Homework 9 - United States Naval Academy

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. Let E[X] = 1, E[X2] = 3, E[XY ] = -4 and E[Y ] = 2. Find Cov(X, 2X + Y ). Solution: By the definition 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. Let X denote the different outcomes you see. (For example (20, 17,18,17,3) would be X = 4).

(a) Find the mean and variance of X.

Solution: We compute the mean and variance much like the die problem from homework 8. 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] + ? ? ? + E[X20] =20E[X1] =20P (number 1 is seen at least once in the five rolls)

=20(1 - P (number 1 is not seen in the five rolls))

19 5

=20 1 -

.

20

To find the variance, we use fact 7.22,

20

V ar(X) =V ar

Xi

i=1

20

= V ar(Xi) + 2 Cov(Xi, Xj).

i=1

i 0 and Y = X2. Show that X and Y are uncorrelated, even though Y is a function of X.

Solution: X is uniformly distributed on [-a, a] and therefore has the probability density

function

pX(x) =

1 2a

,

x [-a, a]

0, x / [-a, a].

To show that X and Y are uncorrelated, we must show that Cov(X, Y ) = 0, or

Cov(X, Y ) = E[XY ] - E[X]E[Y ] = E[X3] - E[X]E[X2] = 0

We compute the third moment of X using the density function,

E[X3] = x3pX (x) dx

-

a x3

=

dx

-a 2a

(a)4 - (-a)4 =

8a

=0.

Because 1/2a is constant in x, and therefore symmetric about x = 0, then every odd moment of X will be zero. That is,

E[X] = E[X3] = ? ? ? = E[X2n+1] = 0

for every n = 0, 1, 2, 3, .... Thus Cov(X, Y ) = E[XY ] - E[X]E[Y ] = E[X3] - E[X]E[X2] = 0 - 0 ? E[X2] = 0.

Therefore, X and Y are uncorrelated.

7. Let IA be the indicator of the event A. Show that for any A, B we have Corr(IA, IB) = Corr(IAc, IBc).

Solution: We start with the left hand side of the above equation. We start with the definition of covariance,

Cov(IA, IB) = E[IAIB] - E[IA]E[IB] = P (A B) - P (A)P (B). Next we use the identity 1 - P (Ac Bc) = P (A B) = P (A) + P (B) - P (A B). This yields,

Cov(IA, IB) = P (A) + P (B) - (1 - P (Ac Bc)) - P (A)P (B).

Next we substitute P (A) = 1 - P (Ac) and P (B) = 1 - P (Bc) to get

Cov(IA, IB) = P (A) + P (B) - P (Ac Bc) - (1 - P (Ac))(1 - P (Bc)).

Simplifying yields,

Cov(IA, IB) = P (Ac Bc) - P (Ac)P (Bc) = E[IAcIBc] - E[IAc]E[IBc] = Cov(IAc, IBc)

To finish the problem, we must show that the variances are equal. For any indicator function we have

V ar(IA) = E[IA2 ] - E[IA]2 = P (A) - P (A)2 = P (A)(1 - P (A)) = P (A)P (Ac).

Thus and therefore,

V ar(IA) = V ar(IAc), Corr(IA, IB) = Corr(IAc, IBc).

8. Suppose that for the random variable X, Y we have E[X] = 2, E[Y ] = 1, E[X2] = 5, E[Y 2] = 10 and E[XY ] = 1.

(a) Compute Corr(X, Y ). Solution: First we compute the covariance,

Cov(X, Y ) = E[XY ] - E[X]E[Y ] = 1 - (2)(1) = -1.

Now the variances,

V ar(X) = E[X2] - E[X]2 = 5 - 4 = 1 V ar(Y ) = E[Y 2] - E[Y ]2 = 10 - 1 = 9.

Now the correlation is,

Cov(X, Y )

-1

1

Corr(X, Y ) =

= =- .

V ar(X) V ar(Y ) 1 9 3

(b) Find a number c so that X and X + cY are uncorrelated. Solution: We want to find a c such that the covariance is zero. That is,

Cov(X, X + cY ) =E[X(X + cY )] - E[X]E[X + cY ] =E[X2] + cE[XY ] - E[X](E[X] + cE[Y ]) =5 + c - 2(2 + c) =1 - c

Thus we solve the equation 1 - c = 0 to find c = 1 will result in X and X - Y being uncorrelated.

9. Let the joint probability density function of (X, Y ) be given by

f (x, y) =

11

10 x

0

for 0 x 10, 0 y x otherwise

Find:

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