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Consider a discrete sequence of values, for lag 1, you compare your time series with a lagged time series, in other words you shift the time series by 1 before comparing it with itself. Proceed doing this for the entire length of time series by shifting it by 1 every time. You now have autocorrelation function.
Time-lagged Pearson Correlation Coefficient. I am a bit confused about the relation between the Pearson Correlation Coefficient (with time-lag) and Cross-Correlation. I used to think that Cross-Correlation IS the Pearson Correlation Coefficient with time-lag. I know that's wrong now, but I'm still a bit confused about the topic.
In your particular case, the recurrence actually breaks up into 15 independent recurrences of the Fibonacci type (but with possibly different initial values). Indeed, for k = 1, …, 15, define a new sequence gk(n) = f(k + 15n). Then your recurrence implies that gk(n) = gk(n − 1) + gk(n − 2). This you already know how to solve.
The mean of a lognormal distribution $\log X \sim \text{Normal}(\mu, \sigma^2)$ is $\exp(\mu+\frac 1 2 \sigma^2)$ so for $$\log(wage_i)\sim \text{Normal}(\beta_0 ...
where, as in the statement of Taylor's theorem, P(x) = f(a) + f ′ (a)(x − a) + f ″ (a) 2! (x − a)2 + ⋯ + f (k) (a) k! (x − a)k. It is sufficient to show that. limx → ahk(x) = 0. The proof here is based on repeated application of L'Hôpital's rule. Note that, for each j = 0, 1,..., k − 1, f (j) (a) = P (j) (a).