5.1. Find the expectation
p,variance 02 and standard deviation
Uof each of the following distributions:
U = d8.25 = 2.9
(iii) p = z i f ( z i ) = 1(.4)
+
3(.1)+
4(.2)+
5(.3) = 32
z:f(zi) = 1(.4)+
9(.1)+
16(.2)+
25(.3) = 12U2 = x s f f ( z i )
-
p2 = 12-
9 = 3U =
fi
= 1.75.2.
A fair die is tossed. Let X denote twice the number appearing, and let Y denote 1or 3 according as an odd or an even number appears. Find the distribution, expectation, variance and standard deviation of (i) X, (ii) Y, (iii) X + Y, (iv) XY.
The sample space is S = {1,2,3,4,5, 6}, and each number appears with probability
9.
(i) X(l) = 2, X(2) = 4, X(3) = 6, X(4) = 8, X(6) = 10, X(6)
=
12. Thus X ( S )= {2,4,6,8,10,12}and each number has probability
8.
Thus the distribution of X is as follows:Accordingly,
89
1
CHAP. 61 RANDOM VARIABLES
(ii) Y(l) = 1, Y(2) = 3, Y(3) = 1, Y(4) = 3, Y(6) = 1, Y(6) = 3. Hence Y(S) = (1,3} and g(1) = P ( Y = l ) = P({1,3,6}) = = and g(3) = P ( Y = 3 ) = P({2,4,6}) = 3 = Thus the distribution of Y is as follows:
Accordingly,
(iii) Using (X+ Y)(s) = X(s)
+
Y(s), we obtain( X + Y ) ( l ) = 2
+
1 = 3 (X+Y)(3) = 6 + 1 = 7 (X+Y)(6) = 10+
1 = 11( X +Y)(2)= 4
+
3 = 7 (X+Y)(4) = 8+
3 = 11 (X+Y)(6) = 12+
3 = 16Hence the image set is (X
+
Y ) ( S )= {3,7,11, IS} and 3 and 16 occur with probabilityf,
and 7 and 11with probability$.
That is, the distribution of X+
Y is as follows:Thus
E ( X + Y ) = 3.:
+
7.:+
11*:+ 16.; =-
546 = 9E((X+Y)2) = 9 * $ + 49.:
+
121.:+
226.;=
!E 6 = 96.7Var(X+ Y) = E ( ( X + Y)2)
-
p2 = 96.7-
92=
14.7ux+y = = 3.8
Observe that E(X)
+
E(Y) = 7+
2 = 9 =E(X+ Y),
but Var ( X )+
Var (Y)=
11.7
+
1 = 12.7 # Var (X+
Y).(iv) Using (XY)(s) = X(s) Y(s),we obtain
(XY)(l) = 2 . 1 = 2 (XY)(3) = 6 . 1 = 6 (XY)(6) = 10.1 = 10 (XY)(2) = 4 . 3 = 12 (XY)(4) = 8.3 = 24 (XY)(6)
=
12.3 = 36 Hence the distribution of XY is as follows:Thus
E(XY) = 2 . 9
+
6 . 9+
1 0 . 9+
12.Q+
2 4 . 9+
3 6 . 9 = = 16Var(XY) = E((XY)2)
-
p2 = 369.3-
162 == 134.390 RANDOM VARIABLES [CHAP. 6 5.3.
A coin weighted so that P(H)
=3 and P(T) =
$is tossed three times. Let X be the
random variable which denotes the longest string of heads which occurs. Find the distribution, expectation, variance and standard deviation of X .
The random variable X is defined on the sample space
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The points in Shave the following respective probabilities:
P(HHH) = 3 . a . l 4 4 4 = 27 6 4 P(THH) =
$ - S * $
=&
P(HHT) = & * 3 * &4 4 = 9 P(THT) = I * & - &= 3
6 4 4 4 6 4
P(HTH) = $ * & * $=
&
P(TTH) = & * & * $ =&
P(HTT) = $ - $ * &= 3 6 4 P(TTT) = & * & * & =
&
Since X denotes the longest string of heads,
X(TTT) =0; X(HTT) =1, X(HTH) =1, X(THT) =1, X(TTH) =1;
X(HHT) =2, X(THH) =2; X(HHH) =3
Thus the image set of X is X ( S ) =(0, 1,2,3}. The probability f ( x i ) of each number xiin X ( S ) is obtained by summing the probabilities of the points in Swhose image is xi:
f ( 0 ) = P(TTT) =
&
f(1) = P(HTT)
+
P(HTH)+
P(THT)+
P(TTH) =f(2) = P(HHT)
+
P(THH) =#
f(3) = P(HHH) = Accordingly, the distribution of X is as follows:
1
1
6 4 6 4 6 4 6 4Thus
- E ( X ) = 0 . h
+ 1.E +
2 . u+
3.27 =-
-- 2.1P - 6 4 6 4 64
E(X2) =
0.A +
1.18+
4.18+
9 . 2 7 = -333 = 5.2b 4 6 4 6 4 64
u2 = Var (X) = E(X2) - p 2 = 5.2
-
(2.1)2 = .8 u = f i = . 95.4.
A fair coin is tossed until a head or five tails occurs. Find the expected number E of tosses of the coin.
Only one toss occurs if heads occurs the first time, i.e. the event H. Two tosses occur if the first is tails and the second is heads, i.e. the event TH. Three tosses occur if the first two are tails and the third is heads, i.e. the event TTH. Four tosses occur if TTTH occurs, and five tosses occur if either TTTTH or TTTTT occurs. Hence
f(1) = P(H) =
+
f(2) = P(TH) =
&
f(3) = P(TTH) =
4
f(4) = P(TTTH) =
f ( 5 ) = P(TTTTH)
+
P(TTTTT) =& + &
=&
- -
91
CHAP. 51 RANDOM VARIABLES
5.5.
Concentric circles of radius 1 and 3 inches are drawn on a circular target of radius 5 inches. A man re- ceives 10,5 or 3 points according if he hits the target, inside the smaller circle, inside the middle annular region or inside the outer annular region respectively.
Suppose the man hits the target with probability 4
and then is just as likely to hit one point of the target, as the other. Find the expected number E of points he scores each time he fires.
The probability of scoring 10, 6, 3 or 0 points follows:
f ( l o ) =
1.
2 area of area of 10 points target- -
1 ~ ( 1 ) 2-.--
2 T(5)2-
- -50 1 1 area of 5 points - -- 1 T(3)2-
T(1)2 --
-8 f(5) = -2 area of target 2 T(5)2 50 -1.
area of 3 points - --
1.
71.(E;)2-
~ ( 3 ) 2 - 16 f(3) = 2 area of target 2 4 5 ) 2 505.6.
A player tosses two fair coins. He wins $1 or
$2according as 1 or
2heads appear.
On the other hand, he loses $5 if no heads appear. Determine the expected value E of the game and if it is favorable to the player.
The probability that 2 heads appear is
&,
that 2 tails (no heads) appear is and that 1 head appears is4.
Thus the probability of winning $2 isB,
of winning $1 is4,
and of losing $6 isi.
Hence E = 2
&- +
19 -
5&-
=-&
= -0.25. That is, the expected value of the game is minus 254, and so is unfavorable to the player.5.7.
A player tosses two fair coins. He wins $5 if
2heads occur,
$2if 1head occurs and
$1
if no heads occur. (i) Find his expected winnings. (ii) How much should he pay to play the game if it is to be fair?
(i) The pro'bability of winning $5 is
a;,
of winning $2 is3,
and of winning $1 is&;
henceE = 6
& +
2+ +
1&-
= 2.50, that is, the expected winnings are $2.50.(ii) If he pays $2.50 to play the game, then the game is fair.
JOINT DISTRIBUTIONS, INDEPENDENT RANDOM VARIABLES 5.8. Suppose X and Y have the following joint distribution:
(i) Find the distributions of X and Y .
(ii) Find Cov ( X , Y ) ,i.e. the covariance of X and Y .
(iii) Find p(X,Y ) ,i.e. the correlation of X and Y .
(iv) Are X and Y independent random variables?
RANDOM VARIABLES [CHAP. 6 92
(i) The marginal distribution on the right is the distribution of X , and the marginal distribution on the bottom is the distribution of Y . Namely,
Distribution of X Distribution of Y (ii)
(iii) First compute uX and ay:
E ( X 2 ) =
2
x : f ( x i ) = (1)(.6)+
(9)(.5) = 5ug = v a r ( x ) = E ( X Z )
-
p i = 5-
( 2 ) ~ = 1 u x = f i = land
E(Y2) =
2
yig(yj) = (9)(.4)+
(4)(.3)+
(16)(.3) = 9.6U; = Var(Y) = E(Yz)
-
p i = 9.6-
(.6)2 = 9.24~y = d9.24 = 3.0 Then
(iv> X and Y are not independent, since P ( X = 1, Y = -3) # P ( X = 1)P ( Y = -3), i.e. the entry h(1,-3) = .1 is not equaI to f(l)g(-3) = (.5)(.4) = .2, the product of its marginal entries.
5.9.
Let X and Y be independent random variables with the following distributions:
Distribution of X Distribution of Y
Find the joint distribution h of X and Y .
Since X and Y are independent, the joint distribution h can be obtained from the marginal distributions f and g. First construct the joint distribution table with only the marginal distributions as shown below on the left, and then multiply the marginal entries to obtain the other entries, i.e.
set h(xi,yj) = f(xi)g(yj), as shown below on the right.
I I I I
I''
..iI
1 .62 .4.4
Sum .2 .5 .3