]   (Calculating for continuous: Math 300. Footnote 17 p.293 (P. N-9))

Law of large numbers:  The average (mean) of the values of X observed in many (independent) trials approaches  µx, as n gets larger . Parallel to probability as long-term proportion.   Applet, called Expected value (Mean of a Random Variable) not Law of Large Numbers as book p.296  says.   Each graphed number on the zigzag line is the average of all the "rolls" up to that point. Choose "Show roll totals" to see blue dot giving that roll's result: below the zigzag line pulls line (average) down, above pulls it up.
"Law of small numbers" = fallacy; run of good must be "made up for" by run of bad, or vice versa.  Runs are "swamped" by mediocre behavior, not compensated for.

µx:  1) balance point, 2) long run average of many independent observations.

Algebra of Means and Variances
   Sometimes we change scale (Fo to Co), sometimes we measure two things in the same experiment (height/weight, red die/black die).
If  X and Y are random variables,  and we do  linear things to them, we can (often) find the mean and variance of the result, just
from knowing the original mean and variance (without having to find the whole distribution of the result.)  I'll call these rules
the Algebra of Means and Variances.  (These rules hold for discrete and continuous models.)
Means: p. 298    Variances: p. 300
Rules 1 are for a linear transformation a+bX of one R.V. (cf. p. 53-55 for data),
Rules 2 (&3)  for a linear combination X + Y

Linear transformation W = a+bX:   Just changing the "ruler" at the base of the histogram or density
    Mean:  µ= µ a+bX  =a +bµx
    Variance: sigmaw2 = sigmaa+bX 2 = b2sigmax2
        S.d. :     sigmaw = sigmaa+bX = |b|sigmax

Example:  X = # of heads on 2 flips of the coin. µx 1     sigmax2 =  .5    sigmax = .71
       x| 2  |    1    |  0  |
  P(X=x) | .25|   .50   | .25 |
Yolanda wins $2 for every head, and pays $1.50 to play.
Y = 2X-1.5µY 2·1-1.5 = .5    sigmaY2 = 22·.5= 2    sigmaY =  |2|·.71 = 1.42
Will runs the game; pays $2 for every head and gets $1.50 from the customer to play..
W = -2X +1.5.  µw-2·1+1.5 = -.5    sigmaw2 =(- 2)2·.5= 2    sigmaw =  |-2|·.71 = 1.42
        w|-2.5|   -.5   | +1.5|
        y| 2.5|    .5   | -1.5|
       x| 2  |    1    |  0  |
  P(X=x) | .25|   .50   | .25 |

Start here Friday:
Linear combination:
If we have two random variables X and Y we will say they are "independent random variables" if every event involving just X is independent of every event involving just Y.  (Usually we know this  from the setup of the probability model.  Roll 2 dice, outcome on red is independent of outcome on black.  Pick a person, weight is not independent of height.)

W =X+Y:   Scenario 1:  Xavier flips a coin twice.  Yancy flips a coin twice.  Wendy counts how many heads total.
               w|  0  |   1  |   2  |  3   |   4  | (p. 280-1: X is first 2 flips, Y second 2)
  P(W=w) | 1/16| 4/16 | 6/16 | 4/16 | 1/16 |
 µw = 2 by symmetry.   = µx + µY
sigmaw2 =(0-2 )2 (1/16)+(1-2 )2 (4/16) +(2-2)2(6/16) + (3-2)2(4/16)  + (4-2)2(1/16)  = (4 + 1·4 + 0 + 1·4 + 4)/16 = 16/16 = 1  = sigmax2 + sigmaY2
 sigmaw = 1 = sqrt(sigmax2 + sigmaY2 )   Pythagorean theorem again...

W =X+YScenario 2:  Wendy flips a coin twice.  Xavier pays her $1 for every head, Yancy pays her another $1 for every head.
  ( Note W has same dist as 2X)     2|       4
                                   1|    2
        w|  0  |   2   |   4  |     0|_0________
  P(W=w) | .25 |  .50  | .25  |       1  2
 µw 2 by symmetry.   = µx + µY
sigmaw2 = same as sigma2X2 = (2)2 .5 = 2, more than Scenario 1.
                 sigmaw = 1.414
Rule 2:  µX + Y µx + µY   whether or not X and Y are independent.
             sigmaX+Y2 = sigmax2 + sigmaY2  IF X and Y are independent.
Rule 3:  sigmaX+Y2 = sigmax2 + sigmaY2 + 2·rho· sigmax · sigmaY in general
   Rho:  theoretical analog of sample correlation coefficient.  In Scenario 2, Rho = 1 (perfectly correlated).
   Check:  sigmax2 + sigmaY2 + 2·rho· sigmax · sigmaY = .5 +.5 + 2·1· .71·.71  = 1+  2·.5 = 2

General linear combinations:
X and Y are independent, µx = 5,  µY = 4, sigmax = 3,  sigmaY =2
 W = 2X - 3Y + 4.  Find mean and s.d.
    Mean:  2·5 - 3·4 + 4 = 10-12+4 = 2
    S.d.:  First find variance.  sigmax2 = 9  ,  sigmaY2 = 4
            sigmaw2 = 22·sigmax2 + (-3)2·sigmaY = 4·9 +9·4 = 72
              sigmaw = sqrt(72) = 6·1.414 = 8.484
Note sigmaX-Y2= sigmax2 + sigmaY2


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