Moments of Distribution - Stochastic Hydrology - Lecture Notes, Study notes of Mathematical Statistics

Download Moments of Distribution - Stochastic Hydrology - Lecture Notes and more Study notes Mathematical Statistics in PDF only on Docsity! Moments of a distribution nth moment about the origin E(X): Expected value of ‘X’ : First moment about the origin µ = E(X) = nth moment about the expected value ( )o nn x f x dxµ ∞ −∞ = ∫ ( )x f x dx ∞ −∞ ∫ ( ) ( )nn x f x dxµ µ ∞ −∞ = −∫ x dx f(x) E(x) 3   Docsity.com Measures of central tendency Mean: Discrete case: Sample estimate of the mean: Mode: Value with highest frequency of occurrence Median: Value such that 50% of area is on either side ( )x f x dxµ ∞ −∞ = ∫ 1 ( ) n i i i x p xµ = =∑ 4   n: Sample size 1 n i i x x n == ∑ Docsity.com σ2 = E[x – µ ]2 = E[x2 – 2xµ + µ2] = E[x2] – 2µE[x] + E[µ2] = E[x2] – 2µ2 + µ2 = E[x2] – µ2 = E[x2] – {E[x]}2 7   Docsity.com Measures of symmetry Coefficient of skewness: Population Sample Estimate 3 3/2 2 3 2 3/2 ( ) ( ) ( ) s x f x dx µ γ µ µ σ ∞ −∞ = − = ∫ ( ) 3 1 3( 1)( 2) n i i s n x x C n n s = − = − − ∑ 8   Negative skew Cs<0 Symmetric Cs=0 Positive skew Cs>0 Docsity.com Measures of Peakedness Kurtosis: Population Sample Estimate 4 2 2 4 2 2 ( ) ( ) ( ) K x f x dx µ µ µ σ ∞ −∞ = − = ∫ 4 2 1 4( 1)( 2)( 3) n i i n x x k n n n s = ⎛ ⎞−⎜ ⎟ ⎝ ⎠= − − − ∑ 9   Lepto kurtic K>3 Normal K=3 Platy kurtic K<3 Docsity.com Obtain the sample estimates of mean, standard deviation, coefficient of variation, coefficient of skewness and kurtosis for the following observed data of annual stream flow for 15 years. Example-2 12   Year 1 2 3 4 5 6 7 8 9 10 Avg. yearly stream flow (Mm3) 150 129 160 152 165 138 149 115 97 154 Year 11 12 13 14 15 Avg. yearly stream flow (Mm3) 168 110 108 105 125 Docsity.com Example-2 (contd.) 13   Mean, Therefore mean, = 2025/15 = 135 Mm3 Variance, 1 n i i x x n == ∑ 1 n i i x = ∑ = 150+129+160+152+165+138+149+115+97+154+ 168+110+108+105+125 = 2025 x ( )2 2 1 1 n i i x x s n = − = − ∑ Docsity.com 14   Year Avg. Stream flow Mm3(xi) 1 150 15 225 3375 50625 2 129 -6 36 -216 1296 3 160 25 625 15625 390625 4 152 17 289 4913 83521 5 165 30 900 27000 810000 6 138 3 9 27 81 7 149 14 196 2744 38416 8 115 -20 400 -8000 160000 9 97 -38 1444 -54872 2085136 10 154 19 361 6859 130321 11 168 33 1089 35937 1185921 12 110 -25 625 -15625 390625 13 108 -27 729 -19683 531441 14 105 -30 900 -27000 810000 15 125 -10 100 -1000 10000 Σ 2025 0 7928 -29916 6678008 ( )ix x− ( ) 2 ix x− ( ) 3 ix x− ( ) 4 ix x− Docsity.com COMMONLY USED DISTRIBUTIONS 17   Docsity.com Normal Distribution Two parameters, µ & σ X ∼ N(µ, σ2) f(x) approaches zero as x → +∞ 21 1( ) exp 22 xf x xµ σπσ ⎧ ⎫−⎪ ⎪⎛ ⎞= − −∞ < < +∞⎨ ⎬⎜ ⎟ ⎝ ⎠⎪ ⎪⎩ ⎭ x f(x) µ +∞ -∞ 18   Bell-shaped Distribution; Gaussian Distribution Symmetric about x = µ Docsity.com Normal Distribution Coefficient of skewness, γs = 0 Kurtosis, K = 3 Y = a + bX – Linear function of ‘X’ Y ∼ N(a+bµ, b2σ2) F(x) = ( ) x f x dx −∞ ∫ 21 21 2 xx e dx x µ σ πσ −⎛ ⎞− ⎜ ⎟ ⎝ ⎠ −∞ = −∞ < < +∞∫ 19   Docsity.com Normal Distribution Obtaining standard variate ‘z’ using tables: z Area P[Z < z] = 0.5+Area from table 22   0 Docsity.com 23   z 0 2 4 6 8 0 0 0.008 0.016 0.0239 0.0319 0.1 0.0398 0.0478 0.0557 0.0636 0.0714 0.2 0.0793 0.0871 0.0948 0.1026 0.1103 0.3 0.1179 0.1255 0.1331 0.1406 0.148 0.4 0.1554 0.1628 0.17 0.1772 0.1844 0.5 0.1915 0.1985 0.2054 0.2123 0.219 0.6 0.2257 0.2324 0.2389 0.2454 0.2517 0.7 0.258 0.2642 0.2704 0.2764 0.2823 0.8 0.2881 0.2939 0.2995 0.3051 0.3106 0.9 0.3159 0.3212 0.3264 0.3315 0.3365 1 0.3413 0.3461 0.3508 0.3554 0.3599 Normal Distribution Tables Docsity.com 24   z 0 2 4 6 8 3.1 0.499 0.4991 0.4992 0.4992 0.4993 3.2 0.4993 0.4994 0.4994 0.4994 0.4995 3.3 0.4995 0.4995 0.4996 0.4996 0.4996 3.4 0.4997 0.4997 0.4997 0.4997 0.4997 3.5 0.4998 0.4998 0.4998 0.4998 0.4998 3.6 0.4998 0.4999 0.4999 0.4999 0.4999 3.7 0.4999 0.4999 0.4999 0.4999 0.4999 3.8 0.4999 0.4999 0.4999 0.4999 0.4999 3.9 0.5 0.5 0.5 0.5 0.5 Normal Distribution Tables Docsity.com Obtain the area under the standard normal curve z < -0.98 From tables: Req. area = 0.5 – area between 0 and +0.98 = 0.5 – 0.3365 = 0.1635 Example-4 z -0.98 0 Req. area 27   z 7 8 9 0.8 0.3078 0.3106 0.3133 0.9 0.334 0.3365 0.3389 1 0.3577 0.3599 0.3621 1.1 0.379 0.381 0.383 Docsity.com Obtain ‘z’ such that P[Z < z]=0.879 Since the value is greater than 0.5, ‘z’ must be +ve area between 0 to z = 0.879 – 0.5 = 0.379 From the table, for the area of 0.379, corresponding z = 1.17 Example-5 z 0 Area=0.879 28   z 6 7 8 1.1 0.377 0.379 0.381 1.2 0.3962 0.398 0.3997 Docsity.com Obtain P[X < 75] if N ∼ (100, 25002) From the table, Req. area = 0.5 – area between 0 and +0.01 = 0.5 – 0.004 = 0.496 P[X < 75] = 0.496 Example-6 29   z 0 1 2 0.0 0 0.004 0.008 0.1 0.0398 0.0438 0.0478 75 100 2500 0.01 xz µ σ − = − = = − z -0.01 0 Req. area Docsity.com Central limit theorem •  For hydrological applications under most general conditions, if Xi‘s are all independent with E[xi]= µi and var(Xi) = σi2, then the sum Sn = X1+X2+…….+Xn as n → ∞ approaches a normal distribution with One condition for this generalised Central Limit Theorem is that each Xi has a negligible effect on the distribution of Sn (Statistical Methods in Hydrology, C.T.Haan, .Affiliated East-West Press Pvt Ltd, 1995, p. 89) 32   [ ] [ ] 1 2 1 & n n i i n n i i E S Var S µ σ = = = = ∑ ∑ Docsity.com Log-Normal Distribution •  ‘X’ is said to be log-normally distributed if Y = lnX is normally distributed. •  The probability density function of the log normal distribution is given by •  γs = 3Cv+Cv3 where Cv is the coefficient of variation of ‘X’ •  As Cv increases, the skewness, γs , increases ( )2 2ln 21( ) e 0 ,0 , 0 2 n nx n n n f x x x µ σ µ σ π σ − −= < <∞ < <∞ > 33   2 2 2 2 1 ln , ln 1 2 1 x y y v v v Sx C whereC C x µ σ ⎡ ⎤ ⎡ ⎤= = + =⎢ ⎥ ⎣ ⎦+⎣ ⎦ Docsity.com 34   Log-Normal Distribution x f(x) •  Positively skewed – skewed to the left with long exponential tail on the right. •  Commonly used for monthly streamflow, monthly/ seasonal precipitation, evapotranspiration etc . µn=0 σn2=1 µn=0.3 σn2=1 µn=0 σn2=0.3 Docsity.com Y = ln X follows log normal distribution P[X > 150] = P [ Y > ln150]; P[Y > ln150] = 1– P[Y < ln150] = 1– P[Z < 0.693] = 1 – 0.25583 = 0.7442 Example-9 (contd.) 37   5.011 4.89 0.1747 0.693 y y yz S − = − = = ln150 = 5.011 Docsity.com Exponential Distribution •  The probability density function is given by •  E[X] = 1/λ •  λ = 1/µ •  Var(X) = 1/λ2 •  γs > 0, therefore positively skewed •  Used for expected time between two critical events (such as floods of a given magnitude), time to failure in hydrologic/water resources systems components ( ) 0, 0xf x e xλλ λ−= > > 38   0 ( ) ( ) 1 0, 0 x xF x f x dx e xλλ λ−= = − > >∫ f(x) x Docsity.com The mean time between high intensity rainfall (rainfall intensity above a specified threshold) events occurring during a rainy season is 4 days. Assuming that the mean time follows an exponential distribution. Obtain the probability of a high intensity rainfall repeating 1.  within next 3 to 5 days. 2.  within next 2 days Mean time (µ) = 4 λ = 1/µ = 1/4 Example-10 39   Docsity.com Gamma Distribution •  Exponential distribution is a special case of gamma distribution with η=1 •  λ → Scale parameter •  η → Shape parameter •  Mean = η/λ •  Variance = η/λ2 → σ = •  Skewness coefficient γ = •  As γ decreases, η increases •  Cdf is given by 42   2 η ( ) 1 0 ( ) 1 ! , , 0jx j F x e x j x η λ λ λ η − − = = − >∑ η λ Docsity.com Gamma Distribution 43   η=3 λ=1 η=3 λ=1/2 η=1 λ=1 η=3 λ=4 η=0.5 λ=1 x f(x) Docsity.com Gamma Distribution •  If ‘X’ and ‘Y’ are two independent gamma rvs having parameters η1, λ and η2, λ respectively then U=X+Y is a gamma rv with parameters η=η1+ η2 and λ •  This property can be extended to sum of ‘n’ number of independent gamma rvs. •  Gamma distribution is generally used for daily/ monthly/annual rainfall data •  Also used for annual runoff data 44   Docsity.com P[X > 3] = 1 – P[X < 3] Example-11 (contd.) 47   3 2 4 0 12 1 32 851 1 0.0005 xx e dx e −= − ⎛ ⎞= − −⎜ ⎟ ⎝ ⎠ = ∫ Docsity.com During the month 2, the mean and standard deviation of the monthly rainfall are 30 and 8.6 cm respectively. 1. Obtain the probability of receiving more than 3cm rain during month 2. 2. Obtain the probability of receiving more than 3cm rain during the two month period assuming that rainfall during both months are independent. Given, µ= 30, σ= 8.66 Initially the parameters λ, η are obtained. µ = η/λ → 30 = η/λ λ = η/30 Example-11 (contd.) 48   Docsity.com σ = → 8.66 = η = 12 λ = 4 Γ(12)= (12-1)!=11! Example-11 (contd.) 49   8.66 / 30η η = ( ) ( ) 1 12 12 1 4 11 4 ( ) , , 0 4 12 0.42 n x x x x ef x x x e x e η λλ λ η η − − − − − = > Γ = Γ = η λ η λ Docsity.com