how to identify turning points in stock price data

Question

This question is a continuation of this one.

My goal is to find the turning points in stock price data.

So far I:

Tried differentiating the smoothed pri

Answer 1

Another approach based on some of the ideas here. For each point in the series, look at n points before and after (the window). If the value of the current point is the highest in the window, make it a peak turning point (if it lowest, make it a trough). Exclude the first and final n points in the series.

Experimented with monthly data and got the following with n=6.

Answer 2

You could take the second derivative into account, meaning you should additionally (to your first derivative) evaluate (y_{i-1} + y_{i+1} - 2y_i) / (dx)². If this is above a certain threshold you have a maximum, if it is below you have a minimum and else you can discard it. This should throw out a lot of points that you keep using your method of finding extrema (y' = 0), because this condition is also valid for saddle points.

Answer 3

That's works Patrick87, Thanks. Following are java function to implement the same:

Assume StockPrices has a map of key date and value StockPrice (price, average where x = 5)

private double getCx(StockPrices stockPrices, LocalDate executionDate, int x, double m) { return Math.abs(getFx(stockPrices, executionDate) - getGx(stockPrices, executionDate)) - m * getHx(stockPrices, executionDate, x); }

private double getGx(StockPrices stockPrices, LocalDate executionDate) {
    return stockPrices.getAvg(executionDate, 5);
}

private double getFx(StockPrices stockPrices, LocalDate executionDate) {
    return stockPrices.getPrice(executionDate);
}

public double getHx(StockPrices stockPrice, LocalDate localDate, int x) {
    //standard deviation
    return Math.sqrt(getVariance(stockPrice, localDate, x));
}

private double getVariance(StockPrices stockPrice, LocalDate localDate, int x) {
    double sum = 0;
    int count = 0;
    for (int i = - (x / 2); i <= (x / 2) ; i++) {
        LocalDate date = localDate.with(BusinessDay.add(localDate, i, stockPrice.getPriceMap(), 2));
        double avg = stockPrice.getAvg(date, 5);
        double price = stockPrice.getPrice(date);
        if (price != 0.0) {
            sum += Math.pow((price - avg), 2);
            count++;
        }
    }
    return sum / count;
}

Answer 4

Here's just an idea, sort of an idea from a different angle, and possibly a very bad idea, but since differentiation isn't working, something like this might be a thought.

First, you need to determine a minimum meaningful X-axis interval. In your figure, if you take this to be too small, you will get false positives from the bumps. This is conceptually similar to the idea of smoothing your data. Call this interval dx.

Next, using a sliding window of size dx, generate a moving average curve corresponding to your curve. There are lots of different ways you could think about doing this (to remove statistical outliers, or to use more or fewer points in the window). Call this curve g(x), and your original curve f(x). Additionally, make a curve h(x) which gives some measure of the variability of data in the sliding window which you use to compute g(x) (standard deviation should work fine if you're using a few points from the interval).

Now, begin computing curves of the form c_m(x) = |f(x) - g(x)| - m * h(x). You can start with m = 1. Any points x for which c_m(x) is positive are candidates for a local min/max. Depending on how many hits you get, you can begin increasing or decreasing m. You can do this in a way similar to binarys search: if you want more points, make m = (min + m) / 2, and if you want fewer points, make m = (max + m) / 2 (adjusting min and max accordingly).

So here's an example of what I'm suggesting. Let's say we have the following series:

f(x) = [  1,   2,   4,   3,   2,   3,   6,   7,   8,   7, 
          5,   4,   3,   2,   2,   3,   2,   3,   5,   8,   9]

We choose dx = 5. We construct g(x) by taking a simple average of the points around x:

g(x) = [2.3, 2.5, 2.4, 2.8, 3.6, 4.2, 5.2, 6.2, 6.6, 6.2, 
        5.4, 4.2, 3.2, 2.8, 2.4, 2.4, 3.0, 4.2, 5.4, 6.3, 7.3]

h(x) = [1.2, 1.1, 1.0, 0.7, 1.4, 2.4, 2.3, 1.7, 1.0, 1.5,
        1.9, 1.7, 1.2, 0.7, 0.5, 0.6, 1.1, 2.1, 2.7, 2.4, 1.7]

With m = 1 we get:

c(x) = [0.1, xxx, 0.6, xxx, 0.2, xxx, xxx, xxx, 0.4, xxx,
        xxx, xxx, xxx, 0.1, xxx, 0.0, xxx, xxx, xxx, xxx, 0.0]

This seems to have worked fairly well, actually. Feel free to share thoughts. Note that this might be more or less the equivalent of differentiation, given the mean value theorem.