Question 1192148: Suppose the probability density of X is given by
f(x) = kxe^(-x^2 ), X>0
= 0, otherwise
(a) Find the value of k.
(b) Find the distribution function of X, i.e., the cumulative density function of X.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **a) Find the value of k**
* **Condition for a valid probability density function (PDF):**
* The total area under the PDF curve must be equal to 1.
* Mathematically, this means: ∫[from 0 to ∞] f(x) dx = 1
* **Set up the integral:**
* ∫[from 0 to ∞] kxe^(-x^2) dx = 1
* **Solve the integral:**
* Let u = -x^2
* du = -2x dx
* dx = -du / (2x)
* Substitute: ∫[from 0 to ∞] kxe^u * (-du / (2x))
= -k/2 ∫[from 0 to -∞] e^u du
= -k/2 [e^u] [from 0 to -∞]
= -k/2 [0 - 1]
= k/2
* **Solve for k:**
* k/2 = 1
* k = 2
**Therefore, the value of k is 2.**
**b) Find the distribution function of X (cumulative density function)**
* **Definition:** The cumulative distribution function (CDF), F(x), gives the probability that the random variable X takes on a value less than or equal to x.
* **Calculate F(x):**
* For x ≤ 0: F(x) = P(X ≤ x) = 0 (since f(x) = 0 for x ≤ 0)
* For x > 0:
* F(x) = P(X ≤ x) = ∫[from 0 to x] f(t) dt
= ∫[from 0 to x] 2te^(-t^2) dt
* Let u = -t^2
* du = -2t dt
* dt = -du / (2t)
* Substitute: ∫[from 0 to -x^2] 2te^u * (-du / (2t))
= -∫[from 0 to -x^2] e^u du
= -[e^u] [from 0 to -x^2]
= -(e^(-x^2) - e^0)
= 1 - e^(-x^2)
* **Combine the results:**
* F(x) =
* 0, for x ≤ 0
* 1 - e^(-x^2), for x > 0
**Therefore, the distribution function of X is:**
F(x) = {
0, for x ≤ 0
1 - e^(-x^2), for x > 0
}
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