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Parameters of distributionPowerPoint Presentation

Parameters of distribution

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Parameters of distribution

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Parameters of distribution

- Location Parameter
- Scale Parameter
- Shape Parameter

Plotting position

- Plotting position of xi means, the probability assigned to each data point to be plotted on probability paper.
- The plotting of ordered data on extreme probability paper is done according to a general plotting position function:
- P = (m-a) / (N+1-2a).
- Constant 'a' is an input variable and is default set to 0.3.
- Many different plotting functions are used, some of them can be reproduced by changing the constant 'a'.
- Gringorton P = (m-0.44)/(N+0.12) a = 0.44
- Weibull P = m/(N+1) a = 0
- Chegadayev P = (m-0.3)/(N+0.4) a = 0.3
- Blom P = (m-0.375)/(N+0.25) a = 0.375

Curve Fitting Methods

- The method is based on the assumption that the observed data follow the theoretical distribution to be fitted and will exhibit a straight line on probability paper.
- Graphical Curve fitting Method
- Mathematical Curve fitting Method.-
- Method of Moments-
- Method of Least squares
- Method of Maximum Likelihood

- Estimation procedures differ
- Comparison of quality by:
- mean square error or its root
- error variance and standard error
- bias
- efficiency
- consistency

- Mean square error in of :

- Consequently:
- First part is the variance of = average of squared differences about expected mean, it gives the random portion of the error
- Second part is square of bias,bias= systematic difference between expected and true mean, it gives the systematic portion of the error

- Root mean square error:
- Standard error
- Consistency:

Mind effective number of data

- Variable is function of reduced variate:
- e.g. for Gumbel:

- Reduced variate function of non-exceedance prob.:
- Determine non-exceedance prob. from rank number of data in ordered set, e.g. for Gumbel:
- Unbiased plotting position depends on distribution

- Procedure:
- rank observations in ascending order
- compute non-exceedance frequency Fi
- transform Fi into reduced variate zi
- plot xi versus zi
- draw straight line through points by eye-fitting
- estimate slope of line and intercept at z = 0 to find the parameters

- Annual maximum river flow at Chooz on Meuse

Graphical estimation

- Gumbel parameters:
- graphical estimation: x0 = 590, = 247
- MLM-method: x0 = 591, = 238

- 100-year flood:
- T = 100 FX(x) = 1-1/100 = 0.99
- z = -ln(-ln(0.99)) = 4.6
- graphical method: x = x0 + z = 590 + 247x4.6 = 1726 m3/s
- MLM method: x = x0 + z = 591 + 238x4.6 = 1686 m3/s

- Graphical method: pro’s and con’s
- easily made
- visual inspection of series
- strong subjective element in method: not preferred for design; only useful for first rough estimate
- confidence limits will be lacking

- Plotting positions should be:
- unbiased
- minimum variance

- General:

- Right censoring: eliminating data from analysis at the high side of the data set
- Left censoring: eliminating data from analysis at the low side of the data set
- Relative frequencies of remaining data is left unchanged.
- Right censoring may be required because:
- extremes in data set have higher T than follows from series
- extremes may not be very accurate

- Left censoring may be required because:
- physics of lower part is not representative for higher values

- Confidence limits become:
- CL diverge away from the mean
- Number of data N also determine width of CL

- Uncertainty in non-exceedance probability for a fixed xp:
- standard error of reduced variate

- It follows with zp approx N(zp,zp):

hence:

Confidence limits for frequency distribution

Normal distribution

FX(z) for

z=(x-877)/357

Ranked observations

T=1/(1-FX(z))

Fi=(i-3/8)/(N+1/4)

T

FX(x) = 1 - 1/T

- Prior to fitting, tests required on:
- 1. stationarity (properties do not vary with time)
- 2. homogeneity (all element are from the same population)
- 3. randomness (all series elements are independent)
- First two conditions transparent and obvious. Violating last condition means that effective number of data reduces when data are correlated
- lack of randomness may have several causes; in case of a trend there will be serial correlation

- HYMOS includes numerous statistical test :
- parametric (sample taken from appr. Normal distribution)
- non-parametric or distribution free tests (no conditions on distribution, which may negatively affect power of test

- On randomness:
- median run test
- turning point test
- difference sign test

- On correlation:
- Spearman rank correlation test
- Spearman rank trend test
- Arithmetic serial correlation coefficient
- Linear trend test

- On homogeneity:
- Wilcoxon-Mann-Whitney U-test
- Student t-test
- Wilcoxon W-test
- Rescaled adjusted range test

- Hypothesis
- F(x) is the distribution function of a population from which sample xi, i =1,…,N is taken
- Actual to theoretical number of occurrences within given classes is compared

- Procedure:
- data set is divided in k class intervals containing at least each 5 values
- Class limits from all classes have equal probability
pj = 1/k = F(zj) - F(zj-1)

e.g. for 5 classes this is p = 0.20, 0.40, 0.60, 0.80 and 1.00

- the interval j contains all xi with: UC(j-1)<xi UC(j)
- the number of samples falling in class j = bj is computed
- the number of values expected in class j = ej according to the theoretical distribution is computed
- the theoretical number of values in any class = N/k because of the equal probability in each class

Chi-squared goodness of fit test

- Consider following test statistic:
- under H0 test statistic has 2 distr, with df = k-1-m
- k= number classes, m = number of parameters
- simplified test statistic:
- H0 not rejected at significance level if:

- Annual rainfall Vagharoli (see parameter estimation)
- test on applicability of normal distribution
- 4 class intervals were assumed (20 data)
- upper class levels are at p=0.25, 0.50, 0.75 and 1.00
- the reduced variates are at -0.674, 0.00, 0.674 and
- hence with mean = 877, and stdv = 357 the class limits become: 877 - 0.674x357 = 636
877 = 877

877 + 0.674x357 = 1118

From the table it follows for the test statistic:

At significance level = 5%, according to Chi-squared distribution for = 4-1-2 df the critical value is at 3.84, hence c2 < critical value, so H0 is not rejected