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AI-GEOSTATS: LAI (leaf area index) fromNDVI

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  • rocchini@unisi.it
    I d like to know if the approach I used to derive LAI from NDVI is correct. STEP 1: I ve got 46 field point values of LAI (leaf area index, namely the cover of
    Message 1 of 3 , Aug 12, 2003
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      I'd like to know if the approach I used to derive LAI from NDVI is correct.
      STEP 1: I've got 46 field point values of LAI (leaf area index, namely the
      cover of plant leafs)
      STEP 2: I derived the NDVI index from a multispectral image.
      STEP 3: For every field plot I calculated the mean NDVI of 3x3 neighbour
      cells
      STEP 4: I made a regression between mean_NDVI and LAI.
      STEP 5: r^2 was low (0.34), r being 0.70, but t, measured as
      r/sqr((1-r^2)/(n-2)) was over the minimum t 2.7, being my t 5.75
      STEP 6: Since the correlation was highly significant p<0.01 I applied the
      equation of the regression line y= 4.9053x + 0.2406 where y was LAI and x
      was NDVI to the NDVI map, obtaining the LAI map
      STEP 7: I made a control on the accuracy of the model by measuring the mbe
      (mean bias error) calculated as the mean of single errors for every plot
      (46 measures): mean of P-O
      where P was the estimated LAI value and O the observed, by obtaining a mbe
      of 0.03249587

      Questions:
      1. Could I apply the equation as in step 6?
      2. Could I control my model by using the same input observed values as in
      step 7?

      Thanks
      Duccio




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    • Chris Hlavka
      You might want to consider the implications of using data with different supports - 3x3 neighborhoods and points. The 3x3 neighborhoods are probably larger
      Message 2 of 3 , Aug 12, 2003
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        You might want to consider the implications of using data with
        different supports -
        3x3 neighborhoods and points. The 3x3 neighborhoods are probably
        larger than the
        areas associated with LAI field values. Thus the 3x3 mean NDVI's can
        be considered
        to be estimates of NDVI's at the points that have associated error.
        Error in the independent
        (x) variable leads to underestimated correlation and (ordinary least
        squares (OLS) regression) slope.
        There are a number of alternatives to OLS, such a MA and RMA
        regression, that might lead to
        improved slope estimates, but I prefer to correct the correlation and
        slope estimates using
        estimates of the precision of the x variable.

        You can roughly estimate the precision of the point NDVI
        by : 1) calculating the standard deviation of the nine pixel NDVI's
        associated with each field observation and 2) plotting the standard
        deviations versus the means to check for dependence of variation with
        magnitude,
        3) estimate standard error of NDVI as the mean standard deviation if
        no dependence, otherwise consider
        regression of LAI versus log(NDVI) or log-log regression. Note that
        if the "point" area is much smaller than a pixel, the x error
        will be underestimated - but fixing this would involve either a
        geostatistical analysis of point values for NDVI
        or estimation involving fractal analysis.

        The error estimate can then be used to correct the estimates of
        correlation and slope (my apologies
        for cutting and pasting a Word file so that subsrcipts and
        superscripts are lost, and maybe also Greek font, and illustrating
        with S commands):

        Preliminaries: First, consider regression of variable x (the
        independent variable) versus y (the dependent variable). The usual
        formula for the slope is:

        S [(xi -mx)*(yi - my)]/S (xi - mx)2
        (1)

        where summation is over the index i for individual data points, and
        the means are mx and my. This formula (section 1.2 in N. Draper and
        H. Smith, Applied Regression Analysis, John Wiley & Sons, Inc., New
        York, 1966) is correct, and computationally simple and accurate, that
        is, works well to preserve floating point accuracy. However,
        formulae involving descriptive statistics (correlation or covariance
        of x and y, and the standard devations of x and y) convey more
        information about the factors related to the slope:

        cor(x,y)*sy/sx or cov(x,y)/sx2
        (2)

        where one can see that the magnitude of the slope increases with the
        correlation and range of the dependent variable y (as measured by the
        standard deviation), and decreases with range of the independent
        variable. If one of the formulae in (2) is used with n data points,
        it will be accurate (unbiased) if multiplied by the square root of
        (n-2)/(n-1) to correct for the effect of using estimated, rather than
        "true", means and if the usual assumptions, including accurate values
        for the indpendent variable, are correct. If the range of the
        independent variable is inflated by errors, the slope will decrease,
        that is. will be biased low.

        Predicting the slope when precise values of independent variable
        variable x are replaced by the estimated or measured values variable
        x, following Section 29.56 in M. Kendall and A. Stuart, The Advanced
        Theory of Statistics: Volume 2: Inference and Relationship, 4th
        Edition, Charles Griffin & Company Limited, London, 1979 (copy in
        your mailbox). Let's assume that the measurements are made without
        bias and with a precision represented as a standard deviation in
        error: the observed measurements (x1,x2, Š) of the dependent variable
        x can be considered as sums of the true values (x1, x2, Š ) with 0
        standard error plus errors (d1,d2, Š.) with average of 0 and standard
        deviation sd. The least squares regression slope is cov(x,y)/sx2 =
        cov(x,y)/( sx2 + sd2), where cov(x,y) is the covariance between x and
        y, i.e. the correlation times the product of the standard deviations
        of x and y. Now if the least squares slope with no errors is
        cov(x,y)/ sx2 = 1, then the slope with the errors is:

        cov(x,y)/( sx2 + sd2) = (cov(x,y)/sx2) * [sx2/ (sx2 + sd2)] =
        sx2/( sx2 + sd2)
        = 1/[1 + (sd/sx)2]
        (3)

        The expression on the right is a function of the relative magnitude
        sd/sx of the measurement error to data range for the independent
        variable, where standard deviation is the metric. The range term sx
        can be approximated with sx, the standard deviation of the measured
        values, if the range of measurements is large compared to the
        measurement errors. Otherwise, correct for the effect of measurement
        error by using , leading (as you have noted) to (sx2 - sd2)/sx2 as
        the predicted slope. The estimate for sd is generally known from an
        independent source, such as instrument specs or calibration analysis.

        You will note that the slope predicted with (1) is always less than
        1. The mathematical cause is due to the inflation of the denominator
        from sx2 to sx2 + sd2. Perhaps what is counter-intuitive is that
        the slope is biased due to mean zero errors. Shouldn't the errors
        just degrade the precision of the least squares slope? No - because
        least squares regression is asymmetrical in the way the independent
        and dependent variables are treated: the sum of squares to be
        minimized are the squared residuals that are distances of points to
        the regression line in the y direction. One way to correct the
        problem, i.e. predict the slope one would have if true x values were
        known, is by multiplying by [1 + (sd/sx)2]. Another approach is to
        use a least squares technique using distances from data points to
        nearest point on line.

        You may also note that inflation of the variance and standard
        deviation of the independent variable leads to degradation in
        correlation between the independent and independent variable:

        cor(x,y) = cov(x,y)/(sx*sy) = cov(x,y)/(sx*sy) = cov(x,y)/( *sy)
        = cor(x,y)*sx/ = cor(x,y)* /sx = cor(x,y)*

        If the dependent variable is a transformed variable, then the
        standard deviations are statistics of the transformed variable rather
        then the original variable. For example, your independent variable
        was the log-transformed predicted leaf area per vine (log(LA)), using
        a calibration equation from regression analysis with this same
        variable (log(LA)) as the dependent variable and pruning weight as
        the independent variable. So the standard deviation of the
        calibration regression residuals is a good estimate of sd. This
        predicted log(LA) is the independent variable in the validation
        regression you are concerned with, that is, the one with a slope of
        less than one. So a good estimate of sx is either the standard
        deviation s of the validation values for log(LA) or the corrected
        value sqrt(s2 - sd2).


        Example calculation:
        measurement std. dev. = 0.326
        std. dev. of independent variable = 0.616
        > 1/(1 + .326^2/.616^2) # predicted slope
        [1] 0.7812045
        > sqrt(.616^2 - .326^2) # corrected std dev of indep var
        [1] 0.5226662
        > 1/(1 + .326^2/.522^2) # predicted slope with corrected std dev
        [1] 0.7194107

        The least squares slope was 0.67 +/- 0.14.

        Chris


        >I'd like to know if the approach I used to derive LAI from NDVI is correct.
        >STEP 1: I've got 46 field point values of LAI (leaf area index, namely the
        >cover of plant leafs)
        >STEP 2: I derived the NDVI index from a multispectral image.
        >STEP 3: For every field plot I calculated the mean NDVI of 3x3 neighbour
        >cells
        >STEP 4: I made a regression between mean_NDVI and LAI.
        >STEP 5: r^2 was low (0.34), r being 0.70, but t, measured as
        >r/sqr((1-r^2)/(n-2)) was over the minimum t 2.7, being my t 5.75
        >STEP 6: Since the correlation was highly significant p<0.01 I applied the
        >equation of the regression line y= 4.9053x + 0.2406 where y was LAI and x
        >was NDVI to the NDVI map, obtaining the LAI map
        >STEP 7: I made a control on the accuracy of the model by measuring the mbe
        >(mean bias error) calculated as the mean of single errors for every plot
        >(46 measures): mean of P-O
        >where P was the estimated LAI value and O the observed, by obtaining a mbe
        >of 0.03249587
        >
        >Questions:
        >1. Could I apply the equation as in step 6?
        >2. Could I control my model by using the same input observed values as in
        >step 7?
        >
        >Thanks
        >Duccio
        >
        >
        >
        >
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        --
        ***************************************
        Chris Hlavka
        NASA/Ames Research Center 242-4
        Moffett Field, CA 94035-1000
        (650)604-3328 FAX 604-4680
        Christine.A.Hlavka@...
        ***************************************

        [Non-text portions of this message have been removed]
      • rocchini@unisi.it
        Sorry, but I didn t give information about spatial resolutions. Plots have a dimension of 10*10 meters. The image (Qickbird) has a resolution of 3 meters. Now,
        Message 3 of 3 , Aug 13, 2003
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          Sorry,
          but I didn't give information about spatial resolutions.
          Plots have a dimension of 10*10 meters. The image (Qickbird) has a
          resolution of 3 meters.
          Now, since I calculated mean NDVI of 3*3 cells, I think that spatial
          resolution is
          almost the same.
          Duccio



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