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 to justify finding
E(kX) using dist. of X.)
Assume P(X = xi) = pi for notational ease.
Proof: 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).
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- - -
Need to do: proof of E (X+Y) = E(X) + E(Y)
(handout, will go over in class)
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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), any x,y.
If X and Y are independent, then E(X · Y) = E(X)
·
E(Y) (will prove, next week.)
This is needed to prove
Var(X+Y) = Var(X)
+ Var(Y).
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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 only): Xi =1 if i'th trial is a Success. E(Xi)
= p
E(X) = Sum[ E(Xi)] = np
Var(X) = Sum[Var(Xi)]
(remaining: variance
of one Bernoulli trial S--was HW) E(S)= p
Var(S) = E(S
- E(S))2 = (0-p)2P(S=0)+ (1-p)2P(S=1)
= p2q + q2p = pq(p+q) = pq
Var(X) = npq
(Re)read
Moore pp.371-3 for derivation of Binomial mean and standard deviation.
Related idea: 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)
Proof of E (X+Y) = E(X) + E(Y)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Read Ash Sec 3.2, 3.4. Skip
3.3.
Next, Ash Ch. 7, pp. 220-235 (Variance. Ignore anything with
integral signs)
HW:
A. Add to your "known distributions" page the means found in sec's 3.1
and 3.2.
B. a. Prove E(kX) = kE(X) as
above, 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.
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.)
C. Use p. 77 #5 to find E(X) for a geometric distribution (another
way)
Hint: p3 + p4
+ p5 +...=P(X>2); you have a simple formula for that.
p.
84, 1 + See addition on Alg of exp. handout p. 4
3, 4, 5,
6 optional
p. 92, 7, 10, 17.
Others Optional, but
good: see the list on the Alg. of exp. handout p. 4.
D)Derivatives
review:
Find
the derivative with respect to
w:
(w3-
h)/(3+w + 2w2),
exp(-w2), ln(2 + 3w2
)
Find the second
derivative with respect to q:
q, q2,
q3,
q4, q5, q6,
qx
E) Graph the four points in
the x,y plane (0,2), (1,0), (1,4), (2,2).
Let each point be equally likely (prob = 1/4). Let X be the value
on the x-coordinate, and Y be the value on the Y coordinate.
a) Find E(X + Y) by
evaluating x+y for each of the 4 points,
multiplying by the probability, and summing.
b) Find E(XY) by
evaluating xy for each of the 4 points, multiplying
by the probability, and summing.
c) Find the probability
distribution of X. Find the probability
distribution of Y. Find E(X). Find E(Y).
d) Check if E (X+Y) =
E(X) + E(Y).
e)
Check if E(X ·
Y) = E(X)
·
E(Y)
f) Check if X and Y
are independent. (For independence,
for any x,y pair, P(x,y) = P(X=x) ·
P(Y=y). So to show
non-independence, you only need to find one x,y for which that is not
true.)
F) Another Data problem: Handout on Bird Calls. Do
all the questions (including 1.6.4, frizzle fowl). Recopy table 1
to have enough room to fill it in. Est. rel. freq. is just the
probabilities.
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