Question

СТ 5. The triangular distribution has pdf 0<<1 f(x) = (2-2) 1<x<2. It is the sum of two independent uniform(0.1) random variables. (a) Find c so that f(x) is a density function. (b) Draw the pdf, and derive the cdf using simple geometry. (c) Derive the cdf from its definition. (d) Derive the mean and variance of a random variable with this distribution.

Answer 1

5. The las distribution has pdf. ,0321 Lemon 1?n ? 2 a » cao [e(2n-ky na triangular f(1) = c(2-2) 5 f is a density frinction , so fror)> 0 v2 s flow de 1 s'enda + f'ec2-a da = 1 C.ź + c{(4-2) – (2-4)} =1 c. £ + c. CANS) Also = st → y = 1 ale = 1 Hence the pdf becomes 03231 f(a)= S x { 2-2 13 212

b fron) 2 When From simple geometry cdf of find 0> x , then edf FOV) - 0 When osat<l , thend edf Flat) is the area odd. by y = x, X= x = x* and F(x) = *x*xx* Ae 2 3 x X y=0 & hence area of the triangle (xtr 2 & therefore when o ļ şi then F(n) = 2 When 13 2* 32 then edf F(2²) is the shaded region & hence F(**) X 2 x1 1 + [+ x(3-1)1 -Ź (260-x*).(2016)] 1 o 2 {+ [Ź - Ź nt 11 (2-*)(2-3 1(2-x)] Ź + Ź - Ź (24-43* +* ?) = 1-2 + 2 ** - 4203 구가 Y = 2-2 = -1 + 2n* sy - 2-24 {2+2 Noa" Hence when 16x12, then F(x) = -1+2a - 2x

when 2 <n<x e then CDF is Fin area of the y=0, you , y= 2.2 lines triangle formed by (2-0) x (1-0) { x 2x1 1, CDF is Fin) - o - or <a so 22 2 9 0 3 2 - + 2n &N 9 1 2 ? 2 ܢܙ • 2< n <a

으 fo) 2 CDF is defined as s few da 2 w Finn) :- (1,1) 1 2 2 3 f r<H30, then F(N) = 0 If 03 X $1, then graph of paf. x x Br F(x) = ſ f(x) dan aan os = and if Isa<2 then 2 a) da F() = Sºpom don = ['x du + gie-x) Ź +[22-2) 11 { + (2x-3-(2-Ź) -1 + 2x Ž 2x2

L F 현 그드의 Co then Fla) = f fins de 이 + / 0 flaram 글 + [ 2 - ] 1 22 + (4-2)-(2-) = , Henee CDF is F(x) = - xo 0 2041 2 ~ + 224 구조 14 0 22 2 24 x < ,

De dhe mean ,u = fäfen dhe Sanda + ļ n (2-2 [*] + [2] Ź (217-( - ) = * + 3-(3-3) 3 + {- + Ź 3+ 2-8 3 3+ -6 3 1 3-2 1, So, mean rean is/M=1. (Am) and variomee is ora اس ۔ Now sas 2² fcal s & flas da ( 22 Säne-a). [ " Lo + [2x] - Page 1 dh 'n du dn * 2 = 1 4 ? (23-13) (2“-1“) 3 4 dl I 4 + M ابيا x 7 1 x 15 4 14. 3 14 4 7 B 14 (3-4) = 15 11 14 12 lu :lo 1 - 1" 7 6 1 = 6 Ams2