# check beta

#### C. Dutang and G.A. Spedicato

#### 18/04/2018

# Formula for beta(a,b)

From loss models with their notation (which differs from the NIST handbook of mathematical functions), we have \[ E(X) = \frac{\Gamma(a+b)\Gamma(a+1)}{\Gamma(a)\Gamma(a+b+1)} = \frac{a}{a+b} \] and \[ E(\min(X,d)) = \frac{a}{a+b}\beta(a+1,b;x) + x(1-\beta(a,b;x)) \] where \(\beta(.,.;.)\) denotes the incomplete beta function \[ \beta(a,b;x)= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} \int_0^x t^{a-1}(1-t)^{b-1}dt = \frac{\int_0^x t^{a-1}(1-t)^{b-1}dt}{\beta(a,b)}. \] Using (8.17.20) of NIST and recurrence relation of the beta function, \[ \beta(a+1,b;x) = \beta(a,b;x) - \frac{x^a(1-x)^b}{a\beta(a,b)} \] Therefore the exposure curve is \[\begin{eqnarray*} G(x) &=& \frac{E(\min(X,x))}{E(X)} = \left(\frac{a}{a+b}\beta(a+1,b;x) + x(1-\beta(a,b;x)) \right)\frac{a+b}{a} = \beta(a+1,b;x) + x(1-\beta(a,b;x)) \frac{a+b}{a} \\ &=& \beta(a,b;x) - \frac{x^a(1-x)^b}{a\beta(a,b)} + x(1-\beta(a,b;x)) \frac{a+b}{a} \end{eqnarray*}\]

# Check by Monte-Carlo

Intermediate result: Equation (8.17.20) of NIST

    deltabetaincomp <- function(a,b,d)
      -d^a*(1-d)^b/a/beta(a,b)
    deltatheo <- function(a,b,d)
      pbeta(d,a+1,b)-pbeta(d,a,b)
    c(deltabetaincomp(pi, 1/pi, 4/5), deltatheo(pi, 1/pi, 4/5))

    ## [1] -0.04677576 -0.04677576

Theoretical value

    library(mbbefd)
    theo <- function(a,b,d)
      d*(1 - pbeta(d,a,b))*{a+b}/{a}+pbeta(d,a,b)-{d^a*(1-d)^b}/{a*beta(a,b)}

    theo2 <- function(a,b,d)
      pbeta(d,a+1,b) + d*(1-pbeta(d,a,b))*(a+b)/a
    emp <- function(n, a, b, d)
    {
      x <- rbeta(n, a, b)
      mean(pmin(x,d))/mean(x)
    }
    c(theo(3,2,1/pi), theo2(3,2,1/pi), emp(1e6, 3,2,1/pi), ecbeta(1/pi, 3, 2))

    ## [1] 0.5166743 0.5166743 0.5166952 0.5166743