MATH 511, Meade HW Solutions

4.2 – 2, 3, 6abcd, 9, **5, **11 4/2/04

2. m = (a + b)/2 = (-1 + 1)/2 = 0

s² = (b - a)²/12 = (1 +1)²/12 = 4/12 = 1/3

p.d.f c.d.f

3. X = U(0,10)

a. f(x) = 1/(b-a) = 1/10, 0<x<10.
Notice that since X is continuous it does not matter if the endpoints are included in the interval.

b. P(X ≥ 8) = ₈∫^∞(1/10)dx = ₈∫¹⁰(1/10)dx = 0.2

c. P(2 ≤ X ≤ 8) = ₂∫⁸(1/10)dx = 0.6

d. E(x) = (0 + 10)/2 = 5

e. Var(X) = (10)²/12 = 100/12 = 25/3

6. X has an exponential distribution such that q = 20. f(x) = (1/20) e^-x/20

a. P( 10 ≤ X ≤ 30) = (1/20)₁₀∫³⁰e^-x/20dx = - e^-x/20 |_(10,30) = e^-1/2 + e^-3/2 = 0.3843

b. P(X > 30) = (1/20) ₃₀∫^∞ e^-x/20dx = lim_b→∞ (- e^-x/20 |_(b,30)) = e^-3/2 – e^-∞/2 = e^-3/2+0
= e^-3/2 = 0.2231

c. P(X>40 | X>10) = P(X > 40)/P(X > 10) = e^-40/20 / e^-10/20 = e^-3/2 = 0.2231

d. Var(X) = q² = 20² = 400

M(t) = 1/(1 - qt) .= 1/( 1 - 20t ), t<1/20

9.

a. M(t) = 1/(1 – 3t)., t < 1/3.

f(x) = (1/3) e^-x/3 , x > 0 , q = 3

m = q = 3

s² = q² = 3² = 9

b. M(t) = 3/(3 – t)., t < 3 , divide by three on the top and bottom to get:

M(t) = 1/(1 – (1/3)t), t < 3.

f(x) = 3e ^-3x , x > 0 , q = 1/3

m = q = 1/3

s² = q² = 3² = 1/9

5. Y = U(0,1)

W = a + (b – a) Y , a < b

a. Find the (cumulative) distribution function of W.

F(w) = P(W < w) = P(a + (b – a)Y < w)
= P((b – a)Y < w – a)
= P(Y < (w-a)/(b-a))
= G( Y = (w – a)/(b – a))
= ( ((w – a)/(b – a)) – 0 ) / (1-0)
= (w – a)/(b – a) , a < w <b

b. How is W distirubted? U(a,b)

11. Let X have an exponential distribution such that q > 0. Show that:

P(X > x+y | X>x) = P(X > y).

P(X > x+y | X>x) = P(X > x+y) / P(X>x) = (1/q)_x+y∫^∞e^-t/qdt / (1/q)_x∫^∞e^-t/qdt
= e^-(x+y)/q / e^-x/q= e^-y/q

and

P(X > y) = (1/q)_y∫^∞e^-t/qdt = e^-y/q □