# Test eq and var cycle

so that {Yt} could be called asymptotically stationary. These two results follow from parts (b) and (c). (e) Suppose now that we alter the initial condition and put Y 1 = e 1 / ( 1 ­ c 2 ) . Show that now {Yt} is stationary. This part can be solved using repeated substitution to express Yt explicitly as ct ­ 1 Y t = c ( cY t ­ 2 + e t ­ 1 ) + e t = ... = e t + ce t ­ 1 + c 2 e t ­ 2 + ... + c t ­ 2 e 2 + ----------------- e 1 1 ­ c2 2 e Then show that Var ( Y t ) = ------------and Corr ( Y t , Y t ­ k ) = c k for k &gt; 0 . 1 ­ c2 Exercise Two processes {Zt} and {Yt} are said to be independent if for any time points t1, t2,..., tm and s1, s2,..., sn, the random variables { Z t , Z t , ..., Z t } are independent of the random variables { Y s , Y s , ..., Y s }. 1 2 m 1 2 n Show that if {Zt} and {Yt} are independent stationary processes, then Wt = Zt + Yt is stationary. First, E(W t ) = E(Z t ) + E(Y t ) = Z + Y. Then Cov(W t ,W t - k ) = Cov(Z t + Y t , Z t - k + Y t - k) = Cov(Z t ,Z t - k ) + Cov(Yt,Yt - k) which is free of t since both {Zt} and {Yt} are stationary. Exercise Let {Xt} be a time series in which we are interested. However, because the measurement process itself is not perfect, we actually observe Yt = Xt + et. We assume that {Xt} and {et} are independent processes. We call Xt the signal and et the measurement noise or error process. If {Xt} is stationary with autocorrelation function k, show that {Yt} is also stationary with k for k 1 Corr ( Y t , Y t ­ k ) = -------------------------2 2 1 + e / X

Fortran has five intrinsic data types : INTEGER , REAL , COMPLEX , LOGICAL and CHARACTER . Each of those types can be additionally characterized by a kind . Kind, basically, defines internal representation of the type: for the three numeric types, it defines the precision and range, and for the other two, the specifics of storage representation. Thus, it is an abstract concept which models the limits of data types' representation; it is expressed as a member of a set of whole numbers (. it may be {1, 2, 4, 8} for integers, denoting bytes of storage), but those values are not specified by the Standard and not portable. For every type, there is a default kind , which is used if no kind is explicitly specified. For each intrinsic type, there is a corresponding form of literal constant . The numeric types INTEGER and REAL can only be signed (there is no concept of sign for type COMPLEX ).

This versions will give correct answers for n less than 341550071728321 and then reverting to the probabilistic form of the first solution. By selecting predetermined values for the a values to use instead of random values, the results can be shown to be deterministically correct below certain thresholds.
For 341550071728321 and beyond, I have followed the pattern in choosing a from the set of prime numbers.
While this uses the best sets known in 1993, there are better sets known , and at most 7 are needed for 64-bit numbers.