Usually we use the first way, as when we find Var(X) =
E(X - E(X))2 , but both work.
E is a linear operator (actually affine)--constants and
-, + pass across E.
Var(X) is not.
mean of Poisson, using formulas on p. 36, Ash
geometric, next time
We'll develop the rules further, prove the ones we haven't.
Handout (homework): Joint distribution, checking Moore's
rules, above rule.
I'll review the following Friday--but look
thru these now:
Handout, p1 arrows-->. If W is a function of X (W=g(X)), you
can find E(W) either by
>> summing g(xi)·P(X=xi)
for all the values of xi, or by
gathering all the probabilities of the x's which
have the same w-value, thus finding the distribution of W, and
·
>> summing wj·P(W=wj)
for all the values of wj.
Usually we use the first way, as when we find Var(X) = E(X -
E(X))2 , but both work.
- - - - - - - - - - - - - - - -
Def: X and Y are independent random variables if
P(X = x and Y = y) = P(X = x)
·P(Y = y) for every possible pair x and y.
What happens on X has no effect on the probabilities of Y:
P(Y = y) = P(Y = y | X = x)
If X and Y are independent, then E(X · Y) = E(X) ·
E(Y) (will prove.) This is needed to prove
Var(X+Y) = Var(X)
+ Var(Y).
- - - - - - - - - - - - - - - - -
Finding E's in complex cases: If X = X1+ X2+...+Xn
then E(X) = E(X1)+E(X2)+...+E(Xn). (Ash
3.2)
(If the Xi's
are independent (not usually true), then the variance can be found the
same way.)
If the Xi's
take on the values 0 or 1, they are called indicator random variables,
and E(Xi) = P(Xi = 1)
Binomial, drawing without replacement (mean
only): Xi =1 if i'th trial is a Success.
Reread
Moore pp.380 for derivation of Binomial mean and standard deviation.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
HW for today:
Handout "Expected values and variances" problems A thru D.
Review Moore hw (p. 4 of handout)
Ash p. 77 : 1, 2, 3, 5
Postponed to next HW: p. 84,
1 + See addition on handout p. 4
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