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(2)(12c)
(3)(15c)
78 42
13 7
a a xy f (x, y)
(0)(0)(0)
(0)(1)(c)
(0)(2)(2c)
(0)(3)(3c) (1)(3)(5c) (2)(3)(7c)
(1)(0)(2c) (2)(0)(4c) 102c (d) E(X2) 102 42
(1)(1)(3c) (2)(1)(5c) 17 7
(1)(2)(4c) (2)(2)(6c)
a a x2 f(x, y)
(0)2(6c) (e) E(Y2) a a y2 f (x, y)
(1)2(14c)
(0)2(6c) (f) (g) (h) (i) sXY r s2 X s2 Y
(1)2(9c)
a y2 B a f (x, y)R
a x2 B a f (x, y)R
(2)2(22c)
102c
102 42
17 7
(2)2(12c)
(3)2(15c) 17 7 32 7 17 7 20
Var (X) Var (Y )
E(X2) E(Y2)
[E(X)]2 [E(Y )]2 E(X )E(Y )
Cov (X, Y ) sXY sXsY
E(XY ) 20>147
2230>441 255>49
2230 255
29 2 21 13 7
192c
192 42 230 441 55 49
32 7
29 13 21 7
20 147
0.2103 approx.
3.26. Work Problem 3.25 if the random variables X and Y are defined as in Problem 2.33, pages 61 63. Using c
(a) (b) (c)
1 > 210, we have:
E(X ) E(Y ) E(XY ) 1 6 210 3x 1 6 210 3x 1 6 210 3x
3 (x)(2x 2 y 0
y) dx dy y) dx dy y) dx dy
268 63 170 63 80 7
3 (y)(2x 2 y 0
3 (xy)(2x 2 y 0
CHAPTER 3 Mathematical Expectation
1 6 210 3x 1 6 210 3x E(X2) E(Y2) E(XY ) 200>3969 25036>3969216,225>7938
5 2 3 (x )(2x 2 y 0 5 2 3 (y )(2x 2 y 0
(d) (e) (f) (g) (h) (i) r sXY sXY sXsY s2 X s2 Y
E(X2) E(Y2) Var (X ) Var (Y)
y) dx dy y) dx dy 1220 63 1175 126 80 7
1220 63 1175 126
[E(X )]2 [E(Y )]2 E(X )E(Y)
Cov(X, Y )
22518 216,225
268 170 63 63
170 63
268 63
5036 3969 16,225 7938 200 3969 0.03129 approx.
Conditional expectation, variance, and moments 3.27. Find the conditional expectation of Y given X 2 in Problem 2.8, pages 47 48.
As in Problem 2.27, page 58, the conditional probability function of Y given X f ( y u2) Then the conditional expectation of Y given X E(Y u X (0) 4 22
2 is
4 22
2 is 2) 1
where the sum is taken over all y corresponding to X E(Y u X 2)
3.28. Find the conditional expectation of (a) Y given X, (b) X given Y in Problem 2.29, pages 58 59.
(a) (b) E(Y u X E(X uY x) 3 y)
yf2 (y u x) dy
5 22
4 y a y 22 y 2 6 22
2. This is given by
3 `xf1(x u y) dx 2(1 3(1 y3) y2)
x 2y 30 y x2 dy 1
7 22
19 11
3y x 1
2(1 y y2) 3(1 y)
dx
2x 3
3.29. Find the conditional variance of Y given X for Problem 2.29, pages 58 59.
The required variance (second moment about the mean) is given by E[(Y m2)2 u X
3 `(y E(Y u X
m2)2f2(y u x) dy x)
where we have used the fact that m2
2x>3 from Problem 3.28(a).
30 y
2 2y 2x 2 dy 3 x
x2 18
Chebyshev s inequality 3.30. Prove Chebyshev s inequality.
We shall present the proof for continuous random variables. A proof for discrete variables is similar if integrals are replaced by sums. If f(x) is the density function of X, then
E[(X
m)2]
3 `(x
m)2f (x) dx
CHAPTER 3 Mathematical Expectation
Since the integrand is nonnegative, the value of the integral can only decrease when the range of integration is diminished. Therefore, s2 3ux
mu P
m)2f (x) dx mu P( uX
P2f (x) dx
P2 3
f (x) dx
But the last integral is equal to P( u X
P). Hence, mu P) s2 P2
1). (b) Use Chebyshev s inequality to ob3.31. For the random variable of Problem 3.18, (a) find P( u X m u 1) and compare with the result in (a). tain an upper bound on P( u X m u (a) From Problem 3.18, m P( uX 1 > 2. Then mu 1) P 2 X
12 2
2x dx
Therefore (b) From Problem 3.18, s2
P 2 X mr 2
12 2 m2
1>4. Chebyshev s inequality with P mu 1) s2 0.25
P(u X
30 2e
1 (1
3 2
e 3)
0.04979 1 then gives
Comparing with (a), we see that the bound furnished by Chebyshev s inequality is here quite crude. In practice, Chebyshev s inequality is used to provide estimates when it is inconvenient or impossible to obtain exact values.
Law of large numbers 3.32. Prove the law of large numbers stated in Theorem 3-19, page 83.
We have Sn E n E(X1) Var (X1) E X1 E(X2) Var (X2) c n Xn c c c E(Xn) m s2 E(Xn)] 1 n (nm) ns2 m
Var (Xn) c
Then
1 n [E(X1) Xn) Var (X1) 1 Var (Sn) n2
Var (Sn)
Var (X1
so that
where we have used Theorem 3-5 and an extension of Theorem 3-7. Therefore, by Chebyshev s inequality with X Sn > n, we have Sn P 2 n m2 P P s2 nP 2
Sn Var n
c s2 n
Var (Xn)
Taking the limit as n S ` , this becomes, as required, Sn lim P 2 n nS` m2
Other measures of central tendency 3.33. The density function of a continuous random variable X is
f (x) e 4x(9 0 x2)>81 0 x 3 otherwise
(a) Find the mode. (b) Find the median. (c) Compare mode, median, and mean.
CHAPTER 3 Mathematical Expectation
(a) The mode is obtained by finding where the density f (x) has a relative maximum. The relative maxima of f(x) occur where the derivative is zero, i.e.,
2 d 4x(9 x ) B R dx 81
12x2 81
Then x !3 1.73 approx., which is the required mode. Note that this does give the maximum since the second derivative, 24x > 81, is negative for x !3. (b) The median is that value a for which P(X P(X a) a) 1>2. Now, for 0 x2) dx 4 9a2 81 2 0 a4 4 a 3,
4 a x(9 81 30
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