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3.9. Find (a) the variance, (b) the standard deviation for the random variable of Problem 3.4.
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(a) As in Problem 3.4, the mean of X is m Var (X) E[(X 30 x
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Then the variance is 1 R 2
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Another method By Theorem 3-4, Var (X)
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E[(X
E(X2)
[E(X)]2 1 A4 1 2
2Var (X)
1 2
CHAPTER 3 Mathematical Expectation
3.10. Prove Theorem 3-4, page 78.
We have E[(X m)2] E(X2 E(X2) E(X2) 2mX 2m2 [E(X)]2 m2) m2 E(X2) E(X2) 2mE(X ) m2 m2
3.11. Prove Theorem 3-6, page 78.
E [(X a)2] E [5(X E [(X E [(X E [(X since E(X (m a)2 m) E(X) m m. E m) m)2 m)2] m)2] (m 2(X 2(m (m a)6 2] m)(m a)E(X a)2 a)2] occurs when a) m) (m (m a)2] a)2
0. From this we see that the minimum value of E[(X
0, i.e., when a
3.12. If X*
m) > s is a standardized random variable, prove that (a) E(X*)
E(X*) X s m 1 s [E(X m)] 1 s [E(X) m]
0, (b) Var(X*)
since E(X) (b)
Var (X*)
using Theorem 3-5, page 78, and the fact that E[(X
3.13. Prove Theorem 3-7, page 78.
Var (X Y) E [5(X E [5(X E [(X E [(X Var (X ) using the fact that E[(X mX)(Y
Var Y) mX) mX)2 mX)2]
1 E[(X s2 s 2.
m)2]
m)2]
(mX (Y 2(X
mY)6 2] mY)6 2] mX)(Y mX)(Y mY) (Y mY)2] E[(Y mY)2]
2E[(X
mY)]
Var(Y )
mY)]
mX)E(Y
since X and Y, and therefore X mX and Y replacing Y by Y and using Theorem 3-5.
mY, are independent. The proof of (19), page 78, follows on
Moments and moment generating functions 3.14. Prove the result (26), page 79.
mr E[(X E B Xr m)r] r Xr 1m 1 r ( 1) j Xr j m j j
( 1)r
r r 1
Xmr
( 1)rmr R
CHAPTER 3 Mathematical Expectation
r E(Xr 1)m 1 ( 1)r 1 r r r ( 1) j E(Xr j)m j j
E(Xr) c
mr r
r mr 1m 1 r
E(X )mr
( 1)rmr
( 1)r 1rmr
( 1) rmr
r ( 1) j mr j mj j r 1)mr.
where the last two terms can be combined to give ( l)r 1(r
3.15. Prove (a) result (31), (b) result (32), page 79.
(a) Using the power series expansion for eu (3., Appendix A), we have MX(t) E(etX) 1 1 E 1 mr 2 tX t2 E(X2) 2! t2 2! mr 3 t3 3! t2X2 2! t3X3 3! c
tE(X ) mt
t3 E(X3) 3! c
(b) This follows immediately from the fact known from calculus that if the Taylor series of f (t) about t
a is
f (t) then cn
a cn(t
1 dn f (t) 2 n! dtn t
3.16. Prove Theorem 3-9, page 80.
Since X and Y are independent, any function of X and any function of Y are independent. Hence, MX
Y (t)
E[et(X
Y )]
E(etXetY )
E(etX )E(etY )
MX(t)MY (t)
3.17. The random variable X can assume the values 1 and 1 with probability 1 each. Find (a) the moment gen2 erating function, (b) the first four moments about the origin.
(a) (b) We have e Then (1) But (2) MX(t) E(etX ) et
1 et(1) 2 1 t 1 t e t) mt
et( t2 2! t2 2! 1
1) 1
1 t (e 2 t4 4! t4 4! c c c t4 4!
e t)
t3 3!
t3 3! t2 2! mr 3
1 t (e 2 1
t4 4! t3 3!
mr 2
t2 2!
Then, comparing (1) and (2), we have m 0, mr 2 1, mr 3 0, mr 4 1, c
The odd moments are all zero, and the even moments are all one.
CHAPTER 3 Mathematical Expectation
3.18. A random variable X has density function given by f (x) e 2e 0
Find (a) the moment generating function, (b) the first four moments about the origin.
(a) M(t) E(etX )
` tx 30 e (2e 2x) dx tx 3 `e f (x) dx `
2 3 e(t
2)x dx
2e(t 2)x 2 ` t 2 0 (b) If | t| 2 we have 2 2 But M(t) t 1 1 1 t>2 mt
1 2,
2 2 t
assuming t
1 t2 2!
1 2,
t 2 mr 3 mr 3
t2 4 t3 3!
3 4,
t3 8 mr 4 t4 4!
t4 16
mr 2 mr 2
Therefore, on comparing terms, m
mr 4
3 2.
3.19. Find the first four moments (a) about the origin, (b) about the mean, for a random variable X having density function f (x)
(a) mr 1 mr2 mr3 mr4
E(X)
4x(9 0
x2)>81
0 x 3 otherwise
x2) dx x2) dx x2) dx x2) dx 8 5 3 216 35 27 2 m
4 3 2 x (9 81 30 4 3 3 x (9 81 30 4 3 4 x (9 81 30 4 3 5 x (9 81 30
E(X2) E(X3) E(X4)
(b) Using the result (27), page 79, we have m1 m2 m3 m4 0 3 216 35 27 2 8 5 4
8 3(3) 5
11 25
s2 8 3 2 5 32 875
Characteristic functions 3.20. Find the characteristic function of the random variable X of Problem 3.17.
The characteristic function is given by E(eivX ) 1 eiv(1) 2
1) 1
216 8 35 5
8 2 6(3) 5
8 4 3 5
3693 8750
eiv(
1 iv (e 2
cos v
CHAPTER 3 Mathematical Expectation
using Euler s formulas, eiu with u cos u i sin u e
cos u
i sin u iv.
v. The result can also be obtained from Problem 3.17(a) on putting t
3.21. Find the characteristic function of the random variable X having density function given by f (x)
The characteristic function is given by
1>2a 0
ZxZ a otherwise
1 a ivx e dx 2a 3 a
E(eivX)
ivx 3 `e f (x) dx
1 eivx 2 a 2a iv a using Euler s formulas (see Problem 3.20) with u
eiav
e 2iav
sin av av
3.22. Find the characteristic function of the random variable X having density function f (x) ` x ` , where a 0, and c is a suitable constant.
Since f(x) is a density function, we must have
ce a|x|,
3 so that c 3 e aZxZ dx
f (x) dx
a(x) dx R
eax 0 c a 2 Then c
cB3 e
0 ` `
` a( x) dx ax `
30 e 2
2c a
a > 2. The characteristic function is therefore given by E(eivX )
0 a B 3 eivxe 2 ` 0 a B 3 e(a 2 ` ivx 3 `e f (x) dx ` a( x) dx ` ivx 30 e e a(x) dx R
iv)x dx
30 e e (a (a a iv)
a e(a iv)x 2 0 2 a iv ` a 2(a iv)
iv)x
(a iv)x dx R
a2 a2 v2
Covariance and correlation coefficient 3.23. Prove Theorem 3-14, page 81.
By definition the covariance of X and Y is sXY Cov (X, Y ) E[XY E(XY ) E(XY ) E(XY ) E(XY ) mXY mXmY mXmY E(X )E(Y ) E[(X mYX mYmX mX)(Y mXmY] mYE(X ) E(mXmY) mXmY mY)]
mXE(Y )
CHAPTER 3 Mathematical Expectation
3.24. Prove Theorem 3-15, page 81.
If X and Y are independent, then E(XY) sXY E(X )E(Y). Therefore, by Problem 3.23, E(XY ) E(X )E(Y ) 0
Cov (X, Y )
3.25. Find (a) E(X), (b) E(Y), (c) E(XY), (d) (e) (f) Var (X), (g) Var (Y), (h) Cov (X, Y), (i) r, if the random variables X and Y are defined as in Problem 2.8, pages 47 48. E(X2), E(Y2),
(a) E(X ) a a xf (x, y)
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