Estimation of Olympic gold medals for Italy

Questions
,
1

Dear All,

According to the data list, Italy has won 10 and 2 Olympic gold medals in 2020 and 2022. I tried to estimate the number of gold medals for these two years using a simple time series model.
Is the method correct? I appreciate your time.

data = {{1952., 8.}, {1956., 8.}, {1960., 13.}, {1964., 10.}, {1968., 
    3.}, {1972., 5.}, {1976., 2.}, {1980., 8.}, {1984., 14.}, {1988., 
    6.}, {1992., 6.}, {1994., 7.}, {1996., 13.}, {1998., 2.}, {2000., 
    13.}, {2002., 4.}, {2004., 10.}, {2006., 5.}, {2008., 8.}, {2010.,
     1.}, {2012., 8.}, {2014., 0.}, {2016., 8.}, {2018., 3.}, {2020., 
    10.}, {2022., 2.}};

year = Map[#[[1]] &, data];

gold = Map[#[[2]] &, data];

Length[gold];

train = Take[gold, 24];

Length[train];

tsm = TimeSeriesModelFit[train];

tsm["BestFit"];

for = TimeSeriesForecast[tsm, {2}];
2

Hi Alex,

I guess you do not need TimeSeriesForecast[] as TimeSeriesModel does already give a forecast. If I really had to, I would do it like so - but under normal circumstances I would not even try! I do not believe that those simple values of the past contain information about the future.

data = {{1952., 8.}, {1956., 8.}, {1960., 13.}, {1964., 10.}, {1968., 
    3.}, {1972., 5.}, {1976., 2.}, {1980., 8.}, {1984., 14.}, {1988., 
    6.}, {1992., 6.}, {1994., 7.}, {1996., 13.}, {1998., 2.}, {2000., 
    13.}, {2002., 4.}, {2004., 10.}, {2006., 5.}, {2008., 8.}, {2010.,
     1.}, {2012., 8.}, {2014., 0.}, {2016., 8.}, {2018., 3.}, {2020., 
    10.}, {2022., 2.}};

dateGold = TimeSeries[{DateObject[{Round@#1}, "Year"], #2} & @@@ data];
train = TimeSeriesWindow[dateGold, {DateObject[{1952}, "Year"], DateObject[{2018}, "Year"]}];
tsm = Quiet@TimeSeriesModelFit[train];
DateListPlot[{tsm["TemporalData"], {#, Around[tsm[#], tsm["PredictionLimits"][#] - tsm[#]]} & /@ {DateObject[{2020}, "Year"], DateObject[{2022}, "Year"]}}, ImageSize -> Large, GridLines -> Automatic]

GoldPrediction.png
GoldPrediction.png576×366 22.8 KB

As you see the “forecast” is basically the mean value - with huge error bars, as there is no trend in the data. If you do give your data some trend, then one can see that this approach is trying the best it can do:

data = MapIndexed[{#1[[1]], #1[[2]] + First[#2]} &, data]

GoldPrediction_t.png
GoldPrediction_t.png576×366 14.7 KB

Does that help? Regards – Henrik

3

Hi Henrik,

I tried to use Markov process with SequencePredict[] for 2020 as follows:

In[13]:=  SequencePredict[{train}, {{}}]

Out[13]= {8.}

What do you think about this approach? Is acceptable?

Thank you for your help.

4

Hello Prof. Ghorbani,

I guess for SequencePredict[] you rather need a bunch of sequences …

Regards – Henrik