Question

1. Determine the Variance V(X) of the triangular distrbution.

Answer 1

The pdf of triangular distribution is, f(x) { 0: x < a 2(x - a)/(b - a)(c - a): a lessthanorequalto x lessthanorequalto c 2(b - x)/(b - a)(b - c): c < x lessthanequal b 0: x > b where, a < c < b so, E(x) = integral_a^c x middot 2(x - a)/(b - a)(c - a) dx + integral^b_c x middot 2(b - x)/(b - a) (b - c) dx = 1/(b - a)(c - a) [2 /3 x^3 - ax^2]^c_a + 1/(b - a)(b - c) [bx^2 - 2/3x^3]^b_c = 1/(b - a)(c - a) [2/3 (c^3 - a^3) -a(c^2 - a^2)] + 1/(b - a)(b-c) [b(b^2 - c^2) - 2/3(b^3 - c^3)] = 1/3(b -a) [2e^2 - ae - a^2] + 1/3(b - a) [b^2 + bc - 2E^2] = 1/3(b - a) [e(b - a) +b^2 -a^2] = a + b +c/3 Again, E(x^2) = integral_a^c x^2 2(x - a)/(b - a)(c - a) dx + integral_c^b x^2 2(b - x)/(b - a)(b -c) dx = 1/(b - a)(c - a) [2x^4/4 - 2ax^3/3]^c_aq + 1/(b - a)(b -a)[abx^3 /3 - 2x^4/4]^b_c