Mini-exam due Monday, Day 40.
DPGraph08 folder is now in Class
Material on all computers in Mac 101 except the 3
closest to the lake. NOT in Mac 110.
HW: Regression: Solve this equation
for yhat. Show that the slope b and the intercept a
(when the line is in the standard form yhat = a + bx) follow the "same"
rules as for data--
b = r sy / sx
and a = ybar - b xbar.
= = = = = = = = = = = = = = = = = = = = = = = = = = =
Continuing with Two random variables X, Y "jointly distributed".
f(x,y) = 2e-x-y , y>x, x>0. Have fX|y(x|y)
=e-x/(1-e-y) , 0<x <y . An
exponential distribution truncated at the right end.
Graph of E(X|y)
Movement to right as y increases appears to "slow down."
E(X|y) = (1 - ye-y -e-y )/(1-e-y) = 1 - [ ye-y
/(1-e-y)]. As y gets large, the [ ] term
--> 0. (Denominator -->1, top --> 0)
So as y gets large, E(X|y) approaches the line x = 1
asymptotically.
This makes sense: The expected value of the
exponential distribution e-x is 1; as the point of
truncation moves farther right, the conditional distribution of X given
y looks more and more like e-x .
E(Y|x) : the "average" y given a particular x. IF it's linear, it coincides with the (abstraction of) the usual least squares best fit line. See Day 38 for details.
Bivariate Normal distribution: ("Joint Normal"). Parameters:
2 means, 2 s.d.'s, and rho (correlation coefficient) Handout
Level lines are concentric ellipses. See Day 38 for details.
Forgot to mention there, that the conditional distributions are also
normal:
and E(Y|x) is the usual regression line (in the
standardized situation, E(Y|x) = rho).
- - - - - - - - - - - - - - - - - - - - - -
Functions of 2 random variables (Ash. 5.3);
especially W = X+Y (Ash Ch. 6.1) a brief overview.
We know the expected value E(W) = E(X) +
E(Y), and,
if X and Y are independent, we know
Var(W) = Var(X) + Var(Y)
But we did that without knowing what the distribution of W
was!
For some distributions, we can tell from their interpretation what
the sum should be (Ash pp. 201-3)
Binomial: If X1 is B(n1, p) and
X is B(n2, p) (same p!) and they are independent, then W = X1
+ X2 is B(n1 + n2, p)
Why? Do the (same) Bernoulli trial n1
times, then another n2 times. # of successes is W.
Ash gives similar arguments for Poisson, Exponential-->Gamma.
(Read the non-computational "proofs")
Normal: You were told in Math 251 that if you sum two
Normal random variables, the result is Normal.
On Day 1 you found the probability function for the sum of 2 6-sided
dice.
You summed along the line x + y = w, to find p(w).
What to do in continuous context? Handout
X + Y
Find FW(wo) = P(W < wo) =
P(X+Y < wo), area below and to left of x + y = wo
line.
To get f(w), take derivative of F.
Try it by geometry on Uniform distribution on square: f(x, y) =
1, 0< x, y < 1.
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Math300-Sp08/Day8p39.htm
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12am
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5/2/08 |