Read Ash Sec 3.2, 3.4. Skip
3.3.
Next, Ash Ch. 7, pp. 220-235 (ignore anything with integral signs)
HW:
Finish handout, M&M problems
A.Find the variance of a Bernoulli random variable (M&M say
"Similarly"
...sigma2S = p(1-p), p. 340, just below µS
formula..) Find it using the distribution of S, the
Bernoulli trial variable.)
B. Finish
the Expected value of Geometric (in class) from : p[deriv of (1/(1-q))]
Take
the derivative of (1/(1-q)) with respect to q, then let 1-q = p
and simplify the formula.
C. a. Prove E(kX) = kE(X) as in class below, only when n = 3.
Write it out with +'s.
Start E(kX) = (kx1)p1+(kx2)p2+(kx3)p3
=...
b.
Prove E (a+X) = a+E(X), first with n = 3 and +'s, then in general with
i's and Sum (big Sigma) notation.
Begin looking at:
From Ash (These are all tricks of one sort or
another.
Don't be discouraged if you can't do most of them without looking.)
Use p. 77 #5 to find E(X) for a geometric distribution (another way)
p.
84 (sec. 3.2), 1 + See addition on Alg. of exp.
Handout p. 4
3, 4, 5,
6 optional
p. 92, 7+, 10, 17+. +See notes on
handout.
(These 3 to hand in) #7 happened to me last week when I
forgot my keys and had to borrow someone's chain of many office keys
Others Optional, but
good: see the list on the handout p. 4.
= = = = = = = = = = = == = = = = = =
Expectation, continued. Day 16
Mean of Poisson, using exponential series, done last time:
(as in Ash)
Mean of geometric, using formulas on p. 36, Ash. Let X
be geometric, parameter p of success.
P(X=k) = qk-1p, k = 1,2,3,4,....
E(X) = 1p+2q1p+3q2p +4q3p+
...
= p(1 + 2q +3q2 +4q3+ ....)
= p[ deriv with respect to q of (1+q + q2
+ q3 + q4+ ..)]
= p[deriv of (1/(1-q))]
Take the derivative with respect to q, then
let 1-q = p and simplify (HW):
= 1/p Expected time to first success is 1/prob of
success
on a single trial.
If prob of "2" on a die is 1/6, expected times to throw a "2" is 6.
..
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.
Prove: Prove E(kX) =
kE(X).
(Use the above.)
E(kX) = Sumi[(kxi) pi] = Sumi[k(xi
pi)] = kSumi[(xi pi)]
(pulling
k out of sum: distributive law)
= kE(X)
by def. of E(X).
- - - - - - - - - - - - - - - -
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, soon.)
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, n drawings without replacement
(mean): Xi =1 if i'th trial is a Success. E(Xi)
= p
E(X) = Sum[ E(Xi)] = np
(Re)read
Moore pp.340-41 for derivation of Binomial mean and standard deviation.
Related idea: Ash p. 77 #5: If the sample space is {0, 1, 2, 3,
4,...}
(or {1, 2, 3, 4,...})
E(X) = sumi=1 to infinity (P(X
> i)) =
p1 + p2 + p3 +......
+ p2 + p3 +.....
+ p3 +......
=
1p1 +2 p2 + 3p3
+....
= sum i = 0 to infinity( i·pi)
- - - - - - - - - - - - - - - -
- - -
Need to do: proof of E (X+Y) = E(X) + E(Y)
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