Let Xi denote the indicator function of the first time i is seen on one of the five die rolls Then, by linearity of expectation and exchangeability, E[X] =E[X1] + ··· +
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[PDF] Math 431 – Spring 2014 Homework 9 Hand in the following problems:
Let Xi denote the indicator function of the first time i is seen on one of the five die rolls Then, by linearity of expectation and exchangeability, E[X] =E[X1] + ··· +
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Math 431 { Spring 2014
Homework 9
Due: April 17th (Sections 2 and 4) or 18th (Sections 1 and 3), 2014, depending upon your section(according to the instructions of your lecturer) Please read the instructions/suggestions on the course webpage.Hand in the following problems:
1. LetE[X] = 1; E[X2] = 3; E[XY] =4 andE[Y] = 2. Find Cov(X;2X+Y).
Solution:By the denition of covariance and linearity of expectation,Cov(X;2X+Y) =E[X(2X+Y)]E[X]E[2X+Y]
=2E[X2] +E[XY]E[X](2E[X] +E[Y]) =2(3) + (4)1(2(1) + 2) =2:2. Suppose you roll a fair 20-sided die 5 times. LetXdenote the dierent outcomes you see.
(For example (20, 17,18,17,3) would beX= 4). (a) Find the mean and variance ofX. Solution:We compute the mean and variance much like the die problem from homework8. LetXidenote the indicator function of the rst timeiis seen on one of the ve die
rolls. Then, by linearity of expectation and exchangeability,E[X] =E[X1] ++E[X20]
=20E[X1] =20P(number 1 is seen at least once in the ve rolls) =20(1P(number 1 is not seen in the ve rolls)) =20 119205!
To nd the variance, we use fact 7.22,
V ar(X) =V ar
20X i=1X i! 20X i=1V ar(Xi) + 2X iV ar(X) = 20V ar(X1) + 2019Cov(X1;X2):
For the indicator function,
V ar(X1) =p(1p) =
119205!1920
5 The covariance is, by exchangeability andX1X2=X1\2,Cov(X1;X2) =E[X1X2]E[X1]E[X2] =E[X1\2]E[X1]2
=P(1 and 2 are seen at least once in the 5 rolls)P(1 is seen at least once in the 5 rolls)2
=(1P(1 or 2 are not seen in the 5 rolls)) (1P(1 is not seen in the 5 rolls))2 118205! 11920
5!2
Putting it all together we have,
V ar(X) = 20
119205!1920
5 + 20192 4 118205! 11920
5!23 5 (b) What is the mean and variance if you roll the dientimes? Solution:In both the mean and variance, the only numbers that depend on the number of rolls are the ves in the exponent. Thus,
E[X] = 20E[X1] = 20
11920n !20 asn! 1. We would expect to see all 20 numbers show up if we are rolling many many times. We would expect the variance to tend to zero. We get,
V ar(X) = 20
11920n1920 n + 2019" 11820
n 11920
n2#
Asn! 1we get,
V ar(X) = 20(10)(0) + 2019(11) = 0
3. LetZ1;Z2;:::;Znbe independent normal random variables with mean 0 and variance 1. Let
=Z21++Z2n: (a) Using thatis the sum of independent random variables, compute both the mean and variance of. Solution:For the mean we use linearity of expectation and the fact that theZiare identically distributed. FurthermoreE[Z2i] =E[Z2i]E[Zi]2=V ar(Z) = 1, becauseE[Zi] = 0. Therefore we have,
E[] =nX
i=1E[Z2i] =nE[Z21] =n:For the variance, by independence,
V ar() =nX
i=1V ar(Z2i) =nV ar(Z21):The variance is dened as
V ar(Z21) =E[Z41]E[Z21]2:
We computed the fourth moment of a standard normal random variable in homework 4 (number 4),E[Z41] = 3. Thus,V ar() =nV ar(Z21) =n(31) = 2n:
(b) Find the moment generating function ofand use it to compute the mean and variance of. Solution:The moment generating function ofis dened to beE[et] =E[et(Z21+Z22++Z2n)]:
By independence ofZiwe use fact 7.13, to write the right hand side as a product of moment generating function. Since theZiare identically distributed, then it is a product of the same moment generating function. That is, M (t) =E[et] =E[et(Z21)]n=MZ21(t)n: We now compute the moment generating function ofZ21using the density of the standard normal distribution. We haveE[etZ21] =1p2Z
1 1 etz2ez2=2dz 1p2Z 1 1 e(2t1)z2=2dz: This integral convergences only fort <1=2. Using the substitution ~z=p12tzwe haveE[etZ21] =1p2Z
1 1 e(2t1)z2=2dz1p12t1p2Z
1 1 e~z2=2d~z1p12tfort <1=2:
Therefore,
M (t) =(12t)n=2fort <1=21fort1=2:
Using the moment generating function we calculate the mean to beE[] =M0(0) =n:
For the variance, we calculate the second moment,
E[2] =M00(0) =n(n2) =n(n2):
The variance is
V ar() =E[2]E[]2=n22nn2= 2n:
4. Let (X;Y) be a uniformly distributed random point on the quadrilateralDwith vertices
(0;0);(2;0);(1;1);and (0;1). Calculate the covariance ofXandY. Solution:To calculate the covariance we need to calculateE[XY]; E[X]; E[Y]:
First the joint distribution of (X;Y) on the quadrilateral, noting that the area is 3/2, is pX;Y(x;y) =
23;(x;y)2D
0;(x;y)=2D:
Then integrating we have,
E[XY] =Z
1 0Z 2y 023xy dx dy Z 1 026
y(2y)2 26
42
y243 y3+14 y4 1 0 =26 1112
=1136
E[X] =Z
1 0Z 2y 023x dx dy Z 1 026
(2y)2 26
4y42 y2+13 y3 1 0 =26 73
=79
E[Y] =Z
1 0Z 2y 023y dx dy Z 1 026
(2y)y 23
22
y213 y3 1 0 =23 23
=49
By the denition of covariance, we have
Cov(X;Y) =E[XY]E[X]E[Y] =1136
7949
=1136 2881
5. Let the joint pmf of (X;Y) be given by the table below.
Y X0123 11 151152
151
15 21
101
101
51
10 31
301
3001
10 (a) Find Cov(X;Y).
Solution:To nd the covariance, we compute
E[XY]; E[X]; E[Y]:
Thus we have
E[XY] =3X
i=13 X j=0ijp(x=i;y=j) =0 115+110
+130
+ 1115 + 2110 +215
+3130
+115
+ 415 + 6110 2:233 ForE[X] andE[Y] we can use the marginal pmfs, by summing over the columns or rows, to simplify the computations. f X(x) X11 3 21
2 31
6 Y f
Y(y)0123
11 5151
34
15
E[X] =3X
i=13 X j=0ip(x=i;y=j) 3X i=1ip X(i) 13 + 212 + 316 =116 1:833E[Y] =3X
j=03 X i=1jp(x=i;y=j) 3X j=0jp Y(j) =015 + 115 + 213 + 3415 1:667Using the above calculations, the covariance is,
Cov(X;Y) =E[XY]E[X]E[Y] = 2:233(1:833)(1:66)
(b) Find Corr(X;Y). Solution:To nd the correlation coecient, we need the covariance from above as well as the variance ofXandY. To nd the variance, we need the second moments.E[X2] =3X
i=13 X j=0i2p(x=i;y=j)
3X i=1i2pX(i)
13 + 2212 + 3216E[Y2] =3X
j=03 X i=1j2p(x=i;y=j)
3X j=0j2pY(j)
=0 215+ 1215 + 2213 + 324:
The correlation coecient is dened as,
Corr(X;Y) =Cov(X;Y)pV ar(X)pV ar(Y)=2:233(1:833)(1:66)p::::::: p::::::6. LetXbe uniformly distributed on [a;a] fora >0 andY=X2. Show thatXandYare
uncorrelated, even thoughYis a function ofX. Solution:Xis uniformly distributed on [a;a] and therefore has the probability density function pX(x) =
12a; x2[a;a]
0; x =2[a;a]:
To show thatXandYare uncorrelated, we must show thatCov(X;Y) = 0, orCov(X;Y) =E[XY]E[X]E[Y] =E[X3]E[X]E[X2] = 0
We compute the third moment ofXusing the density function,E[X3] =Z
1 1 x3pX(x)dx Z a ax 32adx(a)4(a)48a =0: Because 1=2ais constant inx, and therefore symmetric aboutx= 0, then every odd moment ofXwill be zero. That is,