# 2.4. The 95% interval of Gamma(2,1) (Optional question) Jungho Ryu.

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2.4. The 95% interval of Gamma(2,1) (Optional question) Jungho Ryu

The situation We need to find the 95% confidence interval of Gamma distribution f(x) with one of these conditions satisfied: 1.Choose two points x0 and x1 with x0 < x1 such that I.f(x0) = f(x1) II.For all (x’ x1), f(x’) < f(x0). 2.(equivalently) Choose two points x0 and x1 with x0 < x1 such that I.f(x0) = f(x1) II.For all possible x0’ and x1’ with x0’ x1 – x0.

The situation We need to find the 95% confidence interval of Gamma distribution f(x) with one of these conditions satisfied: 1.Choose two points x0 and x1 with x0 < x1 such that I.f(x0) = f(x1) II.For all (x’ x1), f(x’) < f(x0). 2.(equivalently) Choose two points x0 and x1 with x0 < x1 such that I.f(x0) = f(x1) II.For all possible x0’ and x1’ with x0’ x1 – x0.

Pictorically Speaking Case 1: all points not belonging to the interval have lower f(x) biggest f(x) are in the 95% interval All x’s that belong to 95% interval f(x0)f(x1)

Conceptually Speaking Notice that x0 and x1 are on the opposite side of the peak (global maximum) – So we need to find the global maximum To always include only the biggest f(x)’s, we can start x0 and x1 from the top and go down – while f(x0) and f(x1) keeping the same! But the sides look different (function is not symmetrical!)

Into action! First we need to find that global maximum (easy!: Find it analytically). Gamma Distribution (scale = 2, shape = 1) f(x) = x e -x. We take the derivative and set it to 0. f’(x) = e –x – xe -x = 0  e –x = xe -x  x = 1 The global maximum is at x =1. So, x0 1.

Ideally, We would update (decrease) x0 and then find the inverse f -1 to get x1. – Not doable analytically. Then we do it numerically: – Update (decrease) x0  calculate f(x0)  slowly update (increase) x1 so that f(x1) ≈ f(x0) Actually, f(x1) + ɛ = f(x0) We lead with x0 and update with x1 because f(x1) decreases slower (therefore more accurate).

Describe the Update in Picture Start with x0 and x1 close to 1. Possible 95% interval candidate f(x0)f(x1) 1

Describe the Update in Picture Move the x0 to a new position Possible 95% interval candidate f(x0) f(x1) 1 1.Lower x0  lower f(x0)

Describe the Update in Picture Increase x1 little by little biggest f(x) only Possible 95% interval candidate f(x0) f(x1) 2 1.Lower x0  lower f(x0) 2.Increase x1 slowly so that f(x1) ≈ f(x0)

Describe the Update in Picture f(x1) ≈ f(x0) biggest f(x) only Possible 95% interval candidate f(x0) f(x1) 1.Lower x0  lower f(x0) 2.Increase x1 slowly so that f(x1) ≈ f(x0)

Describe the Update in Picture Check to see the interval represents 95% Possible 95% interval candidate f(x0) f(x1) 1.Lower x0  lower f(x0) 2.Increase x1 slowly so that f(x1) ≈ f(x0) 3.Check whether the interval F(x1)-F(x0) is big enough. 3

Describe the Update in Picture All together biggest f(x) only Possible 95% interval candidate (check) f(x0) f(x1) 1 2 3 1.Lower x0  lower f(x0) 2.Increase x1 slowly so that f(x1) ≈ f(x0) 3.Check whether the interval F(x1)-F(x0) is big enough.

the Algorithm To start the algorithm:  f(x0) > f(x1)  F(x1) – F(x0) < 0.95.  Iterates by:  Update x0, and then update x1.  Check for F(x1) – F(x0) > 0.95.  Print result of each iteration  Algorithm Ends when:  F(x1) – F(x0) ≈ 0.95 (actually F(x1) – F(x0)= 0.95 + ɛ’)

Setting up BEFORE the algorithm Positions of f(x0) and f(x1) Definitely smaller than 95% f(x0) f(x1) f(x0) < 1 < f(x1) f(x0) < f(x1) F(x1) – F(x0) < 0.95

Our Algorithm Conceptually Initiate x0 = 0.7, x1 = 1.3 ( f(x0) = 0.3476 and f(x1) = 0.3543. F(x1) – F(x0) = 0.21737) while (F(x1) – F(x0) < 0.95)

Our Algorithm Conceptually Initiate x0 = 0.7; x1 = 1.3; define f(x) while (F(x1) – F(x0) < 0.95) x0 = x0 - 0.001 while ( f(x1) – f(x0) > 0) x1 = x1 + 0.001 (UPDATE)

Our Algorithm Conceptually Initiate x0 = 0.7; x1 = 1.3; define f(x) while (F(x1) – F(x0) < 0.95) interval = F(x1) – F(x0) print x0, x1, f(x0), f(x1), interval x0 = x0 - 0.001 while ( f(x1) – f(x0) > 0) x1 = x1 + 0.001 (UPDATE)

Everything Correct (Final) x0 = 0.7; x1 = 1.3; f = function(x){p = x*exp(-x) return(p)} while (pgamma(x1, shape =2) – pgamma(x0, shape=2) < 0.95){ x0 = x0 – 0.001 while (f(x1) – f(x0) > 0){ x1 = x1 + 0.001} interval = pgamma(x1, shape =2) – pgamma(x0, shape=2) print(c(x0, x1, f(x0), f(x1), interval))}

The Output … x0, x1, f(x0), f(x1), interval

Optional: Improving the algorithm We saw epilson ɛ twice. This implies that the algorithm is biased! The algorithm will always have the right point a little too far out and the interval slightly too big!

Ideally, We would update (decrease) x0 and then find the inverse f -1 to get x1. – Not doable analytically. Then we do it numerically: – Update (decrease) x0  calculate f(x0)  slowly update (increase) x1 so that f(x1) ≈ f(x0) Actually, f(x1) + ɛ = f(x0) We lead with x0 and update with x1 because f(x1) decreases slower (therefore more accurate).

the Algorithm To start the algorithm:  f(x0) > f(x1)  F(x1) – F(x0) < 0.95.  Iterates by:  Update x0, and then update x1.  Check for F(x1) – F(x0) > 0.95.  Print result of each iteration  Algorithm Ends when:  F(x1) – F(x0) ≈ 0.95 (actually F(x1) – F(x0)= 0.95 + ɛ’)

Correcting Bias We can correct the bias by adding or subtracting a small amount to negate ɛ. How much? Look at the output and take the average Where? Right after the updates are done BUT ON NEXT TIME!

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