Math 340 Homework Number 11 _________________________________________________________________________ 1. (#5, page 182) a. This would be a correct judgement if it did take into account that $1 which we may loose. Note the following. b. Let X be the r.v. that denotes our net return on a single dollar bet. Then, no. trials = n = 3 prob of success on a single trial is = p = 1/6 Range of X = R(X) = {-1, 1, 2, 3} Note that the mistake in (a) is in regards to R(X). Had that x = -1 been 0 in R(X), then X would have been binomial and its expectation would have been mu = np = 50%, as asserted. However, in order to calculated the correct value of E(X) we need to use the value x = -1. The dist of X is: i. P( X = -1 ) = Prob for 3 consecutive failures = (5/6)^3 ii. P( X = 1 ) = prob for one S and two F's = "3 choose 1" (1/6) (5/6)^2 iii. P( X = 2 ) = "prob for two S's and one F" = "3 choose 2" (1/6)^2 (5/6) iv. P( X = 3 ) = prob for 3 consecutive successes = (1/6)^3 Finally, Our expected return = E(X) = \sum_{i = 1}^{4} x_{i} P(X = x_{i}), where x_{1} = -1, ..., x_{4} = 3. __________________________________________________ 2. (#11, page 182) p = P(S) = 100 / 1000 = 0.1 means that there is 10% chance for ONE bought ticket to be a prize ticket. Now, let X be the number of prize tickets in 3 trials, namely, we buy three tickets. Note that X is binomial, B(3, .1). Then, the expected value of X is E(X) = np = 0.3. Note that this is a simple upper bound for the prob that we win at least one prize, since expected value of X is 0.3. But we can calculate the actual prob as: P(X \geq 1) = 1 - P(X = 0) = 1 - (0.9)^3 = 0.271. Note that E(X) is very close to 0.271 since \sigma is very small and n is very small. __________________________________________________ 3. (#17, page 182) This problem is based on the discussion on page 171, "Tail Sum Formula," and the relevant example is Example 9 on page 172. This is how it works. There are 13 balls: 3 red, 4 blue, and 6 green. We draw balls without replacement until all red balls are drawn, and D is the number of draws made. Hence, we right away know that R(D) = {3, 4, ..., 13}, Since at least 3 draws are needed and at most it would take 13 draws to draw all red balls. Now let us think about the distribution of D. For example, what is P( D = 3) = ? Well, this is not so bad since we can think about it as: P(D = 3) = (3/13)(2/12)(1/11) = 6/(13.12.11). But, there is a better way to think about this number: (D=3) is the event that we draw 3 consecutive red balls without replacement and without order. So, P(D = 3) = ("3 choose 3") / ("13 choose 3") =(3! 10!) / 13! Thus, you see we would like to play this game in this way. So for example for the next prob we would like to consider [ ("3 choose 3") ("10 choose 1") / ("13 choose 4") ] (*) The denominator in (*) is simple: it is the number of ways that 4 balls can be drawn from a collection of 13 balls without replacement and without order. The numerator in (*) has the interpretation: it is the number of ways that 3 red balls and one other ball are drawn from a collection of 13 balls without replacement and without order. Note that this method of counting includes the case where the three red balls are drawn first, i.e., this method of counting includes the case of D=3. Now, here is the "subtle" point about (*) that connects it to D: (*) is the prob of the event described in the previous two paragraphs. Now, if you think about that carefully you realize that (*) corresponds to the case where D=3 OR D=4. You see (*) does not say that D is EXACTLY equal to 4, it includes both the case where 3 red balls are drawn first (consecutively) and also the case where one of the first three draws is not a red ball. So, we have: i. P(D \leq 4) = ("3 choose 3") ("10 choose 1") / ("13 choose 4") ii. P(D = 4) = P(D \leq 4) - P(D \leq 3) If we apply the procedure in (i) and (ii) to our problem we get the answers in the back of the book. __________________________________________________ 4. (#21, page 182) a. Note that A U A^c = \Omega, and A A^c = \empty, namely, a sample space, \Omega, is the "disjoint" union of A and its complement, A^c. So, take one sample point w \in \Omega and evaluate the indicators of A and A^c at this point, etc. Namely, Let w \in \Omega. Then, only two things can happen: w \in A OR w \in A^c. Hence, i. w \in A => I_A (w) = 1 AND I_A^c (w) = 0 => I_A^c (w) = 1 - I_A (w). ii. w \in A^c => I_A^c (w) = 1 AND I_A(w) = => I_A^c (w) = 1 - I_A (w). Hence, in all cases the value of the function I_A^c = 1 - I_A. b. Simply evaluate the indicator of AB at all w \in \Omega, using the def of the indicator func. Note the following observations. For any two events A and B we have: \Omega = (A^c B) U (B^c A) U (A B) U (A U B)^c. Note that these four subsets of \Omega are disjoint. Hence, any sample point w \in \Omega is a member of exactly (i) Let w \in (A B). Then, I_AB (w) = 1, I_A (w) = 1, I_B (w) = 1 => I_AB(w) = I_A(w) I_B(w). (ii) Let w \not \in (A B). Then, exactly one of the following 3 events is going to occur. 1. w \in (A^c B). In this case I_AB (w) = 0, I_A (w) = 0, I_B (w) = 1 => I_AB(w)=I_A(w) I_B(w). 2. w \in (B^c A). Exactly as above, except that now I_A(w)=1 and I_B(w) = 0. 3. w \in (A U B)^c. In this case I_AB(w) = I_A(w) = I_B(w) = 0 => I_AB(w) = I_A(w) I_B(w). Hence, we see that in all cases the equation is true. c. It is very helpful to do the formula for two sets first. Namely, show that I_{AUB} = 1 - (1 - I_{A}) (1 - I_{B}). If you expand the right hand side and use (a) and (b) you will see why this formula is true. Then, simply generalize. For example 1 - (1 - I_A)(1-I_B) = 1 - (1 - I_B - I_A + I_A I_B) = I_A + I_B - I_A I_B = I_A + I_B - I_AB = I_(AUB). This last part can for example be established by evaluating each function at a sample point. Now you can use an induction argument to generalize this formula. Namely, assume that the formula is true for the case n-1 and prove it is true for the case of n. d. For this part you only need to remember that E(I_{A}) = P(A). So, if you take expectations of both sides in (c) you will get the answer. Namely, expand the right hand side first and then take expectations of both sides. __________________________________________________ 5. (#9, page 219) This problem uses the discussion in Example 3 on page 212. Here is the way it is set up. You pay me $10 up front. Then, we strart tossing a fair coin. We "wait" until heads comes up. Say heads came up on the n-th toss. Then, I pay you $n^2. Let X denote my net return on a game. Then, X has a geometric dist. but its range is: R(X) = { 10 - n^2, n = 1, 2, ...} = {9, 6, 1, -6, ...} We are interested in E(X). If we try to calculate this expectation using the formula for expectation of a discrete r.v. we soon find that it is too hard. So, instead we will try to explore the relation that X has with the standard geometric dist. Recall that if Y has a geometric dist it satisfies: i. R(Y) = {1, 2, 3, ...} ii. P(Y = k) = q^(k-1) p In our problem: i. p = q = 1/2 ii. X = 10 - Y^2 Hence, E(X) = E( 10 - Y^2) = E(10) - E(Y^2) = 10 - p(1+q)/(1-q)^3 (using Example 3, p.212) = 10 - (1+q)/p^2 = 10 - 6 = 4 So, I expect to win $4 per game on the average which means that you should expect to loose $4 per game on the average. __________________________________________________ 6. (#15, page 219) A r.v. F has a geometric dist on {0, 1, ...} if its prob dist is P(F = f) = q^f p. Compare this def with the def of a geometric dist on {1, 2, ...} that is P(F = f) = q^(k-1) p. Thus, we now have a. P(F - k = m | F \geq k) = P(F = m + k, F \geq k) / P(F \geq k) = P(F = m+k) / [sum_{i=k}^{infinity} P(F=i)] = P(F = m+k) / [sum_{i=0}^{infinity} P(F=i+k)] = q^(m+k) p / q^k p [sum_{i=0}^{infinity} q^i] = q^m / [1/ (1-q)] = q^m p = P( F = m). b. Let X be an arbitrary discrete r.v. on {0, 1, ...} with this property. Then, P(X - k = m | X \geq k) = P(X = m) => P(X = m + k, X \geq k) / P(X \geq k) = P(X = m) => P(X = m+k) / P(X = m) = P(X \geq k) (**) Therefore, other than a geometric dist on {0, 1, ...}, tail probabilities of any other discrete r.v. on {0, 1, ...} satisfying the defining property above satisfies (**). As we note in (**) the ratio of the probabilities there are "independent" of m for each fixed m. Namely when we plug in for the distribution of X all terms involving m have to cancel to provide an expression that is independent of m. The independence of tail probabilities from m and the fact that the R.H.S. of (**) is an infinite series imply that the dist of X needs to have the form of a geometric dist, for otherwise we can not get the needed cancellations. __________________________________________________ 7. (#19, page 219) This problem is based on the Example 4 on page 213. In fact the whole problem is worked out there in the general setting that P(S) = p, as opposed to p=1/2 in this problem. b. P(T_r < 2r) = sum_{t=r}^{t=2r-1} P(T_r = t) = sum_{t=r}^{t=2r-1} (t-1 choose r-1) (1/2)^t = sum_{t=r}^{t=2r-1} (t-1 choose t-r) (1/2)^t = (1/2) sum_{k=0}^{r-1} (k+r-1 choose k) (1/2)^k+r-1 By letting k = t-r. Hence, P(T_r < 2r) = (1/2) sum_{k=0}^{n} (k+n choose k) (1/2)^{n+k} by calling r-1 = k. So finally if we define m = n+k, then we get P(T_r < 2r) = (1/2) sum_{m=n}^{2n} (m choose m-n) (1/2)^m = (1/2) sum_{m=n}^{2n} (m choose n) (1/2)^m The sum above is equal to 1 by symmetry in comparison with the binomial exapansion for (1)^n = 1. 1 = 1^n = (1/2 + 1/2)^n = sum_{m=0}^{n} (m choose n) (1/2)^n c. Using the result of part (b) we have: 1/2 = P(T_r < 2r) = sum_{t=r}^{t=2r-1} P(T_r = t) = sum_{t=r}^{t=2r-1} (t-1 choose r-1) (1/2)^t = sum_{t=r}^{t=2r-1} (t-1 choose t-r) (1/2)^t using the fact that (N choose K) = (N choose N-K). Now, if we let i = t - r, then the above identity becomes: 1/2 = sum_{i=0}^{i=r-1} (i+r-1 choose i) (1/2)^{i+r} Note that we can factor (1/2)^r from the above sum. Then, after bringing this (1/2)^r to the left-hand-side of the above identity we get: 2^{r-1} = sum_{i=0}^{r-1} (i+r-1 choose i) (1/2)^i In this last identity r-1 is a constant, so rename it as: n = r-1. Then the last identity becomes: 2^n = sum_{i=0)^(i=n) (n+i choose i) (1/2)^i = sum_{i=0}^{i=n) (n+i choose n) (1/2)^i once again using the fact that: (N choose K) = (N choose N-k). \begin{document}