251: STAT Homework 4 John Peca-Medlin
1. (a) Let N be a non-negative integer-valued random variable.
P (N i) =
i=1 i=1 k=1 k
P (N = k)
=
k=1 i=1
P (N = k) =
k=1
k P (N = k)
= E[N ]
i P (n i) =
i=1 i=1
i
k=i+1
P (N = k)
k
= = 1 2
P (N = k)
k=1 i=1
i=
k=1
k(k + 1) P (N = k) 2
(k 2 + k)P (N = k)
k=1
1 E[N 2 ] + E[N ] = 2 (b) Note P (N i) = q i-1 (p + q) = q i-1 (since the first i - 1 tosses must be tails, which have propability q = 1 - p, and any toss after that doesn't matter). Thus, using (a):
E[N ] =
i=1
P (N i) =
i=1
q i-1 =
i=0
qi =
1 1 = 1-q p
E[N 2 ] = (-E[N ]) + 2
i=1
i P (N i) = (-E[N ]) + 2 d i d (q ) = (-E[N ]) + 2 dq dq
i=1
i q i-1 qi
i=1
= (-E[N ]) + 2
i=1
= (-E[N ]) + 2 -1 1-q
d dq
-1 +
i=0
qi
= (-E[N ]) + 2
d dq
-1 +
1 1-q
2 1+q 2-p = + = = 2 2 (1 - q) (1 - q) p2 1 1-p 2-p - 2 = V ar(N ) = E[N 2 ] - E[N ]2 = p2 p p2 2. (a) Let the sequences W, LW W, LW L, LLW, LLL represent the possible outcomes for my experience of roulette, with W, L corresponding to me winning or losing, respectively. Thus, we see that to determine the values of the probability mass function fX , we note X = 1 when the sequences W or LW W occur; X = -1 when the sequences LW L or LLW occur; and X = -3 when the sequence LLL occurs. Since P (W ) = p = 18 and P (L) = q = 1 - p = 20 , 38 38
1
we have: p + qp2 2 2q p fX (x) = q 3 0 x = 1, x = -1, x = -3, x = 1, -3.
2
18 We see that P (X > 0) = P (X = 1) = fX (1) = p + qp2 = 38 + 20 18 = 38 38 4059 1 6859 = .59171 > 2 . (b) We have E[X] = 1(p + qp2 ) - 1(2q 2 p) - 3(q 3 ) = -39 = -.108033 < 0. 361 Thus, we have an expected loss of about 11 cents. (c) For I{iT } the indicator function of the event {i T } and Xi defined by: 1, if the ith spin of the wheel produces Red, Xi = -1, otherwise,
we see these two random variables are independent since the values of each do not influence the value of the other. This is true since E[Xi ] = p - q = 2p - 1 and E[Xi |i T ] = E[Xi |i > T ] = E[Xi ] since Xi depends only on the ith spin of the wheel which is independent the previous i - 1 spins. A similar argument will show E[i T |Xi = 1] = E[i T |Xi = -1] = E[i T ] since the value of T does not depend on the outcome of the spins. (d) As mentioned earlier, each E[Xi ] = E[Xj ] = E[X1 ] for all i, j since the outcome of each spin does not depend on any previous spins (except the first one for our case T ). Thus, using the previous problem, the linearity of expectation, the independence of our random variables, and also properties of indicator functions, we have: E[X] = E
n i=1
Xi I{iT } =
n i=1 i=1
n i=1
E[Xi ]E[I{iT } ]
n i=1
= E[X1 ]E = E[X1 ]E
I{iT } = E[X1 ]E
P (i T )
P (i T ) = E[X1 ]E[E[T ]] = E[X1 ]E[T ]
Since E[X1 ] = (1)p + (-1)q = p - q and p < q = p - q < 0 we have E[X] = (p - q)E[T ] < 0 for E[T ] > 0 and thus T > 0. Otherwise, for T = 0, E[X] = 0. So, over all values of T (since T is necessarily nonnegative since it is a subset of N), we have E[X] 0, as we wanted. 18 (e) Since E[T ] = p + 3q, we have E[X] = (p - q)(p + 3q) = -2 38 + 3 20 = 38 38 -39 361 = -.108033, as we got before. 3. (a) We see for each value of X and thus Y , that the only numbers that matter are Y and Y + 1 in calculating the probability. That is, (using 1,2,3,... to represent each player with respect to the order they meet player 1, so player 1 is 1, his first opponent is 2, etc., and then using i > j to mean player i had a higher card than player j), so we see: P (X = 0) = P (Y = 1) = P (2 > 1) = 1 1 1 1 2 , P (X = 1) = P (Y = 2) = P (3 > 1 > 2) = 3 2 = 6 , P (X = 2) = P (Y = 1 1 1 3) = P (4 > 1 > max(2, 3)) = 4 3 = 12 , . . . , P (X = 8) = P (Y = 9) = 1 1 1 1 10 9 = 90 , P (X = 9) + P (Y = 10) = 10 (for a check that this works, see 1 1 1 1 1 1 1 1 1 1 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90 + 10 = 1). Thus, we can define the
2
probability mass function fY by: 1 x(x+1) , 1 fY (x) = P (Y = x) = x , 0,
10 i-1
x {1, 2, . . . , 9}, x = 10, otherwise. = = =
So P (Y i) = n=i fY (n) = 1 - n=1 fY (n). So we have P (Y 1) 1 1 1, P (Y 2) = 2 , P (Y 3) = 1 , P (Y 4) = 4 , . . . , P (Y 9) = 1 , P (Y 3 9 1 -1 10) = 10 . Thus, P (Y i) = i for all i. 10 10 7381 (b) Using Problem 1, we have E[Y ] = i=1 P (Y i) = i=1 i-1 = 2520 2.92897. 9 10 1 2 2 (c) Note, we have E[Y 2 ] = i=1 i i(i+1) i=1 i P (Y = i) = 10 +
9 i 10 + i=1 i+1 = 10 + 17819 = 43019 = 27.071. 2520 2520 43019 53928719 7381 2 2 E[Y ] = 2520 - 2520 = 6350400 = 8.49218. 7381 2520
Thus,
2 Y
= V ar(Y ) = E[Y ] -
2
(d) By the linearity of the expectation, we see: = E[X] E[Y -1] = E[Y ]-1 = - 1 = 4861 = 1.92897. 2520 2 Also, X = V ar(X) = V ar(Y - 1) = E[(Y - 1 - Y + 1)2 ] = E[(Y - Y )2 ] = V ar(Y ) = 53928719 = 8.49218. 6350400 4. (a) Let Xi be a random variable defined by: Xi = 1, 0, if the ith pair has a man and woman, otherwise,
-1
Thus, we have E[Xi ] = P (Xi = 1) = 10 10 20 = 10 for all i {1, 2, . . . , 10}. 19 1 1 2 (b) For all i = j, we have E[Xi Xj ] = P (Xi = 1, Xj = 1) = P (Xi = 1)P (Xj = 1|Xi = 1) =
10 i=1 10 1 10 1 20 -1 2 9 1 9 1 18 -1 2
=
90 323
= .278638.
(c) Let X = Xi . Note for all i = j, we have Cov(Xi , Xj ) = E[Xi Xj ] - 2 10 90 E[Xi ]E[Xj ] = E[Xi Xj ] - E[Xi ]2 = 323 - 10 = 6137 = .001629. Also, since 19
2 2 E[Xi ] = E[Xi ], we have V ar(Xi ) = E[Xi ] - E[Xi ]2 = .249307. Thus, we have: 10 19
-
10 2 19
=
90 361
=
100 10 = = 5.26316 19 19 10 10 10 V ar(X) = V ar Xi = V ar(Xi ) + 2 Cov(Xi , Xj ) i=1 1 2 90 10 16200 = 10 + 10 9 = = 2.63973 361 6137 6137 16200 90 34 X = V ar(X) = = = 1.62472 6137 323 E[X] = E
10 i=1
Xi =
10
i=1
E[Xi ] =
10
i=1
5. (a) Let Y be the random variable that takes the value of 1 or 3 on the graph, and X = 1 + Y + 1 = Y + 2. Thus, E[X] = 2 + E[Y ] = 2 + (1 + 3)/2 = 4. 2 2 Note E[Y 2 ] = (1 + 9)/2 = 5 and E[Y ] = 2. Thus, we have X = Y = 2 2 E[Y ] - E[Y ] = 5 - 4 = 1 = X = 1. (b) Again, let Y denote the random variable that takes the values of 1, 2, 3, 4, or 5, each with probability 1/5, and let X = 2 + Y . Thus, E[X] = 2 + E[Y ] = 2 + (1 + 2 + 3 + 4 + 5)/5 = 5. 3
Note E[Y 2 ] = (1 + 4 + 9 + 16 + 25)/5 = 11 and E[Y ] = 3. Thus, we have 2 2 X = Y = E[Y 2 ] - E[Y ]2 = 11 - 9 = 2 = X = 2 = 1.41421. (c) Once again, let Y denote the random variable that takes the values of 1, 2 and 3 (each with multiplicity 3), and 4, with each path having probability 1/8, and let X = 2 + Y . Thus, E[X] = 2 + E[Y ] = 2 + (1 + 3(2 + 3) + 4)/8 = 9/2. Note E[Y 2 ] = (1 + 3(4 + 9) + 16)/8 = 7 and E[Y ] = 5/2. Thus, we have 2 2 X = Y = E[Y 2 ] - E[Y ]2 = 7 - (5/2)2 = 3/4 = X = 3/2 = .866025. (d) Let Y denote the random variable taking on the values 1, 2, 3, 4, and 5, each with probability 1/5 (the same Y from part (b)), and let X = 4 + 3Y (which would be the graph shown for this part). Thus, E[X] = 4 + 3E[Y ] = 4 + 3(1 + 2 + 3 + 4 + 5)/5 = 13. Note E[Y 2 ] = 11 and E[Y ] = 3. Thus, we have X = V ar(X) = V ar(4 + 2 3Y ) = V ar(3Y ) = 9V ar(Y ) = 9 2 = 18 = X = 3 2 = 4.24264. (e) Let Y denote the random variable taking on the values 0 with probability 1/2, 1 with probability 1/4, 2 with probability 1/8, etc. (so we have P (Y = i) = 2-(i+1) ) and let X = 2 + Y . Thus, E[X] = 2 + E[Y ] = 2 + i=1 i 2-(i+1) = 2 + 1 = 3. 2 2 Note E[Y 2 ] = i=1 i2 2-(i+1) = 3 and E[Y ] = 1. Thus, we have X = Y = E[Y 2 ] - E[Y ]2 = 3 - 1 = 2 = X = 2 = 1.41421. 6. Let Y be ...