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

Download Exponential Distribution - Stochastic Hydrology - Lecture Notes and more Study notes Mathematical Statistics in PDF only on Docsity! Exponential Distribution •  The probability density function of the exponential distribution is given by •  E[X] = 1/λ •  λ = 1/µ •  Var(X) = 1/λ2 ( ) 0, 0xf x e xλλ λ−= > > 3   0 ( ) ( ) 1 0, 0 x xF x f x dx e xλ λ−= = − > >∫ f(x)     x     Docsity.com Exponential Distribution •  γs > 0; positively skewed •  Used for expected time between critical events (such as floods of a given magnitude), time to failure in hydrologic/water resources systems components 4   f(x)     x     Docsity.com Gamma Distribution •  The probability density function of the Gamma distribution is given by •  Γ(η) is a gamma function •  Γ(η) = (η-1)!, η = 1,2,… •  Γ(η+1) = η > 0 •  Γ(η) = η > 0 ( ) 1 ( ) , , 0 xx ef x x η η λλ λ η η − − = > Γ 7   η η 1 0 tt e dtη ∞ − −∫ Two parameters λ & η Γ(1) = Γ(2) =1; Γ(1/2)= π 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 = •  Gamma distribution is generally used for daily/ monthly/annual rainfall data 8   2 η η λ Docsity.com Gamma Distribution 9   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)     λ → Scale parameter η → Shape parameter Gamma distribution is in fact a family of distributions   Docsity.com σ1 = → 0.3535 = η1 = 2 λ1 = η1/0.5 = 4 Γ(2)= 1 Example-2 (contd.) 12   1/2 1 1.4144η = ( ) ( ) 1 1 1 1 1 1 1 1 1 2 2 1 4 4 ( ) , , 0 4 2 16 x X x x x ef x x x e xe η η λλ λ η η − − − − − = > Γ = Γ = 1 1η λ 1 1 1 10.5η λ η η= × Docsity.com P[X > 1] = 1 – P[X < 1] Example-2 (contd.) 13   1 4 0 4 1 16 51 1 0.092 xxe dx e −= − ⎛ ⎞= − −⎜ ⎟ ⎝ ⎠ = ∫ Docsity.com During the month 2, the mean and standard deviation of the monthly rainfall are 1 and 0.5cm respectively. 1. Obtain the probability of receiving more than 1cm rain during month 2. 2. Obtain the probability of receiving more than 1cm rain during the two month period assuming that rainfalls during the two months are independent. Given, µ2= 1, σ2 = 0.5 The parameters λ2 and η2 are estimated. µ2 = η2 /λ2 → 1 = η2 /λ2 λ2 = η2 Example-2 (contd.) 14   Docsity.com Since λ1= λ2 and the rainfalls during the two months are independent, the distribution of (X1+X2) will have the parameters η , λ as η = η1+ η2 = 2+4 = 6 λ = 4 Therefore Γ(6)= (6-1)!=5!=120 Example-2 (contd.) 17   ( ) ( ) 1 2 1 6 6 1 4 5 4 ( ) , , 0 4 6 34.13 X X x x x x ef x x x e x e η η λλ λ η η+ − − − − − = > Γ = Γ = Docsity.com P[X1+X2 > 1] = 1 – P[X1+X2 < 1] •  The values of cumulative gamma distribution can be evaluated using tables with χ2=2λx and ν=2η Example-2 (contd.) 18   1 5 4 0 12 1 34.13 42.8621 0.9999 0.785 xx e dx e −= − ⎛ ⎞= − −⎜ ⎟ ⎝ ⎠ = ∫ Docsity.com Extreme Value Distributions •  Extreme events: –  Peak flood discharge in a stream –  Maximum rainfall intensity –  Minimum flow •  The extreme value of a set of random variables is also a random variable •  The probability of this extreme value depends on the sample size and parent distribution from which the sample is obtained 19   Docsity.com Extreme Value Type-I Distribution (Gumbel’s Extreme Value Distribution) •  Referred as Gumbel’s distribution •  Pdf is given by •  ‘–’ applies for maximum values and ‘+’ for minimum values •  α and β are scale and location parameters •  β = mode of distribution •  Mean E[x] = α + 0.577 β (Maximum) = α - 0.577 β (Minimum) 22   ( ) ( ){ }( ) exp exp ; ; 0 f x x x x β α β α α β α ⎡ ⎤= − − −⎣ ⎦ −∞ < <∞ −∞ < <∞ > m m Docsity.com 23   α=1   β=2   x     f(x)     α → Scale parameter β → Location parameter α=0.1   β=2   α=2   β=2   α=1   β=3   α=1   β=5   Extreme Value Type-I Distribution (Gumbel’s Extreme Value Distribution) Docsity.com Extreme Value Type-I Distribution (Gumbel’s Extreme Value Distribution) •  Variance Var(x) = 1.645 α2 •  Skewness coefficient γ = 1.1396 (maximum) = -1.1396 (minimum) •  Y = (X – β)/ α → transformation •  Pdf becomes •  Cdf – (maximum) (minimum) 24   [ ]{ }( ) exp expf y y y= −m m ( ){ } ( ){ } ( ) exp exp 1 exp exp F y y y = − − = − − Fmin(y)  =  1  –  Fmax(-­‐y)   –∞  <  y  <  ∞     Docsity.com 1.  To obtain P[X > 18000], the parameters α and β are obtained initially α = σ/1.283 = 4/1.283 = 3.118 β = µ – 0.45 σ = 9-0.45*4 = 7.2 P[X > 18] = 1 – P[X < 18] = 1 – F(18) = 1 – exp{-exp(-y)} Example-3 (contd.) 27   Docsity.com y = (x – β)/ α = (18-7.2)/3.118 = 3.464 P[X > 18] = 1 – exp{-exp(-y)} = 1 – exp{-exp(-3.464)} = 1 – 0.9692 = 0.0308 Example-3 (contd.) 28   Docsity.com 2.  To obtain P[X < 15], y = (x – β)/ α = (15-7.2)/3.118 = 2.502 F(y) = exp{-exp(-y)} = exp{-exp(-2.502)} = 0.9213 P[X < 15] = 0.9213 Example-3 (contd.) 29   Docsity.com