Austin Wen
STAT4260 HW9
7.1 Suppose that a home mortgage company has N mortgages numbered serially in the
order that they were granted over a period of 20 years. There is a generally increasing trend
Note that finiteexp performs its calculations on the sample values sx and probabilities px
using the column vectors sx(:) and px(:). As a result, we can use the same finiteexp
function when the random
PY |X (1|1) $ Y =1
$
PX (1)
d PX (0)
d
d
d
PX (1)
d
d
$
X =1 $
Y =0
PY |X (0|1)
1/4
PY |X (1|0) Y =1
X =0
X =1
rr (0|0)
PY |X
rr
PY |X (1|0) r
r
P
(0|1)
Y |X
PY |X (1|1)
d
Y =0
Y =1
Y =0
3/4 $ Y =1
Since the trials needed to generate successes 2 though 100 are independent of the trials that yield
the rst success, N and T are independent. Hence
PN |T (n|t) = PN |T (n t|t) = PN (n t)
(3)
Applying
We can calculate the requested moments.
E [X ] = 3/4 0 + 1/4 20 = 5
(2)
Var[X ] = 3/4 (0 5) + 1/4 (20 5) = 75
2
2
E [X + Y ] = E [X ] + E [X ] = 2E [X ] = 10
(3)
(4)
Since X and Y are independent, The
The complete expressions for the marginal PMFs are
(1 p)m1 p m = 1, 2, . . .
0
otherwise
PM (m) =
(n 1)(1 p)n2 p 2 n = 2, 3, . . .
0
otherwise
PN (n) =
(9)
(10)
Not surprisingly, if we view each voice
(c) It is possible to nd the marginal PMFs by summing the joint PMF. However, it is much
easier to obtain the marginal PMFs by consideration of the experiment. Specically, when a
bus arrives, it is bo
Problem 4.9.10 Solution
This problem is fairly easy when we use conditional PMFs. In particular, given that N = n pizzas
were sold before noon, each of those pizzas has mushrooms with probability 1/3.
Problem 4.9.12 Solution
We are given that the joint PDF of X and Y is
1/(r 2 ) 0 x 2 + y 2 r 2
0
otherwise
f X,Y (x, y) =
(1)
(a) The marginal PDF of X is
r 2 x 2
f X (x) = 2
0
1
dy =
r 2
2
0
r 2 x 2
Problem 4.9.5 Solution
Random variables X and Y have joint PDF
Y
1
2 0yx 1
0 otherwise
f X,Y (x, y) =
1
(1)
X
For 0 x 1, the marginal PDF for X satises
f X (x) =
x
f X,Y (x, y) dy =
2 dy = 2x
(2)
0
No
Problem 4.9.3 Solution
(x + y) 0 x, y 1
0
otherwise
f X,Y (x, y) =
(1)
(a) The conditional PDF f X |Y (x|y) is dened for all y such that 0 y 1. For 0 y 1,
f X |Y (x) =
f X,Y (x, y)
=
f X (x)
(x + y)
1
Problem 4.9.8 Solution
First we need to nd the conditional expectations
1
E [B|A = 1] =
b PB|A (b| 1) = 0(1/3) + 1(2/3) = 2/3
(1)
b PB|A (b|1) = 0(1/2) + 1(1/2) = 1/2
(2)
b=0
1
E [B|A = 1] =
b=0
Keep
(b) Since PA (1) = PA,B (1, 0) + PA,B (1, 1) = 2/3,
PB|A (b|1) =
PA,B (1, b)
=
PA (1)
1/2 b = 0, 1,
0
otherwise.
(3)
If A = 1, the conditional expectation of B is
1
b PB|A (b|1) = PB|A (1|1) = 1/2.
E
Problem 4.9.1 Solution
The main part of this problem is just interpreting the problem statement. No calculations are necessary. Since a trip is equally likely to last 2, 3 or 4 days,
PD (d) =
1/3 d =
Problem 4.8.5 Solution
The joint PDF of X and Y is
(x + y)/3 0 x 1, 0 y 2
0
otherwise
f X,Y (x, y) =
(1)
(a) The probability that Y 1 is
P [A] = P [Y 1] =
Y
f X,Y (x, y) d x d y
y1
1
1
2
=
Y 1
1
1
0
1
Note that further along in the problem we will need E[N 2 |B] which we now calculate.
E N 2 |B = Var[N |B] + (E [N |B])2
17
2
+ 81
= 2+
p
p
(9)
(10)
For the conditional moments of K , we work directly
Problem 4.10.6 Solution
We will solve this problem when the probability of heads is p. For the fair coin, p = 1/2. The
number X 1 of ips until the rst heads and the number X 2 of additional ips for th
Problem 4.10.9 Solution
Since X and Y are take on only integer values, W = X + Y is integer valued as well. Thus for an
integer w,
PW (w) = P [W = w] = P [X + Y = w] .
(1)
Suppose X = k, then W = w if
(b) Before we nd E[B], it will prove helpful to nd the marginal PMFs PB (b) and PM (m).
These can be found from the row and column sums of the table of the joint PMF
PB,M (b, m) m = 60 m = 180
b=1
0.3
function [SX,SY,PXY]=circuits(n,p);
%Usage: [SX,SY,PXY]=circuits(n,p);
%
(See Problem 4.12.4)
[SX,SY]=ndgrid(0:n,0:n);
PXY=0*SX;
PXY(find(SX=n) & (SY=n)=pn;
for y=0:(n-1),
I=find(SY=y) &(SX>=SY) &(SX<
function covxy=finitecov(SX,SY,PXY);
%Usage: cxy=finitecov(SX,SY,PXY)
%returns the covariance of
0inite random variables X and Y
%given by grids SX, SY, and PXY
ex=finiteexp(SX,PXY);
ey=finiteexp(SY,P
Problem 4.11.1 Solution
f X,Y (x, y) = ce(x
2 /8)(y 2 /18)
(1)
The omission of any limits for the PDF indicates that it is dened over all x and y. We know that
f X,Y (x, y) is in the form of the bivar
The marked integral equals 1 because for each value of x, it is the integral of a Gaussian PDF of
one variable over all possible values. In fact, it is the integral of the conditional PDF fY |X (y|x)
Given T = t, the conditional probability the person is declared ill is
P [I |T = t] = P [W > 10|T = t]
W (7 + 2(t 37)
10 (7 + 2(t 37)
>
=P
2
2
3 2
3 2(t 37)
=Q
=P Z>
(t 37)
2
2
(11)
(12)
(13)
Problem
(b) In this case, the joint PDF of X and Y is inversely proportional to the area of the target.
f X,Y (x, y) =
1/[ 502 ] x 2 + y 2 502
0
otherwise
(4)
The probability of a bullseye is
P [B] = P X 2 +
p
$
$
$
$
1 p
$
q $ W >10
$
$
$
W 10
$
$ T >38 $
1q
T 38
The probability the person is ill is
P [I ] = P [T > 38, W > 10] = P [T > 38] P [W > 10] = pq = 0.0107.
(b) The general form of the bivariate G
0
cfw_X<xcfw_X<Y<X+w
x +w
x
FX,W (x, w) =
Y
x
=
2 ey d y d x
ey
0
x
=
w
(3)
x
x +w
x
(4)
dx
e(x +w) + ex
dx
(5)
0
X
= e(x +w) ex
x
x
(6)
0
w
= (1 ex )(1 e
)
We see that FX,W (x, w) = FX (x)FW (w). Mor
Homework 4
Mayank Gandhi
STAT 4260
1. 6.18 Did not use JMP
2. 6.27 Did not use JMP
3. 6.34
a. X-Bar Chart
R Chart
b.
c.
d.
4. 6.61
a.
Did not use JMP
Did not know how to find normality
Did not use JMP
Homework 3
Mayank Gandhi
STAT 4260
Z =
1.
a.
i.
P ( Z <3.556 )=
ii.
P ( Z <4.667 ) =
iii.
P ( Z <5.778 )=
b. Did not use JMP
c. Calculated Probability using JMP
2.
Homework 3
Mayank Gandhi
STAT 4260
Z
STAT 4260/6260
Homework #1 Solutions
Problem Set
1. Use the DATA from Table 3E.4 on page 104 in the text. Read the data across, then down.
a.
Compute the mean, median, standard deviation, and interqua
Homework 1
Mayank Gandhi
STAT 4260
1.
a.
Mean
Std Dev
Std Err Mean
Upper 95%
Mean
Lower 95%
Mean
N
b. Histogram:
Boxplot:
Time Series Plot:
8700.25
6156.81
58
1376.70
59
11581.7
28
5818.77
15
20
Homew
Homework 4
Mayank Gandhi
STAT 4260
1. Type I error occurs when the process is in control but the sample indicates out of control.
This can cause a false alarm and lead to producer risks because there