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why the part I circled is k-1 rather just k? I made my assumption using red words, please tell me whether my assumption is true. The k-1 problem really bothers m

why the part I circled is k-1 rather just k? I made my assumption using red words, please tell me whether my assumption is true. The k-1 problem really bothers m

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section: Chapter Questions
Problem 41RE
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Could u please tell me why the part I circled is k-1 rather just k? I made my assumption using red words, please tell me whether my assumption is true. The k-1 problem really bothers me, thanks a lot! :)

12:01 Sun Apr 30 < Probability Notes Date C A child buys coupons to try to complete a coupon book. There are N different types of coupons. They are sold in sealed envelopes; each envelope contains a single coupon, which is equally likely to be of any of the N types. Let X be the number of coupons bought until the collection is completed. We will compute the expectation of X. E[X] = E[Y₁+ Y₂+...+YN] lectures_week3 We define auxiliary random variables Y₁,..., YN. We let Y₁ 1 (that is, Y₁ is actually de- terministic, always equal to 1). For k = {2,..., N}, we define Y as the number of coupons bought from (and not including) the moment when her collection reached k − 1 different types to (and including) the moment when her collection reached k different types. To illustrate this, assume that the coupon types are A, B, C, and the sequence of coupons bought by the child new are: new A, A, A, B, B, A, A, B, C. In this case, we have that Y₁ = 1, Y₂ = 3 and Y3 5. We now make two observations. = First, we have = = E[Y₁] + E[Y₂] + ... + E[YN] N N N N + N - 1 N - 2 N k=1 X = Y₁+ + YN. • Second, for k ≥ 2, the distribution of Y is Geometric (N=k+1) To see this, assume that the child just bought the coupon that made her collection grow to k - 1 different types. •E(X) of geometric distribution Geo (p) is 1/p: why is k-1 ? + + Probability Notes N 1 = T N k=1 exercises_week3 1 k N-(k-1) لر number of type The number is very close to f₁1dx = log(N). Hence, E[X] is very close to N log(N). è a Calculus Notes we don't have. total number of type Ys, If Ys there are 10 types in total. We obviously still have 10-5=5 types to be collected why 10- (5-1) = 6 ? 72% need
Transcribed Image Text:12:01 Sun Apr 30 < Probability Notes Date C A child buys coupons to try to complete a coupon book. There are N different types of coupons. They are sold in sealed envelopes; each envelope contains a single coupon, which is equally likely to be of any of the N types. Let X be the number of coupons bought until the collection is completed. We will compute the expectation of X. E[X] = E[Y₁+ Y₂+...+YN] lectures_week3 We define auxiliary random variables Y₁,..., YN. We let Y₁ 1 (that is, Y₁ is actually de- terministic, always equal to 1). For k = {2,..., N}, we define Y as the number of coupons bought from (and not including) the moment when her collection reached k − 1 different types to (and including) the moment when her collection reached k different types. To illustrate this, assume that the coupon types are A, B, C, and the sequence of coupons bought by the child new are: new A, A, A, B, B, A, A, B, C. In this case, we have that Y₁ = 1, Y₂ = 3 and Y3 5. We now make two observations. = First, we have = = E[Y₁] + E[Y₂] + ... + E[YN] N N N N + N - 1 N - 2 N k=1 X = Y₁+ + YN. • Second, for k ≥ 2, the distribution of Y is Geometric (N=k+1) To see this, assume that the child just bought the coupon that made her collection grow to k - 1 different types. •E(X) of geometric distribution Geo (p) is 1/p: why is k-1 ? + + Probability Notes N 1 = T N k=1 exercises_week3 1 k N-(k-1) لر number of type The number is very close to f₁1dx = log(N). Hence, E[X] is very close to N log(N). è a Calculus Notes we don't have. total number of type Ys, If Ys there are 10 types in total. We obviously still have 10-5=5 types to be collected why 10- (5-1) = 6 ? 72% need
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