Question 1177295: Let X be a random variable with pdf f (x) = kx^2 where 0 ≤ x ≤ 1.
(a) Find k.
(b) Find E(X) and Var(X).
(c) Find MX(t). Using the mgf, find E(X).
Thank You
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **(a) Find k**
Since f(x) is a probability density function (pdf), the integral of f(x) over its domain must equal 1:
* ∫[0, 1] kx² dx = 1
Integrate:
* [ (kx³/3) ] from 0 to 1 = 1
* (k/3) - (0) = 1
* k/3 = 1
* k = 3
Therefore, f(x) = 3x² for 0 ≤ x ≤ 1.
**(b) Find E(X) and Var(X)**
1. **Find E(X):**
* E(X) = ∫[0, 1] x * f(x) dx
* E(X) = ∫[0, 1] x * (3x²) dx
* E(X) = ∫[0, 1] 3x³ dx
* E(X) = [ (3x⁴/4) ] from 0 to 1
* E(X) = (3/4) - (0)
* E(X) = 3/4
2. **Find E(X²):**
* E(X²) = ∫[0, 1] x² * f(x) dx
* E(X²) = ∫[0, 1] x² * (3x²) dx
* E(X²) = ∫[0, 1] 3x⁴ dx
* E(X²) = [ (3x⁵/5) ] from 0 to 1
* E(X²) = (3/5) - (0)
* E(X²) = 3/5
3. **Find Var(X):**
* Var(X) = E(X²) - [E(X)]²
* Var(X) = 3/5 - (3/4)²
* Var(X) = 3/5 - 9/16
* Var(X) = (48 - 45) / 80
* Var(X) = 3/80
**(c) Find MX(t) and Use the mgf to Find E(X)**
1. **Find MX(t) (Moment Generating Function):**
* MX(t) = E(e^(tX)) = ∫[0, 1] e^(tx) * f(x) dx
* MX(t) = ∫[0, 1] e^(tx) * (3x²) dx
We use integration by parts twice. Let u = 3x², dv = e^(tx) dx.
Then du = 6x dx, v = e^(tx)/t.
* MX(t) = [ 3x²e^(tx)/t ] from 0 to 1 - ∫[0, 1] 6xe^(tx)/t dx
* MX(t) = 3e^t/t - (6/t) ∫[0, 1] xe^(tx) dx
Now, integrate ∫xe^(tx) dx by parts again. Let u = x, dv = e^(tx) dx.
Then du = dx, v = e^(tx)/t.
* ∫xe^(tx) dx = [ xe^(tx)/t ] from 0 to 1 - ∫[0, 1] e^(tx)/t dx
* ∫xe^(tx) dx = e^t/t - [ e^(tx)/t² ] from 0 to 1
* ∫xe^(tx) dx = e^t/t - (e^t/t² - 1/t²)
* ∫xe^(tx) dx = e^t/t - e^t/t² + 1/t²
Substitute back into MX(t):
* MX(t) = 3e^t/t - (6/t) [ e^t/t - e^t/t² + 1/t² ]
* MX(t) = 3e^t/t - 6e^t/t² + 6e^t/t³ - 6/t³
* MX(t) = (3te^t - 6e^t + 6e^t/t - 6/t²)/t
* MX(t) = (3t^2e^t - 6te^t + 6e^t - 6)/t^3
2. **Find E(X) using MX(t):**
* E(X) = MX'(0)
* MX'(t) = d/dt [(3t²e^t - 6te^t + 6e^t - 6)/t^3]
We can use L'Hopital's rule or series expansions to find the limit.
* E(X) = lim(t→0) MX'(t)
* Using L'Hopital's rule three times, or the series expansion of e^t, we find:
* E(X) = lim(t→0) (3t^2e^t+6te^t+3e^t-6e^t-6te^t+6e^t)/3t^2
* E(X) = lim(t→0) (3t^2e^t+3e^t)/3t^2
* E(X) = lim(t→0) (3t^2e^t+6te^t+6e^t)/6t
* E(X) = lim(t→0) (3t^2e^t+12te^t+6e^t)/6
* E(X) = 6/8 = 3/4
**Answers**
* **(a) k = 3**
* **(b) E(X) = 3/4, Var(X) = 3/80**
* **(c) MX(t) = (3t²e^t - 6te^t + 6e^t - 6)/t^3, E(X) = 3/4**
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