MATH 251, Probability and Statistics I, Fall 2007, Mon Oct. 15, Day 22.after class.

Projects back; mostly quite nice. 
    Notes I didn't discuss the notes in class. Read if you like; I'll mention briefly WED.
   Presentations: Sign up on the little pad: Wed. or Fri. or Mon. 10 min. max.  Show and tell.
   What your data set "is" and 1 or 2 results:  The most interesting, or the most curious or surprising thing, or the most clever or sophisticated or different thing you did to the data to look for meaning. AND (if applicable) the thing that caused you the most trouble with the dataset and what you did to deal with it...
Sampling distributions:  Add your data to circulating lists.
    Notice x-bars  (updated) and p-hats (link to old collection now works) roughly normal looking, centered around population parameter.  Distribution of x-bars is narrower than distribution of individual x's. 
Actual sampling distribution for the 4- grades:  Solutions manual.

Homework questions?  Day 21
   Dice?  Compare to triangular.
Sec. 4.4, continued.  Mean and variance of probability distributions of R.V. 's:
mean µ_x  p.292 , variance sigma_x² & std. dev. sigma_x 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  µ_x = sum (x_i p_i)    sigma_x² = sum[ (x_i -µ_x )² p_i]   (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  µ_x, as n gets larger . Parallel to probability as long-term proportion. Law of Large Numbers  Applet,  book p.296.     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.

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

Algebra of Means and Variances
   Sometimes we change scale (F^o to C^o), 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_w  = µ _a+bX  =a +bµ_x
    Variance: sigma_w² = sigma_a+bX ² = b²sigma_x²
        S.d. :     sigma_w = sigma_a+bX = |b|sigma_x

Example:  X = # of heads on 2 flips of the coin. µ_x=  1     sigma_x² =  .5    sigma_x = .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-1.5.  µ_Y=  2·1-1.5 = .5    sigma_Y² = 2²·.5= 2    sigma_Y =  |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.  µ_w=  -2·1+1.5 = -.5    sigma_w² =(- 2)²·.5= 2    sigma_w =  |-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 |

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 |
 µ_w = 2 by symmetry.   = µ_x + µ_Y
sigma_w² =(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  = sigma_x² + sigma_Y²
 sigma_w = 1 = sqrt(sigma_x² + sigma_Y² )   Pythagorean theorem again...

W =X+Y:  Scenario 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  |        0  1  2
 µ_w =  2 by symmetry.   = µ_x + µ_Y
sigma_w² = same as sigma_2X² = (2)² .5 = 2, more than Scenario 1.
                 sigma_w = 1.414
Rule 2:  µ_{X + Y} =  µ_x + µ_Y   whether or not X and Y are independent.
             sigma_X+Y² = sigma_x² + sigma_Y²  IF X and Y are independent.
Rule 3:  sigma_X+Y² = sigma_x² + sigma_Y² + 2·rho· sigma_x · sigma_Y in general
   Rho:  theoretical analog of sample correlation coefficient.  In Scenario 2, Rho = 1 (perfectly correlated).
   Check:  sigma_x² + sigma_Y² + 2·rho· sigma_x · sigma_Y = .5 +.5 + 2·1· .71·.71  ₌ 1+  2·.5 = 2

General linear combinations:
X and Y are independent, µ_x = 5,  µ_Y = 4, sigma_x = 3,  sigma_Y =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.  sigma_x² = 9  ,  sigma_Y² = 4
            sigma_w² = 2²·sigma_x² + (-3)²·sigma_Y²  = 4·9 +9·4 = 72
              sigma_w = sqrt(72) = 6·1.414 = 8.484
Note (sigma_X-_Y)²= sigma_x² + sigma_Y²
finished this in class.