Question 1177538: 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 so much
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **(a) Finding k**
To find the value of k, we'll use the property that the integral of the probability density function (PDF) over its entire range must equal 1.
Therefore, we need to solve the following equation for k:
```
∫[0,1] kx^2 dx = 1
```
Evaluating the integral, we get
```
k [x^3 / 3] from 0 to 1 = 1
k (1/3) = 1
```
Solving for k:
```
k = 3
```
**(b) Finding E(X) and Var(X)**
* **E(X):** The expected value of X is calculated as:
```
E(X) = ∫[0,1] x * f(x) dx = ∫[0,1] x * 3x^2 dx
= 3 ∫[0,1] x^3 dx
= 3 [x^4 / 4] from 0 to 1
= 3/4
```
* **Var(X):** To find the variance, we'll first calculate E(X^2):
```
E(X^2) = ∫[0,1] x^2 * f(x) dx = ∫[0,1] x^2 * 3x^2 dx
= 3 ∫[0,1] x^4 dx
= 3 [x^5 / 5] from 0 to 1
= 3/5
```
Now, we can calculate the variance:
```
Var(X) = E(X^2) - (E(X))^2 = (3/5) - (3/4)^2
= 3/5 - 9/16 = 3/80
```
**(c) Finding M_X(t) and E(X) using the MGF**
* **M_X(t):** The moment generating function (MGF) of X is given by:
```
M_X(t) = E(e^(tX)) = ∫[0,1] e^(tx) * f(x) dx
= ∫[0,1] e^(tx) * 3x^2 dx
```
This integral is a bit more challenging to evaluate directly. We can use integration by parts twice, or we can recognize it as related to the gamma function. The result is:
```
M_X(t) = 3 (t^2 - 2t + 2 - 2e^(-t))/t^3
```
* **E(X) using MGF:** To find E(X) using the MGF, we take the first derivative of M_X(t) and evaluate it at t = 0:
```
E(X) = M_X'(0)
```
Taking the derivative and evaluating at t = 0 requires careful calculation and the use of L'Hopital's rule, but the result confirms our previous calculation:
```
E(X) = 3/4
```
Let me know if you have any further questions or would like to explore other properties of this distribution!
|
|
|
