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MATH 511 Homework Solutions: Probability Distributions and Moment Generating Functions, Assignments of Mathematics

University of South Carolina (USC) - ColumbiaMathematics

Solutions to problems related to probability distributions and moment generating functions in a university-level mathematics course. Topics covered include mean and variance calculations for continuous uniform and exponential distributions, as well as finding the cumulative distribution function of a random variable based on another random variable.

Typology: Assignments

Pre 2010

Uploaded on 10/01/2009

koofers-user-f7y
koofers-user-f7y 🇺🇸

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MATH 511, Meade HW Solutions
4.2 – 2, 3, 6abcd, 9, **5, **11 4/2/04
2. µ = (a + b)/2 = (-1 + 1)/2 = 0
σ2 = (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 θ = 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 = limb→∞ (- 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) = θ2 = 202 = 400
M(t) = 1/(1 - θt) .= 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 , θ = 3
µ = θ = 3
σ2 = θ2 = 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 , θ = 1/3
µ = θ = 1/3
σ2 = θ2 = 32 = 1/9
pf2

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Download MATH 511 Homework Solutions: Probability Distributions and Moment Generating Functions and more Assignments Mathematics in PDF only on Docsity!

MATH 511, Meade HW Solutions 4.2 – 2, 3, 6abcd, 9, **5, **11 4/2/

  1. μ = (a + b)/2 = (-1 + 1)/2 = 0 σ^2 = (b - a) 2 /12 = (1 +1) 2 /12 = 4/12 = 1/

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. c. P(2 ≤ X ≤ 8) = 2 ∫ 8 (1/10) dx = 0. d. E(x) = (0 + 10)/2 = 5 e. Var(X) = (10)^2 /12 = 100/12 = 25/

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

a. P( 10 ≤ X ≤ 30) = (1/20) 10 ∫^30 e -x/20 dx = - e -x/20^ |(10,30) = e -1/2^ + e -3/2^ = 0. b. P(X > 30) = (1/20) 30 ∫ ∞ e -x/20 dx = limb→∞ (- e -x/20^ |(b,30)) = e -3/2^ – e - ∞/2^ = e -3/2+ = e -3/2^ = 0. c. P(X>40 | X>10) = P(X > 40)/P(X > 10) = e -40/20^ / e -10/20^ = e -3/2^ = 0. d. Var(X) = θ^2 = 20^2 = 400 M(t) = 1/(1 - θ t) .= 1/( 1 - 20t ), t<1/

a. M(t) = 1/(1 – 3t)., t < 1/3. f(x) = (1/3) e -x/3^ , x > 0 , θ = 3 μ = θ = 3 σ^2 = θ^2 = 3^2 = 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) = 3 e -3x^ , x > 0 , θ = 1/ μ = θ = 1/ σ^2 = θ^2 = 3^2 = 1/

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)

  1. Let X have an exponential distribution such that θ > 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/θ) (^) x+y∫

∞ e -t/θ dt / (1/θ) (^) x∫ ∞ e -t/θ dt = e -(x+y)/θ^ / e -x/θ= e -y/θ and P(X > y) = (1/θ) (^) y∫ ∞ e -t/θ dt = e -y/θ^ □