TS Module 8: Non-stationary time series basics HW

(The attached PDF file has better formatting.)

Homework assignment: Stationarity through differencing and logarithms

Automobile liability claim severities have a geometric trend of +8% per annum.

The average claim severity in year t is the average claim severity in year t-1 adjusted for the geometric trend, plus or minus a random error term.

Assume the error term is added to the logarithm of the average claim severities.

The average claim severities are multiplied by a random error term.

　

Is the time series of average claim severities stationary?

Is the first difference of this time series stationary?

Is the second difference of this time series stationary?

Is the logarithm of this time series stationary?

What transformation makes the time series stationary?

Jacob:

What is the form of this time series?

Rachel:

Actuaries write: Y_t = 1.08 Y_t-1. The error term is multiplicative: Y_t = 1.08 Y_t-1 × (1 +

å).

A separate discussion forum posting shows the solution.

It is me or is the homework for module 9 look to be affiliated with module 8?  And vice versa, the homework for module 8 looks like it should be affiliated with module 9?

[NEAS: Both homework assignments deal with ARIMA processes. Answer them after working through both modules.]

 

Tom McNamara III

I'm having the same problems getting started with this module. I originally thought that the answer to A. was "Yes" since "the probability laws that govern the behavior of the process do not change over time" (page 16 of the textbook) is the very definition of stationarity, and this process has the same equation over time. But then E. asks what transformation makes this time series stationary, leading me to believe that my answer to A. is incorrect.

rmconrad, think about what is happening over time.  We are given that the average claim severity increases by 8% a year.  So, the average claim severity process does change over time.  Looking back, section 5.1 covered this process at a reasonable level.  The final answer, to my understanding, follows page 99 almost verbatum.  Best of luck!

[NEAS: Yes. Some candidates say this is hard to grasp at first. To see the form of this time series, make a sequence in Excel. First value is 1,000, second value is 1,080, etc. The time series is clearly not stationary, since it increase without bound. Now take first differences, second differences, logarithms, and first difference of logarithms. see which time seires is constant.]

"....Jacob: What is the form of this time series?

Rachel: Actuaries write: Yt = 1.08 Yt-1. The error term is multiplicative: Yt = 1.08 Yt-1 × (1 + å)...."

What does mean the sentences above in the question?
We use Yt = 1.08 Yt-1 × (1 + å)as the times series in this question? but, I don't think it's right.
Even if the time series we should use is like that, both the first difference and the second difference are nonstationary, right? because:
D1(Yt)=0.08Y(t-1)+sigma(t) its mean is not constant. And,
D2(Yt+1)=(Yt+1-2*Yt+Yt-1), its mean is not constant also.

But from the text part 5.1, we can see the derivation results of first and second differences are stationary, so I can't be sure the results above are true.

Please help anybody.

I have been reading the posts for module 8 and 9 and trying to do the hw on my own and am still very confused as to how to compute first and second differences, and how to determine if they are stationary or nonstationary.

I have that Y_t = 1.08Y_t-1 x (1 + e), and I understand why this series is nonstationary. Is this the correct equation, or should it be Y_t = 1.08Y_t-1 + e_t? Also, I noticed that these equations are similar to the percentage change transformations in section 5.4, but I don't understand how to find the first and second differences that way either. All it says is that "notice that we take logs first and then computer first differences..."..there is nothing about HOW to compute first differences.

If we try to use the percentage changes with Y_t = 1.08Y_t-1, we get that log(Y_t) - log(Y_t-1) = log(Y_t/Y_t-1) = log(1.08) = 0.0334. What does this number represent?

I agree. I think I understand the concepts of this module, but what I don't understand is the quote "Rachel: Actuaries write: Y = 1.08 Y . The error term is multiplicative: Y = 1.08 Y × (1 + e)." Which equation are we supposed to use?

The wording the problem is also somewhat confusing..."
" Assume the error term is added to the logarithm of the average claim severities.
" A The average claim severities are multiplied by a random error term"

The first assumption leads me to the equation log(Y_t) = log(1.08Y_t-1) + e
The second assumption leads me to the equation Y_t = 1.08Y_t-1 × (e)

Can someone please clarify?

Thanks.

If you look under the 'old textbook' section for the module 12 homework I think it helps a little bit with the reasoning for the difference equation parts.

For part A I just thought of it intuitively. If Y_0 = 1, then
E(Y_1)=1.08...
E(Y_t)= 1.08^t
Here the limit as t increases without bound is infinite, so the series is not stationary.

For part b, we want to know if the sequence of first differences is stationary. Again assuming Y_0 = 1, we have:
E(Y_1 - Y_0) = 1.08 - 1
...
E(Y_t - Y_t-1) = 1.08^t - 1.08^(t-1) = (.08)*1.08^(t-1)
as t increases without bound, this again will increase without bound, so it is not stationary.

And so on for the second difference.

Here I'm using that the expected value of the error term is zero, so as a side note you have to use the form Y_t = 1.08*Y_t-1 *(1+e)

So my understanding thus far is that if there is no trend, then any arithmetic or logarithmic differencing will still yield a stationary time series.

If the series is arithmetically increasing, then either a differencing or logarithmic adjustment will yield a stationary time series.

If the series is geometrically trending, then only a logarithm can make the series stationary.

My reasoning is that you could add zero or multiply by a logarithm with the base equal to the stationary mean, and still have get the stationary mean as the correct answer.

I'm thinking of it as (and correct me if I'm wrong):

Geometric Trending Adjustment = Square

Arithmetic Trending Adjustment= Rectangle

No Trend Adjustment = Quadrilateral

Such that any type of adjustment can be made to a no-trend adjustment, but a geometric trending adjustment requires a more specific type of adjustment.

If anyone could help me clarify the following, I would appreciate it. If one performs a logarithmic transformation on Y_t = 1.08*Y_t-1*(1+e_t) it seems the algebra would be as follows:
log(Y_t) = log[1.08*Y_t-1*(1+e_t)]
log(Y_t) = log(Y_t-1) + log[1.08*(1+e_t)]
log(Y_t) - log(Y_t-1) = log[1.08*(1+e_t)]
log(Y_t / Y_t-1) = log[1.08*(1+e_t)]

Now if one performs the logarithmic difference transformation per the example on page 99 of the text:
log(Y_t) - log(Y_t-1) = log(Y_t / Y_t-1) where Y_t / Y_t-1 = 1.08*(1+e_t)
then log(Y_t / Y_t-1) = log[1.08*(1+e_t)]

It seems to me the result of the transformation mentioned in part D of the homework assignment is algebraically equivalent to the logarithmic difference transformation because the error term is multiplicative rather than additive. I my reasoning sound?