Question 1172355: A random variable X has pdf = 1 - x/2 in interval (0, 2). What is the expected value of random variable X²
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's find the expected value of X².
**Understanding Expected Value**
The expected value of a function g(X) of a continuous random variable X with probability density function (PDF) f(x) is given by:
E[g(X)] = ∫ g(x) * f(x) dx
**Applying the Formula**
In this case:
* g(X) = X²
* f(x) = 1 - x/2 for 0 ≤ x ≤ 2
* f(x) = 0 elsewhere
So, we need to calculate:
E[X²] = ∫(from 0 to 2) x² * (1 - x/2) dx
**Calculating the Integral**
1. **Expand the expression:**
* E[X²] = ∫(from 0 to 2) (x² - x³/2) dx
2. **Integrate:**
* E[X²] = [x³/3 - x⁴/8](from 0 to 2)
3. **Evaluate the integral at the limits:**
* E[X²] = [(2)³/3 - (2)⁴/8] - [(0)³/3 - (0)⁴/8]
* E[X²] = [8/3 - 16/8] - [0]
* E[X²] = 8/3 - 2
* E[X²] = 8/3 - 6/3
* E[X²] = 2/3
**Result**
Therefore, the expected value of X² is 2/3, which is approximately 0.6667.
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