Statistics and computer science

On January 1, 2018, Slacks' recorded book values were equal to fair values for all items except for an unrecorded trademark with a fair value of $150,000 and a 10 year useful life. Any remaining excess acquisition-date fair value was assigned to goodwill. Slacks routinely sells merchandise inventory to Primadonna.

During 2018, Slacks sold inventory to Primadonna for $280,000. The cost of this inventory, to Slacks, was $200,000. Finally, $40,000 remains in the ending inventory of Primadonna on December 31, 2018. ?

During 2019, Slacks sold inventory to Primadonna for $300,000. The cost of this inventory, to Slacks, was $200,000. $60,000 remains in the ending inventory of Primadonna on December 31, 2019. ?

During 2020, Slacks sold inventory to Primadonna for $320,000. The cost of this inventory, to Slacks, was $240,000. $136,000 remains in the ending inventory of Primadonna on December 31, 2020.

On January 1, 2019, Primadonna sold equipment to Slacks for $160,000. On this date, the equipment was carried on Primadonna's books at a cost of $200,000 with accumulated depreciation of $112,000. Both companies estimated that the equipment has a remaining life of 10 years on the intercompany sale date, with no salvage value. The asset is depreciated using the straight-line method

On January 1, 2020, Primadonna sold land to Slacks for $100,000. That amount was considered to be the fair value of the land on that date. The cost of the land on Primadonna's books was $44,000. As of December 31, 2020, Slacks still owed Primadonna the entire $100,000 for this land. This liability is non-interest bearing and is included in accounts receivable of Primadonna and accounts payable of Slacks. Primadonna accounts for its investment in Slacks using the equity method. Unconfirmed profits are allocated pro-rata.

1.76. Consider a branching process as defined in Example 1.8, in which each family has exactly three children, but invert Galton and Watson's original mo- tivation and ignore male children. In this model a mother will have an average of 1.5 daughters. Compute the probability that a given woman's descendents will die out. Example 1.8. Branching processes. These processes arose from Francis Galton's statistical investigation of the extinction of family names. Consider a population in which each individual in the nth generation independently gives birth, producing k children (who are members of generation n + 1) with proba- bility px. In Galton's application only male children count since only they carry on the family name. To define the Markov chain, note that the number of individuals in genera- tion n, X,, can be any nonnegative integer, so the state space is {0, 1, 2, .. . ). If we let Y1, Y2. .. . be independent random variables with P(Ym = k) = p, then we can write the transition probability as p(i, j) = P(Yi + . . . + Mi=j) for i > 0 and j 2 0 When there are no living members of the population, no new ones can be born, so p(0, 0) = 1. Galton's question, originally posed in the Educational Times of 1873, is Q. What is the probability that the line of a man becomes extinct?, i.e., the branching process becomes absorbed at 0? Reverend Henry William Watson replied with a solution. Together, they then wrote an 1874 paper entitled On the probability of extinction of families. For this reason, these chains are often called Galton-Watson processes.Consider a Galton-Watson process (Xn)n where we assume that initially, there is exactly one ancestor, i.e. X0 = 1. Assume that the probability for each individual in each generation to have exactly j descendants is given by (3-1) pj = 19(1 p)j, where 0 0) of the species. 0 Find the expected number ,u of descendants of each individual. 0 Determine no for each possible value of p. . Consider a zero-mean WSS complex Gauss-Markov process of order 1 defined by the following equation r =-ar _tv, n=0,1,.... where v ~ /(0, o') and Ela, v* ] = 0 for 1 = +1, 12, .... Multiply both sides by * and take expectation values to find Er, *]. Note that E[rv*] = Ex*v,]. Multiply both sides by r* and take expectation values to show R., [0] = -aR_ [-1] + 0) = -aR* [1]+02. This process is also known as the auto-regressive process of order 1 and is denoted by AR(1). . Now multiply both sides by e* , I = 1, 2...., and take expec- tation values to show R, [!] = -aR,, [[ - 1], 1 =1, 2. .... This result, for a Gauss-Markov process of order P, or an AR(P) process. forms the basis for high resolution spectral estimation of zero-mean WSS random signals.48 Exam Revision Questions Question #1 The offspring distribution in a Galton-Watson branching process is binomial with mean E[Y] = 1.2, and variance 0.96. (i) The Newton-Raphson method is used to solve x = [Ty(x) where IT(s) is the probability generating function for random variable Y. Show that the iterating formula is given by 0.8 - X (0.8 + 0.2x,)" - 1.2 (ii) Hence solve for the probability of ultimate extinction of the branching process, giving your answer to 4 decimal places.Time sequences: Generate a sequence of 1000 equally spaced samples of a Gauss- Markov process using the recursive relation: Xn = ax- + Wnin = 1,2, ..., 1000. Here assume that Xo = 0, and {w,} is a sequence of independent identically distributed Gaussian random variables. You can use the randn function in MATLAB to generate zero-mean, unit-variance random variables. Below is given a low-pass filter (a>0) excited by a white noise sequence, and the filter has a real pole at z- a. Plot the output waveform for the following values of a: a= 0.5, a = 0.95, a = 0.995. Comment on the effect of the selection of a on the resulting time sequence